1.

Complex Sequences and Series

Let C denote the set {(x, y): x, y real} of complex numbers and i denote the number (0, 1). For
pany real
number t, identify t with (t, 0). For z = (x, y) = x+iy, let Re z = x, Im z = y, z = x−iy and |z| = x2 + y2 .
The distance between z and w is then given by |z − w|. For z 6= 0, arg z denotes the polar angle of (x, y) in
radian (modulo 2π). Every nonzero complex number has a polar representation z = r cis θ, where r = |z|,
θ = arg z and cis θ = cos θ + i sin θ.

Properties. Let z, w be complex numbers and n be an integer. Then

z
(1) z + z = 2 Re z, z − z = 2i Im z, zz = |z|2, = cis(2 arg z) for z 6= 0,
z
(2) z + w = z + w, z − w = z − w, zw = z w, z/w = z/w for w 6= 0,
 
(3) |z + w| ≤ |z| + |w|, |z − w| ≥ |z| − |w|, |zw| = |z||w|, |z/w| = |z|/|w| for w 6= 0,
z
(4) for z, w 6= 0, arg(zw) = arg z + arg w, arg( ) = arg z − arg w, arg(z n ) = n arg z.
w

In analysis, reasoning involving limits are very common and important. We will begin with the concept
of the limit of a sequence.
(For convenience, we will abbreviate “if and only if” by “iff” or “ ⇐⇒ ” in the sequel.)

Definition. A sequence {z1, z2 , z3, . . .} (or in short, {zn }) converges to z ∈ C (denoted by zn → z) iff for
each ε > 0, there is Nε such that n ≥ Nε implies |zn − z| ≤ ε (in short, lim |zn − z| = 0.) Otherwise, the
n→∞
sequense is said to diverge.

 converges to 0 if |z| < 1,
Examples. (1) zn = z n converges to 1 if z = 1, (For |z| < 1, |zn − 0| = |z|n → 0 as n → ∞. For

diverges otherwise.
|z| > 1, |zn| = |z|n → ∞. For |z| = 1 and z 6= 1, we have z = cis θ with θ 6= 2kπ, z n = cis(nθ) “spins” around
the unit circle.)
  p
n  n  |z| x2 + y 2
(2) lim = 1, since  − 1 = = p → 0 as n → ∞.
n→∞ n + z n+z |n + z| (n + x)2 + y2

Often the limit of a sequence is difficult or impossible to find. We now introduce a criterion that allows
us to conclude a sequence is convergent without having to identify the limit explicitly.

Definition. A sequence {zn } is a Cauchy sequence iff for each ε > 0, there is Nε such that m, n ≥ Nε implies
|zm − zn | ≤ ε (in short, lim |zn − zm | = 0).
m,n→∞

Lemma. A Cauchy sequence {an } of real numbers must converge to some real number.
Proof. For ε = 1, there is N1 such that m, n ≥ N1 implies |am − an | ≤ 1 (i.e. an − 1 ≤ am ≤ an + 1). Taking
n = N1 and letting p = min{a1 , . . . , aN1 } − 1 and q = max{a1, . . . , aN1 } + 1, we get p ≤ am ≤ q for all m.
Next, let S = {x ∈ R : x ≤ ak for infinitely many k}. Note p ∈ S, but q 6∈ S. Also, if x0 ≤ x and x ∈ S,
then x0 ∈ S. This implies S is an interval of the form (−∞, a) or (−∞, a], where a is the right endpoint of
S. We will show an → a.

1
 ε ε ε
Given ε > 0, there is Nε such that m, n ≥ Nε implies |am − an | ≤ . Since a + 6∈ S, a + ≤ ak for
2 2 2
ε ε
only finitely many k. However, a − ∈ S, so a − ≤ ak for infinitely many k. Hence, we can find some
2 2
ε ε ε
m ≥ Nε such that a − ≤ am ≤ a + (i.e. |am − a| ≤ ). Then for all n ≥ Nε , we have
2 2 2
ε ε
|an − a| ≤ |an − am | + |am − a| ≤ + = ε.
2 2
QED.

Theorem. A sequence {zn } converges if and only if it is a Cauchy sequence.

ε
Proof. Suppose zn → z, then for any ε > 0, the closed disk with z as center and radius contains
2
zK , zK+1 , zK+2 , . . . for some K. Let Nε = K, then m, n ≥ Nε implies |zm − zn | = |(zm − z) + (z − zn )| ≤
ε ε
|zm − z| + |z − zn | ≤ + = ε.
2 2
Conversely, suppose {zn } is a Cauchy sequence. Since | Re zm − Re zn | ≤ |zm − zn|, {Re zn} is also a
Cauchy sequence. Similarly, {Im zn } is also a Cauchy sequence. By the lemma above, {Re zn } converges to
some real x and {Im zn } converges to some real y. Then {zn} converges to x + iy. QED.

X
∞
Definitions. A series zk = z1 + z2 + z3 + . . . converges iff the sequence z1 , z1 +z2 , z1+z2 +z3 , . . . converges
k=1
(iff Sn = z1 + z2 + . . . + zn is a Cauchy sequence). Otherwise, the series is said to diverge.
There are two simple tests for checking convergence of series, namely the term test and the absolute
convergence test. The former provides a necessary condition for convergence and the latter provides a
sufficient condition for convergence.
X
∞
Term Test. If zk converges, then lim zn = lim (Sn − Sn−1 ) = lim Sn − lim Sn−1 = 0.
n→∞ n→∞ n→∞ n→∞
k=1
∞
X ∞
X
Absolute Convergence Test. If |zk | converges, then zk converges (because Tn = |z1|+|z2 |+. . .+|zn|
k=1 k=1
is a Cauchy sequence and for m > n, |Sm −Sn | = |zn+1 +zn+2 +. . .+zm | ≤ |zn+1|+|zn+2 |+. . .+|zm | = Tm −Tn ,
forcing Sn to be a Cauchy sequence.)

∞ ∞   ∞ ∞
!
X 1 X  1  X 1 X 1
Examples. (1) converges because   √
k2 + i  k2 + i  = k4 + 1
≈
k2
converges.
k=1 k=1 k=1 k=1
! ! !
X
∞
1 X∞
1 X
∞
k−i X∞
k X
∞
1
(2) diverges because Re = Re = ≈ diverges.
k+i k+i k2 + 1 k2 + 1 k
k=1 k=1 k=1 k=1 k=1

Exercises

1. For what complex values z will the following series converge

X∞  n ∞
X
z zn
(a) ; (b) ?
n=0
1+z n=0
1 + z 2n

2. When will equality occur in the triangle inequality? That is, under what conditions on w and z will
|w + z| = |w| + |z|?

2
  2
Xn  n
X n
X X
 
3. Establish the identity  α k βk  = |αk|2 |βk |2 − |αkβj − αj βk |2 for the case n = 2.
 
k=1 k=1 k=1 1≤k<j≤n
(This implies the Cauchy-Schwarz inequality
  v v
Xn  u uXn u n
uX
 
 α k βk  ≤ t |αk |2t |βk |2 .)
 
k=1 k=1 k=1

4. Suppose 0 < a0 ≤ a1 ≤ · · · ≤ an . Prove that the polynomial P (z) = a0 z n + a1z n−1 + · · · + an has no
root in the open unit disk D = {z : |z| < 1}. [Hint: Consider (1 − z)P (z).]

5. Prove that if 11z 10 + 10iz 9 + 10iz − 11 = 0, then |z| = 1. [Hint: Solve for z 9 .] (This problem came from
the 1989 William Lowell Putnam Mathematical Competition.)

6. Let P (z) = z n + c1 z n−1 + · · · + cn with c1 , · · · , cn real. Suppose |P (i)| < 1. Prove that there is a root
x + iy of P (z) satisfying (x2 + y2 + 1)2 − 4y2 < 1. (This problem came from the 1989 USA Mathematical
Olympiad.)

3
2. Set Descriptions and Terminologies

For
 a, b∈ C, 
the line
 L through a in the direction b consists of all z = a + tb (−∞ < t < ∞), so
z −a
L = z: Im = 0 . The circle C with center c ∈ C and radius r is given by C = {z: |z − c| = r}.
b
y y y
Imaginary y=x≤
y=-x
axis

1 z= ( x,y )
1 i
4 (1,1)
Real x
O axis O x x
_ 1i O
-i 4
S

Examples. (1) S = {z: |z + i| ≤ |z − 1|} consists of complex numbers that are closer to −i than 1, i.e. the
closed half plane below y = −x.
z+z z −z z −z
(2) To describe the parabola y = x2 in complex variable, we set x = , y = and get =
 2 2 2i 2i
z+z 1 1 1
. Alternatively, using the fact that the focus is at (0, ) = i and the directrix is y = − , the
2     4 4 4
 i   1 

equation of the parabola is z −  = Im z + .
4 4
n πo
(3) z: 0 < arg(z − 1 − i) < is the open quarter-plane lying in the first quadrant with vertex at (1, 1)
2
and edges parallel to the x, y axes.

Next, we will introduce some notations and terminologies often used in the sequel.
The open disk {z: |z − z0 | < r} is denoted by B(z0 , r) and is referred to as the r-neighborhood of z0.
The closed disk {z: |z − z0 | ≤ r} is denoted by B(z0 , r). The circle {z: |z − z0 | = r} is denoted by C(z0 , r).
The boundary of a set S is denoted by ∂S and consists of all points z such that every neighborhood of
z contains a point in S and a point not in S. In particular, we note that a set S and its complement C \S
have the same boundary.
A set is open iff it does not contain any boundary point. A set is closed iff it contains all boundary
points. A set containing some, but not all, boundary points is neither open nor closed. The interior of a set
S is S \ ∂S and the closure of S is S ∪ ∂S.
A set is bounded iff it is contained inside a neighborhood of O. A set is compact iff it is closed and
bounded. A set S is disconnected iff it is contained in the union of two disjoint, open sets A, B each of which
contains at least one point of S, i.e. there are open sets A and B such that S ⊂ A ∪ B, A ∩ B = ∅, S ∩ A 6= ∅
and S ∩ B 6= ∅. A set is connected iff it is not disconnected. A set is a region or a domain iff it is open and
connected.

Remarks. (1) S is open ⇐⇒ every point in S has a neighborhood containing in S.

(2) S is closed ⇐⇒ C \S is open ⇐⇒ zn ∈ S, zn → z implies z ∈ S.

Theorem. Every region S is polygonally connected, i.e. any two points in S can be joined by a polygonal
line lying totally in S. In fact, the polygonal line can be chosen to consist of horizontal or vertical segments.

4
Proof. Suppose p ∈ S. Let A = {z ∈ S: z can be connected to p by a polygonal line lying totally in S} and
B = S \ A. Then A must be open because for each z ∈ A, there is B(z, r) ⊂ S and every point in B(z, r)
can be connected to z (hence to p) by such a polygonal line, i.e. B(z, r) ⊂ A. Similarly B is open because
for each w ∈ B, there is B(w, t) ⊂ S and every point in this B(w, t) can be connected to w (hence not to p)
by such a polygonal line, i.e. B(w, t) ⊂ B.
Since S is connected, S = A ∪ B and A is nonempty, B must be empty. Therefore A = S. QED.

The theorem requires S open! A circle is connected, but not polygonally connected. In advanced courses,
it will be showed that a polygonally connected set is connected. So an open and polygonally connected set
is a region.
y
y

1
x x

Examples. (1) Let S = B(0, 1) \ {0}. Then ∂S = C(0, 1) ∪ {0} (0 is a boundary point because every
neighborhood of 0 contains points in S and 0, which is not in S). So S is open, bounded, connected, not
closed and not compact.
(2) Let S = {z: Re z ≥ 0}, the right half plane. Then ∂S is the imaginary axis. So S is closed, connected,
not open, not bounded and not compact.
(3) Let S = C(0, 1), the unit circle. Then ∂S = S. So S is closed, bounded, compact, connected and not
open.
 
1
(4) Let S = z: |z| = for n = 1, 2, 3, . . . . Observe that 0 is not in S, but it is a boundary point of S. So
n    
3 3
∂S = S ∪ {0}. S is not connected because for instance, S ⊂ z: |z| > ∪ z: |z| < , where the two sets
4 4
| {z } | {z }
open open
are disjoint.

Definitions. (1) A right-angled polygon is a polygon (without self intersections)

whose edges consist of horizontal and vertical segments.
(2) A simply connected region is a region such that if a right-angled polygon is in
the region, then the inside of the polygon is also in the region.

Remarks. Simply connected regions have no “holes”. A region is simply connected iff every polygon can
be “shrunk” to a point without leaving the region.

Examples. (1) The following are examples of simply connected regions: (i) open convex sets (in particular,
disks, half-planes, infinite strips), (ii) complex plane with nonintersecting infinite slits removed, (iii) open
unit disk with a spiral joining the center to the boundary removed.

o o

(2) The following are examples of non-simply connected regions: (i) punctured disk, (ii) annulus, (iii)
complement of a segment in C.

5
 o o

Exercises

1. Describe the sets whose points satisfy the following relations

 
z − 1

(a)   = 2; (b) |z + 1| − |z − 1| < 2;
z + 1
z −1 π
(c) |z 2 − 1| = 1; (d) arg = .
z+i 3

2. Given distinct complex numbers α, β, γ. Show that they are collinear if and only if Im(αβ+βγ +γα) = 0.

3. Show that the numbers α, β, γ are the vertices of an equilateral triangle if and only if α + ωβ + ω2 γ = 0,
where ω 6= 1 is a cube root of unity.

z2 − z3 1 z2
4. Prove that if z1 , z2 , z3 are distinct complex numbers and |z1| = |z2 | = |z3|, then arg = arg .
z1 − z3 2 z1

5. What is the boundary of the set {z : Re z and Im z are rational} ?

6
3. Continuity and Uniform Convergence

Definitions. A function f: D ⊂ C → C has a limit c at z ∈ D ∪ ∂D (i.e. lim f(w) = c) iff for every
w→z
sequence zn ∈ D → z, we have f(zn ) → c (iff for each ε > 0, there is δ > 0 such that 0 < |w − z| < δ
implies |f(w) − c| < ε.) We say f is continuous at z ∈ D iff lim f(w) = f(z); f is continuous on D iff f is
w→z
continuous at every point of D.

The ε-δ definition of limit at z can be rephrased as “for each ε-neighborhood B(c, ε) of c, there is a
δ-neighborhood B(z, δ) of z such that w ∈ B(z, δ) \ {z} implies f(w) ∈ B(c, ε).” Later we will discuss about
“neighborhoods of ∞” and use this version to extend the notion of limit to the case at ∞.
Of course, as usual, the sum, difference, product, quotient (where the denomiator is nonzero) and
composition of continuous functions are continuous.

Examples. (1) Polynomials P (z) = anz n + an−1z n−1 + . . . + a0 are continuous everywhere on C.
P (z)
(2) Rational functions (where P and Q are polynomials) are continuous at z, where Q(z) 6= 0.
Q(z)
N
X
(3) If P (x, y) = amn xm yn , then f(z) = P (z, z) is continuous everywhere on C. In fact, compositions
m,n=1
of continuous functions are continuous.

For functions defined by taking limits or summations, to check continuity we usually rely on uniform
convergence.

Definition. A sequence of functions Sn : D ⊂ C → C converges uniformly to a function S: D → C iff for

X
∞
every ε > 0, there is Nε such that m ≥ Nε implies |S(z) − Sn (z)| < ε for all z ∈ D. (For series fk (z),
k=1
n
X
consider Sn = fk (z).)
k=1

X
∞
Weierstrass M-test. If for each k, there is a number Mk such that |fk (z)| ≤ Mk for all z in D and Mk
k=1
∞
X
converges, then fk (z) converges (absolutely and) uniformly on D.
k=1
∞
X ∞
X
Proof. (Absolute convergence follows by comparing |fk (z)| with Mk .) For uniform convergence, let
k=1 k=1
∞
X X∞
ε > 0. Since Mk converges, there is Nε such that n ≥ Nε implies Mk < ε. Then for every z in D,
k=1 k=n+1
   
X∞ X
n   X ∞  X
∞ X
∞
   
 fk (z) − fk (z) =  fk (z) ≤ |fk (z)| ≤ Mk < ε.
   
k=1 k=1 k=n+1 k=n+1 k=n+1

QED.

7
Properties of Uniform Convergence.
(1) If Sk ’s are continuous on D and Sk converges uniformly to S on D, then S is continuous on D and
lim lim Sk (w) = lim S(w) = S(z) = lim Sk (z) = lim lim Sk (w). (For series, if fk ’s are continuous on
w→z k→∞ w→z k→∞ k→∞ w→z
∞
X ∞
X
D and fk converges uniformly to f on D, then f is continuous on D and lim fk (w) = lim f(w) =
w→z w→z
k=1 k=1
X∞ X
∞
f(z) = fk (z) = lim fk (w).)
w→z
k=1 k=1
Z Z
(2) If Sk ’s are integrable and converges uniformly, then lim Sk = lim Sk . (For series, if fk ’s are
k→∞ k→∞
X∞ Z X
∞ XZ
∞
integrable and fk converges uniformly, then fk = fk .)
k=1 k=1 k=1

(Proof. We will prove (1) and postpone (2) until complex integration is defined. (In (1), the parenthetical
n
X
statement follows from the first statement by taking Sn = fk .) Given ε > 0. Since Sk converges uniformly
k=1
ε
to S, there is N = Nε/3 such that k ≥ Nε/3 implies |S(z) − Sk (z)| < for all z ∈ D. For a fixed w ∈ D,
3
ε
since SN is continuous at w, there is δ > 0 such that |w − z| < δ ⇒ |SN (w) − SN (z)| < . Then
3

|S(w) − S(z)| = |S(w) − SN (w) + SN (w) − SN (z) + SN (z) − S(z)|

ε ε ε
≤ |S(w) − SN (w)| + |SN (w) − SN (z)| + |SN (z) − S(z)| < + + = ε.
3 3 3
So S is continuous at any arbitrary w ∈ D.)

X
∞
zk X
∞
Rk
Example. Consider f(z) = . Suppose z ∈ B(0, R). The series converges by the ratio test
k! k!
k=0 k=0
  k ∞
Rk+1 Rk 1  z  Rk X zk
because lim = R lim = 0. Since   ≤ = Mk , by Weierstrass M-test,
k→∞ (k + 1)! k! k→∞ k + 1 k! k! k!
k=0
zk
converges uniformly on B(0, R). Since are continuous on B(0, R), f is continuous on B(0, R). Finally,
k!
since R is arbitrary, f is continuous everywhere on the complex plane.

Exercises

X
∞
1
1. Show that f(z) = is continuous on the right half plane H = {z : Re z > 0}.
k2 +z
k=1

X
∞
2. Show that f(z) = kz k is continuous on D = {z : |z| < 1}.
k=0

8
4. Stereographic Projection

In many situations, it is best to treat infinity as a point. Stereographic projection explains how this can
be done.

Let S 2 = {(ξ, η, ζ): ξ 2 +η2 +ζ 2 = 1} be the unit sphere

and N = (0, 0, 1) be the north pole. There is a one-to-one
correspondence between S 2 \ N and the complex plane C
N
given by Z = (ξ, η, ζ) ↔ z = x + iy = (x, y, 0) as in
the figure. The equation of the line through N, z implies
x y 1 ξ η
= = . It follows that x = , y = .
ξ η 1−ζ 1−ζ 1−ζ Z
2 2 2 2 2 2 2 2
Conversely, 1 = ξ +η +ζ = (x +y )(1−ζ) +ζ ⇒ x + z
1+ζ 2 2x O
y2 = = − 1. So ξ = x(1 − ζ) = 2 ,
1−ζ 1−ζ x + y2 + 1 z Z
2y x2 + y 2 − 1
η = y(1 − ζ) = 2 , ζ = .
x + y2 + 1 x2 + y 2 + 1

We see that if z = x + iy → ∞, then x2 + y2 = |z|2 → ∞, so (ξ, η, ζ) → (0, 0, 1). Thus N = (0, 0, 1) ↔ ∞.

The set C ∪{∞} is usually referred to as the extended complex plane. Because of the correspondence,
some like to use S 2 to denote the same set. If S 2 is used, then the extended complex plane is sometimes also
referred to as the Riemann sphere.

Now to analyze how the neighborhoods of a point are transformed under stereographic projection,
we study the correspondence of the boundary circles. A circle on S 2 is the intersection of S 2 with a
plane Aξ + Bη + Cζ = D. This corresponds to (C − D)(x2 + y2 ) + 2Ax + 2By = C + D, which is a
n
circle on C if C 6= D
. Conversely, a circle on C is given by x2 + y2 + ax + by = c, which corresponds to
line on C if C = D
1 + ζ + aξ + bη = c(1 − ζ) on S 2 , i.e. a circle on S 2 .
1 + ζ0
In particular, the circle on S 2 coming from the plane ζ = ζ0 < 1 corresponds to x2 + y2 = , i.e.
1 − ζ0
the boundary of a neighborhood of N corresponds to a circle centered at the origin. Then neighborhoods of
N ↔ ∞ correspond to sets of the form {z: |z| > r}. By abuse of notation, we will let B(∞, r) = {z: |z| > r}.
The remark following the definition of lim f(w) suggests how limits involving ∞ can be defined.
w→z

Definitions. (1) zn → ∞ iff |zn | → ∞ (iff for any ε > 0, there is Nε such that n ≥ Nε ⇒ |zn| > ε.)
(2) lim f(z) = c ∈ C iff for any ε > 0, there is δ > 0 such that |z| > δ ⇒ |f(z) − c| < ε (iff lim f(z) =
z→∞ z→∞
1
lim f( ) = c.)
w→0 w
1
(3) lim f(z) = ∞ iff for any ε > 0, there is δ > 0 such that 0 < |z − c| < δ ⇒ |f(z)| > ε (iff lim = 0.)
z→c∈C z→c f(z)

(4) lim f(z) = ∞ iff for any ε > 0, there is δ > 0 such that |z| > δ ⇒ |f(z)| > ε.
z→∞

9
 1
(5) f(z) is continuous at ∞ iff g(z) = f( ) is continuous at 0.
z

Definitions. (1) Let f be defined in a neighborhood of z ∈ C. Then f is differentiable at z iff

f(z + h) − f(z)
f 0 (z) = lim
h→0,h∈C h

1
exists; f is differentiable at ∞ iff f( ) is differentiable at 0.
z
(2) f is holomorphic or analytic (or regular) on a set S iff f is differentiable at every point of some open set
containing S.
(3) f is univalent (or schlicht) on an open set iff f is differentiable and one-to-one there.
(4) f is entire (or integral) iff f is differentiable on the complex plane.

Of course, as usual, the sum, difference, product, quotient and chain rules are valid in (complex)
differentiation. Some common functions are differentiable, e.g. polynomials are entire functions and (z n)0 =
nz n−1 for integer n. However, some are not, e.g. the conjugate function z is not differentiable anywhere
z +h−z h
because lim = lim doesn’t exist, as can be seen by the following computations :
h→0,h∈C h h→0,h∈C h

h h h −it
lim = lim = 1 and lim = lim = −1.
h→0,h∈R h h→0,h∈R h h=it→0,t∈R h t→0 it

Exercises

1. The chordal metric d(w, z) of two complex numbers w and z is the distance in R3 between the points
corresponded to w and z under stereographic projection.
(a) Express d(w, z) in terms of w and z only.
(b) Express d(w, ∞) = lim d(w, z) in terms of w only.
z→∞
(c) Show that zn → z ∈ C ∪{∞} if and only if d(zn , z) → 0.

2. Prove that f(z) = |z| is not differentiable anywhere, but g(z) = |z|2 is differentiable at 0 only.

10
5. Power Series

X
∞
Definition. A power series centered at c is a function of the form f(z) = an(z − c)n . (Note that
n=0
f(c) = a0 , so the series converges for at least the point z = c.)

To understand what the domains of power series (as functions) look like, we introduce the concept of
the upper limit of a sequence and recall the root test.

Definition. Let {xn } be a sequence of real numbers. The upper limit or limit superior of {xn } (denoted by
lim xn or limsup xn is
n→∞ n→∞
(a) the number L if L has the properties that for each ε > 0,
(i) xn < L + ε for all except finitely many n and
(ii) xn > L − ε for infinitely many n.
(b) +∞ if for each real number r, there is xn > r.
(c) −∞ if for each real number r, only a finite number of the xn’s are greater than r.

Essentially, by taking ε → 0, items (i) and (ii) imply L is the limit of some subsequence. Furthermore,
item (i) implies no larger number is also a limit of a subsequence. So the upper limit of {xn } is the largest
limit of any subsequence of {xn}. Now if lim xn exists, then all subsequences converge to the same limit.
n→∞
In that case, lim xn = lim xn.
n→∞ n→∞

1
Examples. (1) lim = 0.
n→∞ n

(2) lim [n + (−1)n n] = lim {0, 4, 0, 8, 0, 12, . . .} = lim {4, 8, 12, . . .} = +∞.
n→∞ n→∞
nπ
(3) lim sin = lim {1, 0, −1, 0, 1, 0, . . .} = lim {1, 1, 1, . . .} = 1.
n→∞ 2 n→∞

(
P p P converges absolutely if ρ < 1,
Root Test. Given a series an, let ρ = lim n |an|, then an diverges if ρ > 1,
n→∞
is inconclusive if ρ = 1.

Proof. (Case ρ p < 1) Take x such that ρ < x < 1. Let ε = x − ρ > 0.PThen by property (i), all but finitely
P
many n satisfy n |an | < ρ + ε = x, then |an | < xn . Since x < 1, xn converges, which implies |an|
converges.
p
n
(Case ρ > 1) Take x such that 1 < x < ρ. Let Pε = ρ − x > 0. Then by property (ii) |an| > ρ − ε = x > 1
for infinitely many n, then lim |an| 6= 0, so an diverges.
n→∞
X∞
1 X∞
1
(Case ρ = 1) Consider and 2
. For both cases, ρ = 1, but the first series diverges and the second
n=1
n n=1
n
series converges. QED.

X
∞
Theorem 1. The power series f(z) = an (z − c)n converges absolutely for |z − c| < R and diverges for
n=0
1
|z − c| > R, where R = p . (In case R = 0, the series converges at z = c only.) R is called the
lim n |an|
n→∞

11
radius of convergence and the disk |z − c| < R is called the disk or domain of convergence. Furthermore,
X∞
an (z − c)n converges uniformly on any smaller disk |z − c| ≤ R0 < R.
n=0

Proof. The first statement follows from the root test. The last statement follows by Weierstrass M-test.
Since |an (z − c)n | ≤ |an| R0n for all |z − c| ≤ R0, so we set Mn = |an | R0n and use the absolute convergence
of the power series at z = R0 + c. QED.
   
 an+1  p  an+1 

In a course on real analysis, it is usually shown that if lim   n 
exists, then lim |an| = lim  .
n→∞ an  n→∞ n→∞ an 
This can be used in computing the radius of convergence.

The convergence on the boundary circle |z − c| = R can be arbitrary as the following examples show.

X∞ X∞  n ∞
zn z  X 1
Examples. (1) For , R = 1. If |z| = 1, then  = converges by p-test.
n 2  n2  n 2
n=1 n=1 n=1

X
∞ X
∞
(2) For z n , R = 1. If |z| = 1, then z n diverges because lim z n 6= 0.
n→∞
n=0 n=0

X∞ X∞ X∞
zn 1 (−1)n
(3) For , R = 1. If z = 1, diverges. If z = −1, converges by alternating series test.
n=1
n n=1
n n=1
n

Later we will show that every holomorphic function can be locally represented by power series (i.e.
∞
X
f holomorphic at c ⇒ f(z) = an (z − c)n in a neighborhood of c). Therefore, the local properties of
n=0
holomorphic functions can be understood by studying power series. First, we will see that power series can
be differentiated (term-by-term) in their domains of convergence.

X
∞ X
∞
Theorem 2. If f(z) = an(z − c)n converges on |z − c| < R, then f 0 (z) = nan(z − c)n−1 converges
n=0 n=1
on |z − c| < R, too. Consequently, power series are infinitely differentiable on their disk of convergence.
Proof. The proof is divided into two steps. The first step is to observe that

p  1
 n−1
n  1 1
 p
n−1
lim n |an| = lim (n |an |) n = lim n n |an| n = lim n |an|,
n→∞ n→∞ n→∞ n→∞

so the power series have the same disk of convergence. The second step is to obtain an inequality of the
 f(z + h) − f(z) X ∞ 
 
form  − nan(z − c)n−1  ≤ A|h|.
 h 
n=1
Fix z such that |z − c| < R. Suffices to consider small h (say |h| < δ < R − |z − c|.) Then

n  
!
f(z + h) − f(z) X X X
∞ ∞
n k−1
− nan(z − c)n−1 = an h (z − c) n−k
.
h n=1 n=2
k
k=2
| {z }
bn

Observe that for k ≥ 2,

     
n n (n − k + 2) (n − k + 1) n
= ≤ n(n − 1).
k k−2 k−1 k k−2

12
So, for z 6= c,

X∞   n  
n k−1 n−k n(n − 1)|h| X n
|bn| ≤ |h| |z − c| ≤ |h|k−2|z − c|n−k+2
k |z − c|2 k−2
k=2 k=2
n  
n(n − 1)|h| X n n(n − 1)|h| n
≤ |h|j |z − c|n−j = (|z − c| + |h|) .
|z − c|2 j |z − c|2
j=0

Therefore, as h → 0,
  ∞ 
 f(z + h) − f(z) X ∞  X  X ∞
 n−1   
 − nan(z − c)  =  anbn  ≤ |an||bn|
 h   
n=1 n=2 n=2
1 X∞
≤ |h| n(n − 1)|an|(|z − c| + δ)n → 0.
|z − c|2 n=2
| {z }
Az
P∞
(Note that the series in Az converges because the first step above implies f 00 (w) = n=2 n(n−1)an(w−c)n−2
converges absolutely on |w − c| < R, in particular at w = c + |z − c| + δ by the condition on δ.) QED.

X
∞
Taylor’s Theorem for Power Series. If f(z) = an(z − c)n has a nonzero radius of convergence, then
n=0
X∞
f (n) (c) f (n) (c)
the coefficient an = , i.e. f(z) equals its Taylor series (z − c)n .
n! n=0
n!

Proof. Apply theorem 2 repeatedly (i.e. inductively) and evaluate at z = c. QED.

∞
X
Identity Theorem for Power Series. If f(z) = an(z − c)n = 0 for z = zk 6= c, (k = 1, 2, . . .) and {zk }
n=0
converges to c, then an = 0 for all n.
Proof. We are given a power series f(z) = a0 + a1 (z − c) + a2 (z − c)2 + · · ·. Now a0 = f(c) = lim f(zk ) = 0,
k→∞
f(z) f(zk ) f(z) f(zk )
a1 = lim = lim = 0, . . ., an = lim n
= lim = 0, . . .. QED.
z→c z − c k→∞ zk − c z→c (z − c) k→∞ (zk − c)n

∞
X ∞
X
Uniqueness Theorem for Power Series. If an (z − c)n = bn (z − c)n for z = zk 6= c, (k = 1, 2, . . . )
n=0 n=0
and {zk } converges to c, then an = bn for all n.
X
∞
Proof. Consider f(z) = (an − bn)(z − c)n and apply the identity theorem for power series. QED.
n=0

Exercises

1. Find lim xn , where

n→∞
 
1 nπ
(a) xn = n sin ; (b) xn = cos ;
n 4
2n
(c) xn = 1 + (−1)n .
n+1

13
2. Find the radius of convergence for the following power series:
∞
X ∞
X
(a) z n! ; (b) (n + 2n)z n .
n=1 n=0

∞
X ∞
X
3. Give an example of two power series an z n and bnz n with radii of convergence R1 and R2,
n=0 n=0
X
∞
n
respectively, such that the power series (an + bn )z has a radius of convergence greater than R1 +R2.
n=0

X
∞
1 1
4. Show that if |z| < 1, then zn =
. Find a power series for with centered at 0. What
n=0
1−z (1 − z)2
is the radius of convergence for this series?

X
∞  
n 1
5. Explain why there is no power series f(z) = cn z such that f = 1 for k = 2, 3, 4, . . . and
n=0
k
f 0 (0) > 0.

X
∞    
1 1 1 1
6. Does there exist a power series f(z) = cn z n such that f = 2 and f − = 3 for k =
n=0
k k k k
1, 2, 3, . . .?

X
∞  
1 k2
7. If f(z) = cnz n satisfies f = 2 , k = 1, 2, 3, . . ., compute the values of the derivatives
n=0
k k +1
f (n) (0), n = 1, 2, 3 . . ..

14
6. Cauchy−Riemann Equations

f(z0 + h) − f(z0 )
Recall f is differentiable at z0 means lim = f 0 (z0 ) exists and f is holomorphic at
h→0,h∈C h
z0 means f is differentiable in a neighborhood of z0 in which case f 0 (w) exists in some disk B(z0 , r) about
z0 .

Question. Given a function f: R2 → R2 with two real variables, say f(x, y) = (u(x, y), v(x, y)) = u(x, y) +
iv(x, y), how can we tell if it is a differentiable function of z = x + iy?

Example. If f(z) = z 2 , z = x + iy, then f(x, y) = (x + iy)2 = (x2 − y2 ) + i(2xy) = (x2 − y2 , 2xy).

real real
z }| { z }| {
Theorem. If f(z) = f(x, y) = (u(x, y), v(x, y)) = u(x, y) + iv(x, y) = u(z) + iv(z) is differentiable at z0,
∂u ∂u ∂v ∂v ∂u ∂v ∂u ∂v
then , , , exist at z0 , (z0 ) = (z0 ) and (z0 ) = − (z0 ).
∂x ∂y ∂x ∂y ∂x ∂y ∂y ∂x
∂f ∂u ∂v ∂f ∂u ∂v
[Notation. If we define fx = = +i and fy = = + i , then the above Cauchy-Riemann
∂x ∂x ∂x ∂y ∂y ∂y
∂u ∂v ∂u ∂v
equations = and =− are equivalent to fy = ifx .]
∂x ∂y ∂y ∂x
Proof. Taking h → 0 along the real axis, we have

f(x0 + iy0 + h) − f(x0 + iy0 ) f(x0 + h, y0 ) − f(x0 , y0)

f 0 (z0 ) = lim = lim = fx (x0, y0 ).
h→0 h h→0 h

Taking h → 0 along the imaginary axis, say h = it, we have

f(x0 + iy0 + it) − f(x0 + iy0 ) 1 f(x0 , y0 + t) − f(x0 , y0) 1

f 0 (z0 ) = lim = lim = fy (x0, y0 ).
t→0 it i t→0 t i
QED.

Theorem. If fx and fy exist in a neighborhood of z0 , are continuous at z0 and fy (z0 ) = ifx (z0 ), then f is
differentiable at z0 .
Proof. Write f(z0 ) = u(x0, y0) + iv(x0 , y0), h = s + it. Then
u(z0 + h) − u(z0 ) u(x0 + s, y0 + t) − u(x0 , y0)
=
h s + it
u(x0 + s, y0 + t) − u(x0 + s, y0 ) u(x0 + s, y0 ) − u(x0, y0 )
= +
s + it s + it
   
t ∂u s ∂u
= (x0 + s, y0 + αt) + (x0 + βt, y0 ) , 0 < α, β < 1
s + it ∂y s + it ∂x
(by the mean value theorem).
Similarly,
   
v(z0 + h) − v(z0 ) t ∂v s ∂v
= (x0 + s, y0 + γt) + (x0 + δt, y0) , 0 < γ, δ < 1.
h s + it ∂y s + it ∂x

15
 t s
Now, fy (z0 ) = ifx (z0 ) implies fx (z0 ) = fy (z0 ) + fx (z0 ). Using these equations and the fact
    s + it s + it
 t   s 
   
 s + it  ,  s + it  ≤ 1, it follows that as h = s + it → 0,
    
 f(z0 + h) − f(z0 )   t ∂u ∂v
 − f (z )  =  (x + s, y + αt) + i (x + s, y + γt) − f (z )
 h
x 0   s + it ∂y 0 0
∂y
0 0 y 0
 
s ∂u ∂v 
+ (x0 + βt, y0 ) + i (x0 + δt, y0) − fx (z0 ) 
s + it ∂x ∂x

   
 ∂u ∂v   ∂u ∂v 

≤  (x0 + s, y0 + αt) + i (x0 + s, y0 + γt) − fy (z0 ) +  (x0 + βt, y0 ) + i (x0 + δt, y0) − fx (z0 ) → 0.
 
∂y ∂y ∂x ∂x
So f is differentiable at z0 and f 0 (z0 ) = fx (z0 ). QED.

Examples. (1) If a holomorphic function G(z) = u(z) + iv(z) = u(x, y) + iv(x, y) has the real part u(x, y) =
x2 − y2 , what can v(z) be?
Solution. 
∂v ∂u
=− = 2y ⇒ v(x, y) = 2xy + C1(y) 

∂x ∂y
⇒ v(x, y) = 2xy + constant
∂v ∂u 

= = 2x ⇒ v(x, y) = 2xy + C2 (x) 
∂y ∂x

(2) If f = u + iv is holomorphic on a region D and u ≡ constant, show that f ≡ constant.

Solution. 
∂v ∂u
=− = 0 ⇒ v(x, y) ≡ C1(y) 


∂x ∂y
⇒ v(x, y) ≡ constant ⇒ f ≡ constant
∂v ∂u 
= = 0 ⇒ v(x, y) = C2(x) 

∂y ∂x

(3) If f is holomorphic on a region D and |f(z)| ≡ constant, then f ≡ constant.

∂u
Solution. If |f| ≡ 0, then f ≡ 0. Otherwise u2 + v2 ≡ constant, so taking partial derivatives, we get 2u +
   ∂x
∂v ∂u ∂v ∂u ∂v ∂v ∂u ∂u ∂v ∂v ∂u
2v ≡ 0, 2u + 2v ≡ 0. Since = − , = , we get = ≡ 0, =− ≡ 0.
∂x ∂y ∂y ∂y ∂x ∂y ∂x ∂x ∂y ∂x ∂y
Then u ≡ constant, v ≡ constant, so f ≡ constant.
Examples (2) and (3) are special cases of the open mapping theorem, to be proved later.

Exercises

1. If f = u + iv is holomorphic on some domain, given u(x, y) below, find the possibilities of v(x, y):

(a) u(x, y) = x3 − 3xy2 ; (b) u(x, y) = e−y cos x;

y
(c) u(x, y) = ln(x2 + y2 ); (d) u(x, y) = .
(1 − x)2 + y2

2. Show that there is no holomorphic function f = u + iv with u(x, y) = x2 + y2 .

3. Suppose f is an entire function of the form f(x, y) = u(x) + iv(y). Prove that f is a polynomial of
degree at most one.

16
4. Suppose that f and f are holomorphic on a region D. Show that f is a constant function on D.

5. Let G be a region and G∗ = {z : z ∈ G} is the mirror image of G across the x-axis. If f : G → C is

holomorphic, show that f ∗ : G∗ → C defined by f ∗ (z) = f(z) is holomorphic.

6. Write z in polar coordinates. Then f(z) = u(z) + iv(z) = u(r, θ) + iv(r, θ). Establish the polar form of
the Cauchy-Riemann equations
∂u 1 ∂v ∂v 1 ∂u
= and =− .
∂r r ∂θ ∂r r ∂θ

17
7. Definitions of Common Functions

In this section, we will enlarge our collection of complex-valued functions by extending the common
functions, such as exponential, logarithm and trigonometric functions, to complex domains.
To define ez , we want it to satisfy (1) ew+z = ew ez , (2) ex is the same as before for x real, and (3) ez
is entire.
Suppose f(w + z) = f(w)f(z) and f(x) = ex for x real. Then f(z) = f(x + iy) = ex f(iy) = ex (A(y) +
iB(y)). For f to be entire (differentiable everywhere), the Cauchy-Riemann equations require fy = ifx , i.e.
ex (A0 (y) + iB 0 (y)) = iex (A(y) + iB(y)). Then A0 (y) = −B(y) and B 0 (y) = A(y). So A00(y) = −A(y) and
B(y) = −A0 (y). Now f(0) = 1, so A(0) = 1 and B(0) = 0. Therefore, A(y) = cos y and B(y) = sin y.

Definition. For z = x + iy, ez = ex (cos y + i sin y) = ex cis y. (In particular, eiy = cis y.)

In passing, we mention the famous Euler’s equation eiπ + 1 = 0, which relates five of the most important
constants e, i, π, 1, 0 in mathematics.

Properties.
d z
(1) ew+z = ew ez (2) |ez | = ex (3) ez 6= 0 for all z (4) e = ez
dz

eiy + e−iy eiy − e−iy

Since eiy = cos y + i sin y, e−iy = cos y − i sin y, so cos y = and sin y = . We will
2 2i
use these identities to define trigonometric functions of complex numbers.

eiz + e−iz eiz − e−iz sin z 1 1 cos z

Definitions. cos z = , sin z = , tan z = , sec z = , csc z = ,cot z = .
2 2i cos z cos z sin z sin z
The derivatives of these functions are the same as before. Also, cos2 z + sin2 z ≡ 1 for all z and the
usual trigonometric identities are still valid. However, it is not true that | cos z| ≤ 1 and | sin z| ≤ 1 for all z,
e−10 + e10
e.g. cos 10i = > 1000.
2

Now we turn our attention to defining logarithm. This is done by inverting exponentiation.
Definitions. If U, V are open sets, f: U → V is one-to-one and onto, then the inverse of f is f −1 : V → U
defined by f −1 (v) = u whenever f(u) = v. Also we say g is the inverse of f at z ∈ V if g is the inverse of f
in some neighborhood of z in V .

Consider the equation ez = w with w 6= 0. Let z = x + iy be a solution of ez = w. Then ez = ex cis y =

w ⇒ ex = |w|, y = arg w ⇒ z = ln |w| + i arg w. However, arg w is multivalued. Let Arg w denote the
so-called “principal branch” of argument, which is defined by −π < Arg w ≤ π. The general solution of
ez = w is z = ln |w| + i(Arg w + 2kπ), where k is any integer.

Definitions. For w 6= 0, log w = ln |w| + i arg w, where arg w = Arg w + 2kπ, k any integer. The choice
Log w = ln |w| + i Arg w is the principal branch of the logarithm function.

To make log w continuous, we need to choose arg w continuously.

18
 Examples. (1) On the annulus A = {w: 1 < |w| < 2}, arg w cannot be defined con-
3
tinuously. This can be seen by looking at the values of arg w on the circle |w| = .
2
3 3
0 1 2 At w = , arg w = 2kπ. As w moves on the circle |w| = counterclockwise, the
2 2
3
value of arg w increases. When it get back to w = at the end, the value of arg w
2
have increased by 2π from the starting 2kπ. So arg w cannot be made continuously.
(2) For the region on the left, arg w can be defined continuously. We can define
arg w by choosing arg 1 = Arg 1 = 0 and for any other w in the region, draw a
polygonal path joining 1 to w and define arg w continuously along the path by
0 1
starting at 1. It is visually clear that arg w will be well-defined (even if different
paths are used) and continuous. It is bad to simply define arg w by taking the
principal branch Arg w because the region “wraps” around the origin a little more
than once.

On the word “branch” If arg w is defined continuously on a region, then arg w + 2kπ (for any integer k)
is also continuous on the same region and gives also the argument of points on the region. We refer to both
of these functions as different “branches” of the argument function.
Similarly, if log z is defined continuously on a domain, then log z + 2πi is also a continuous inverse of ez
(since elog z+2πi = z = elog z ). We simply say that both are different “branches” of the logarithm function.

dw dz
To differentiate the logarithm function, we will apply the inverse rule “ =1 ”.
dz dw
Inverse Rule. Let g be the inverse of f at z0 and g be continuous near z0. If f is differentiable at g(z0 )
1
and if f 0 (g(z0 )) 6= 0, then g is differentiable at z0 , g0(z0 ) = 0 .
f (g(z0 ))
Proof. Since f(g(z)) = z in a neighborhood of z0 , and g(z) → g(z0 ) as z → z0 in that neighborhood, we
g(z) − g(z0 ) 1 1
have = → 0 by the existence of f 0 (g(z0 )) 6= 0.
z − z0 f(g(z)) − f(g(z0 )) f (g(z0 ))
g(z) − g(z0 )
QED.

From the theorem, we see that if arg w can be defined continuously on a region, then log w will be
d 1 1
continuous and even differentiable. Just take g(z) = log z, f(z) = ez , then log z = g0 (z) = log z = .
dz e z

Definitions. For α 6= 0, define αβ = e(β log α) and the principal value of αβ is e(β Log α) .
π π
Example. ii = e(i log i) = ei(ln |i|+i arg i) = ei(0+i( 2 +2kπ)) = e− 2 −2kπ (which is amazingly all real-valued!).
π
The principal value of ii is e− 2 .
d z d (z log α)
Remarks. (1) α = e = αz log α (for each fixed value of log α.)
dz dz
d α d (α log z) α
(2) If arg z can be defined continuously on a region, then z = e = e(α log z) = αz α−1.
dz dz z
√
Example. We will show that, on Ω = C \ ((−∞, −1] ∪ [1, +∞)) , there is a continuous branch of z 2 − 1.
Observe that p 2 2 2
z 2 − 1 = (z 2 − 1)1/2 = e 2 log(z −1) = e 2 (ln |z −1|+i arg(z −1)) .
1 1

Since z 2 − 1 6= 0 on Ω, ln |z 2 − 1| is continuous on Ω. Next we will consider arg(z 2 − 1) as the composition

of w = f(z) = z 2 − 1 on Ω with g(w) = arg w on f(Ω), where the argument is to be chosen later. Now f is
clearly continuous on Ω and f(Ω) = C \[0, +∞). Next we will choose 0 < arg w < 2π on f(Ω) so as to make

19
g continuous
√ on f(Ω). Therefore, arg(z 2 − 1) = g (f(z)) is continuous on Ω, log(z 2 − 1) is continuous on Ω
2
and z − 1 is continuous on Ω.

Exercises

1. Find all z = x + iy such that ez = 1. Use this to find all the roots of cos z and sin z.

2. Prove that sin 2z = 2 sin z cos z for all complex number z.

3. Discuss
 if it
 is possible to define log(z − 1) continuously on C \[−1, 1]. Also, discuss the possibility for
z+1
log to be continuously defined on C \[−1, 1].
z−1

20
8. Conformal Mappings

Definitions. (1) f is conformal at z0 ∈ C iff f is holomorphic in a neighborhood of z0 and f 0 (z0 ) 6= 0.

(2) f is a conformal mapping from a region U onto a region V iff f is conformal at each point of U , one-to-one
on U and onto V . In that case, we also say U and V are conformally equivalent.

Observation. If f is conformal at z0 , then f is angle preserving (in direction and magnitude).

(For γ(t) = (x(t), y(t)), γ 0 (t) = (x0 (t), y0 (t)). In the figure, we have
γ ( t) dy y0 (t) y0 (t)
tan θ = = 0 . So, θ = tan−1 0 = arg γ 0 (t).
dx x (t) x (t)
θ
Suppose γ1 (t0) = γ2 (t0 ) = z0 .
γ ( t)
2
Then the angle between γ1 , γ2 at z0 is given by arg γ20 (t0 ) − arg γ10 (t0 ).

z0 γ ( t)
1

So, angle between f ◦ γ1 , f ◦ γ2 at w0 = f(z0 )

γ
2 w=f ( z )
= arg(f ◦ γ2 )0(t0 ) − arg(f ◦ γ1 )0(t0 )
θ θ
γ
1
f γ
1
z0 = arg(f 0 (z0 )γ20 (t0 )) − arg(f 0 (z0 )γ10 (t0 ))
z0
= arg f 0 (z0 ) + arg γ20 (t0 ) − arg f 0 (z0 ) − arg γ10 (t0 )
f γ
= angle between γ1 , γ2 at z0 .)
2

The observation has an interesting interpretation: if you lives at z0 and f is a typhoon that blows your
home from z0 to f(z0 ), then you would not know that you have been moved because all the streets near your
home intersect at the same angles as before!

Examples. [It will be convenient to treat lines as circles of infinite radius and a “circle” will mean either a
line or a circle.]
(1) w = az + b, a 6= 0 is a conformal mapping from C to C. Writing a = Reiα, b = u + iv), we see that
az + b = R(eiα z) + u + iv rotates z (with respect to the origin) by angle α, then expands (or contracts)
by a factor of R, then translate by (u, v). This mapping takes a line to a line and a circle to a circle,
i.e. takes “circles” to “circles”.
1
(2) w = is a conformal mapping of C \{0} onto C \{0}. The equation α(x2 + y2 ) + βx + γy + δ = 0 (line if
z
β − iγ
α = 0, circle if α 6= 0) is equivalent to αzz + Dz + Dz + δ = 0, where D = . Under the mapping
2
1
it becomes δww + Dw + Dw + α = 0. So the map w = sends “circles” to “circles”.
z
1
w= z
.w
w=az+b 1
r
z θ z -θ
w
θ
r

21
(3) w = z α , α > 0 is conformal at all z 6= 0.
w=zα

z
w
ϕ αϕ
θ αθ
r rα

(4) w = Log z is conformal at z ∈ C \(−∞, 0].

w= Logz
z w
πi

θ θi

R LogR

(5) Möbius transformations (or linear fractional transformations or bilinear transformations) are functions
az + b
of the form w = T (z) = , where ad − bc 6= 0. (Note if ad − bc = 0, then T (z) ≡ constant.)
cz + d

Properties of Möbius Transformations.

az + b
(1) (Basic Property) T (z) = (ad − bc 6= 0) is a one-to-one map from C ∪{∞} onto C ∪{∞}, where
cz + d(
a −d
T (∞) = if c 6= 0 , ∞ = T ( ) if c 6= 0 ; T −1 (z) = dz − b , T (T −1(z)) = z = T −1 (T (z));
c c −cz + a
∞ if c = 0 T (∞) if c = 0
ad − bc −d
T 0 (z) = 6= 0 for z 6= . So for c = 0, T (z) is a conformal mapping from C onto C and for
(cz + d)2 c  
d nao
c 6= 0, T (z) is a conformal mapping from C \ − onto C \ .
c c
a1 z + b1 a2 z + b2
(2) (Algebraic Property) If T1 (z) = (a1 d1 − b1c1 6= 0) and T2 (z) = (a2 d2 − b2c2 6= 0),
c1 z + d1 c2 z + d2
(a1a2 + b1c2 )z + (a1b2 + b1d2)
then T1 ◦ T2 (x) = T1 (T2 (z)) = .
(c1 a2 + d1c2 )z + (c1b2 + d1d2)
    
a1 b1 a2 b2 a1 a2 + b1 c2 a1b2 + b1d2
(Observe that = . This shows that there is a group
c1 d1 c2 d2 c1a2 + d1 c2 c1 b2 + d1d2
homomorphism from the Möbius transformations to the two-by-two matrices.)
  
 a ad − bc 1
az + b  − if c 6= 0
(3) (Geometric Property) T (z) = = c c cz +d is a composition of map-
cz + d 
 az + b if c = 0
d d
1
pings of the form Az + B and . From examples (1) and (2), we conclude that Möbius transformations
z
take “circles” to “circles”.
(4) (Analytic Property) A fixed point of T (z) is a point z0 such that T (z0 ) = z0 . The only Möbius trans-
formation having more than two fixed points is the identity mapping.
az + b
(Proof. T (z) = = z ⇐⇒ cz 2 + (d − a)z − b = 0, which has at most two roots, unless c = 0,
cz + d
d − a = 0, b = 0.)
Hence, if two Möbius transformations take the same values on three points of C ∪{∞}, then they must
be identical.

22
 (Proof. S(z1 ) = T (z1 ), S(z2 ) = T (z2 ), S(z3 ) = T (z3 ) ⇒ S ◦ T −1 has three fixed points T (z1 ), T (z2 ),
T (z3 ) ⇒ S ◦ T −1 is the identity mapping ⇐⇒ S(z) ≡ T (z).)


z − z3 z2 − z3
Definitions. For distinct z2 , z3, z4 ∈ C ∪{∞}, let Sz2 ,z3 ,z4 (z) = . (Then Sz2 ,z3 ,z4 is a Möbius
z − z4 z2 − z4
transformation that takes the “circle” through z2 , z3 , z4 to the real axis because Sz2 ,z3 ,z4 sends z2 , z3 , z4 to
1, 0, ∞, respectively.)
z1 − z3 z2 − z4
The cross ratio of z1 , z2 , z3, z4 is (z1 , z2, z3, z4 ) = Sz2 ,z3 ,z4 (z1 ) = × . (If one of z1 , z2, z3 , z4
z1 − z4 z2 − z3
is ∞, then we take limit to get the cross ratio.)

Properties of Cross Ratio.

z(1 − w)
(1) (z, 1, 0, ∞) = lim (z, 1, 0, w) = lim = z.
w→∞ w→∞ (z − w)
(2) (z 1 , z2 , z 3, z 4) = (z1 , z2, z3 , z4).
(3) (T z1 , T z2, T z3 , T z4) = (z1 , z2, z3, z4 ) for any Möbius transformation T .
T z2 → 1
(Proof. Sz2 ,z3 ,z4 ◦ T −1: T z3 → 0 . So by the holomorphic property, Sz2 ,z3 ,z4 ◦ T −1 = ST z2 ,T z3 ,T z4 . Hence,
T z4 → ∞
evaluating both sides at T (z1 ), we get Sz2 ,z3 ,z4 (z1 ) = ST z2 ,T z3 ,T z4 (T z1 ).)
(4) z1 , z2, z3 , z4 are distinct points on a “circle” if and only if (z1 , z2, z3, z4 ) is real. (This is because Sz2 ,z3 ,z4
takes the circle through z2 , z3 , z4 to the circle through 1, 0, ∞, which is the real axis.)

Below we shall show that cross ratio is a useful device that allows us to transform geometrical informa-
tions (such as, symmetry of points and inside/outside of circles) into algebraic quantities for computation
and derivation of important principles.

Definition. For a given circle Γ: |z − c| = R, points w, w∗ ∈ C ∪{∞}

are symmetric with respect to Γ iff (w∗ −c)(w−c) = R2 . This is denoted
b R2
Γ by w ∼Γ w∗ . Equivalently,w ∼Γ w∗ ⇐⇒ w∗ = + c. (Observe
w−c
R 2
R
that c, w, w∗ are collinear because w∗ − c = (w − c). Also,
c |w − c|2
w w*
|w − c| R
= ∗ implies ∠cbw∗ = 90◦ . ) In the case Γ is a line, w∗
R |w − c|
is taken to be the mirror image of w with respect to Γ.

Properties of Symmetric Points.

(1) The line ww∗ is orthogonal to Γ (because c, w, w∗ are collinear).
(2) A point is the center of Γ if and only if it is symmetric to ∞ with respect to Γ, i.e. w = c ⇐⇒ w∗ = ∞.
(3) w ∼Γ w∗ ⇐⇒ for any distinct z2 , z3 , z4 on Γ, (w∗ , z2, z3, z4 ) = (w, z2 , z3, z4).
R2 R2
(Proof. Consider the Möbius transformation T (z) = + c, then T ( + c) = w. Since z2 on Γ
z−c w−c
2
implies (z2 − c)(z 2 − c) = R , it follows that T (z2 ) = z 2 . Similarly, T (z3 ) = z 3 and T (z4 ) = z 4 .
(⇒) If w ∼Γ w∗ , then (w∗ , z2, z3 , z4) = (T (w∗ ), T (z2), T (z3 ), T (z4 )) = (w, z 2 , z3 , z4 ).
(⇐) We have

R2 R2
(w∗ , z2 , z3, z4) = (w, z2, z3 , z4) = (T ( + c), T (z2 ), T (z3), T (z4 )) = ( + c, z2, z3 , z4).
w−c w−c

23
 R2
Since Sz2 ,z3 ,z4 is one-to-one, the definition of cross ratio implies w∗ = + c, i.e. w ∼Γ w∗ .)
w−c
Möbius transformations send pairs of symmetric points to pairs of symmetric points as the following
shows.
Symmetry Principle. w ∼Γ w∗ ⇐⇒ T (w) ∼T (Γ) T (w∗ ) for every Möbius transformation T .
Proof. This follows easily by property (3) above because

(w∗ , z2, z3, z4 ) = (w, z2 , z3, z4) if and only if (T (w∗ ), T (z2), T (z3 ), T (z4 )) = (T (w), T (z2 ), T (z3 ), T (z4 )).

QED.

Möbius transformations need not send the inside of “circles” to the inside of the image “circles” (e.g.
1
consider w = inside the unit circle). However, there is an orientation principle which says Möbius trans-
z
formations do send the right sides of oriented “circles” to the right sides of their image “circles”.

Definition. An orientation of a circle Γ is an ordered triple points (z2 , z3 , z4) on Γ.

By property (4) of cross ratio, Γ = {z: Im(z, z2 , z3, z4) = 0}. Now, the right side of Γ with respect to
(z2 , z3, z4 ) (i.e. the side the right hand “touches” as one “walks” on Γ from z2 to z3 to z4 ) is given by the
set Rz2,z3 ,z4 Γ = {z: Im(z, z2, z3 , z4) > 0}.
To see this, take the case (z2 , z3, z4 ) is counterclockwise, then the right side is the outside of Γ, which
includes ∞. Since Sz2 ,z3,z4 (z) = (z, z2, z3 , z4) is one-to-one and onto and sends Γ to the real axis, it must
send the outside of Γ onto either the upper or the lower half plane because of connectivity. (To be precise,
suppose p1, p2 are outside Γ such that Sz2 ,z3 ,z4 sends them to q1, q2, one on the upper half plane and the
other on the lower half plane. Take a (continuous) curve joining p1 and p2 without intersecting Γ, then the
image of the curve under Sz2 ,z3 ,z4 will not intersect the real axis. However, all curves joining q1 and q2 must
intersect the real axis, which yields a contradiction.)
By definition, Rz2 ,z3 ,z4 Γ is the side of Γ that is sent to the upper half plane. We intend to show, for the
counterclockwise case, that this is the outside of Γ. For this, it suffices to see ∞ ∈ Rz2 ,z3,z4 Γ.

z4 z2 − z4 z4 − z2
Now observe that (∞, z2, z3, z4 ) = = = Reiθ , 0 < θ < π. So,
Γ z2 − z3 z3 − z2
Im(∞, z2, z3 , z4) > 0, which implies ∞ ∈ Rz2,z3 ,z4 Γ.
z2 θ
oo The clockwise case is similar. Also, the case Γ is a line can be checked, e.g.
R1,0,∞ R = {z: Im z > 0} is the upper half plane)
z3

Similarly, the left side of Γ with respect to (z2 , z3, z4) is given by Lz2 ,z3 ,z4 Γ = {z: Im(z, z2, z3 , z4) < 0}.

Orientation Principle. For an orientation (z2 , z3 , z4) on a circle Γ and T a Möbius transformation, we
have T (Rz2,z3 ,z4 Γ) = RT (z2),T (z3 ),T (z4 ) T (Γ) and T (Lz2 ,z3 ,z4 Γ) = LT (z2 ),T (z3 ),T (z4 ) T (Γ). In particular, T must
send each side of Γ onto a side of T Γ.
Proof. z ∈ Rz2 ,z3,z4 Γ ⇐⇒ Im(z, z2 , z3, z4 ) = Im(T z, T z2 , T z3, T z4 ) > 0 ⇐⇒ T z ∈ RT z2 ,T z3 ,T z4 T (Γ).
Similarly, z ∈ Lz2 ,z3 ,z4 Γ ⇐⇒ T z ∈ LT z2 ,T z3 ,T z4 T (Γ). QED.
 
z−a
Theorem. For real θ, |a| < 1, T (z) = eiθ is a conformal map of the open unit disk D onto itself.
1 − az
(Later we will show these are the only conformal maps from D onto D.)
 
 z − a 
Proof. Observe that T takes the unit circle ∂D to itself because for |z| = 1, |T (z)| = eiθ =
1 − az 
|z − a| |z − a|
= = 1. Since T (a) = 0, the orientation principle implies T maps D onto D. QED.
|zz − az| |z − a||z|

24
 In an advanced course, a theorem called the Riemann Mapping Theorem is usually proved, which asserts
that for every simply connected region Ω 6= C (i.e. Ω has no “holes” in its interior), there is a conformal
mapping f from the open unit disk D onto Ω. That is, every proper simply connected region is conformally
equivalent to D. The mapping between D and Ω is called a Riemann mapping.

Remarks. By composing with another conformal mapping from D onto D, we see that Riemann mappings
are not unique. In fact, if there is one Riemann mapping, there are infinitely many. However, if the values
f(0) ∈ Ω and f 0 (0) > 0 are prescribed, then there is a second half of the Riemann mapping theorem, which
asserts that there is exactly one Riemann mapping satisfying these additional conditions.
Later on we will prove many theorems about the open unit disk or functions on it. Because of the
Riemann mapping theorem, they then can be viewed as theorems about arbitrary proper simply connected
regions or functions on them. So, the open unit disk is a “canonical object” in complex analysis.

Let us compute some conformal mappings between D and some regions. Examples (1),(2),(3),(4) and
(6) are Riemmann mappings between the open unit disk D and some common regions. Example (8) shows
an interesting application of the symmetry principle and the orientation principle in conformal mappings.

Examples.  
1−z
i T (1) T (z) = Si,1,−1(z) = i ,
UHP 1+z
Upper Half i−z
Plane T −1 (z) = .
-1 D 1 oo i+z
0 1

S RHP 1 1−z
Right Half (2) S(z) = T (z) = ,
Plane i 1+z
D 1−z
0 S −1 (z) = .
1+z
S -1
 2
R 1−z
2
(3) R(z) = (S(z)) = .
1+z
D
C \ (-oo,0]

Q s  
p 1−z
(4) Q(z) = T (z) = i .
D 1+z
1 st Quadrant (the principal square root)

i z P i 7→ 1
 
i 1−z 1 7→ 0
(5) P (z) = i ,P: ,
1+z 0 7→ i
-1 0 1 0 1 w −1 7→ ∞
i−w
P −1(w) = . (c.f. Example (1))
i+w
i  2 !
ϑ 1−z
−1 2 −1
(6) ϑ(z) = T (P (z) ) = T −
D 1+z
 2
-1 0 1 0 − 1−z
1+z
−i z 2 + 2iz + 1
=  2 = −i 2
P -1 1−z z − 2iz + 1
z z≤ T 1+z
−i

25
  2  2
N 1 1 − (− 1z ) (z + 1)2
(7) N (z) = P − =− = − .
z 1 + (− 1z ) (z − 1)2

z 1 P( z )≤
z i
-1 1

-1 1
-i z -z

Γ2
(8) Let U be the interior of the annulus bounded by Γ1: |z + 1| = 9, Γ2: |z + 6| = 16.
Γ1
. . . . Find a conformal map from U onto a concentric annulus V centred at the origin.
-6 -1 w w*
Key. Find a pair of points w and w∗ symmetric with respect to Γ1 and Γ2. (Such a
pair must be collinear with the centers, so they must be real.)

For such a pair, (w + 1)(w∗ + 1) = 81, (w + 6)(w∗ + 6) = 256. Solving we get w = 2, w∗ = 26 (or
z −2
converse). Consider T (z) = . This is a Möbius transformation, hence takes circles to circles and right
z − 26
sides to right sides.
Γ1 ,Γ2 T Γ1 ,T Γ2
Since w = 2 ∼ w∗ = 26, the symmetry principle implies T (w) = 0 ∼ T (w∗ ) = ∞. It follows
1 1
T (w) = 0 is the center of T Γ1, T Γ2. Since T (8) = − and T (10) = − , the orientation principle implies
3 2
that 
T sends the inside
 of Γ2 to the inside of T Γ2 and the outside of Γ1 to the outside of T Γ1 . So,
1 1
V = w: < |w| < .
3 2

Exercises

1. Find conformal mappings from the open unit disk D = {z : |z| < 1} onto the following regions:
i
(a) the infinite strip {z : 0 < Im z < 1};

(b) the upper semidisk {z: |z| < 1 and Im z > 0}.
-1 0 1
[Hint: Find a map from 1st quadrant onto the upper half semidisk.]

(c) the slit disk D \ [0, 1) 0 1


(d) the finite-slit plane C ∪ ∞ \ [−1, 1] -1 1

[Hint: Find a map from C \(−∞, 0] to (C ∪{∞} \ [−1, 1].]

26
 π
2. What is the range of ez if we take z to lie in the infinite strip | Im z| < ? What are the images of
2
π
the horizontal lines and vertical segments in | Im z| < under the ez mapping? Give an example of a
2
holomorphic function from the open unit disk D onto C \{0} with derivative never equal to zero.

3. Find r such that there is a one-to-one conformal mapping from {z : Im z < 0 and |z + 2i| > 1} onto
{w : r < |w| < 1}.

z − ia
4. Let a < b and T (z) = . Define L1 = {z: Im z = b}, L2 = {z: Im z = a}, L3 = {z: Re z = 0}.
z − ib
Determine which of the regions A, B, C, D, E, F in Figure 1, are mapped by T onto the regions U , V ,
W , X, Y , Z in Figure 2.
L3 T
A ib D U W
L1
V
B E 1
ia 0
L2 Y
C F X Z
Figure 1 Figure 2

[Hint: Orient L3 by (∞, ia, ib). ]

27
9. Contour Integrals

Definition. If γ: [a, b] → C is continuous and is given by γ(t) = u(t) + iv(t), then we define
Z b Z b Z b
γ(t) dt = u(t) dt + i v(t) dt.
a a a

Z π Z π Z π
π it
2 2 2
Example. For γ(t) = e , 0 ≤ t ≤ , γ(t) dt = cos t dt + i sin t dt = 1 + i.
2 0 0 0

Definitions. A continuous curve z: [a, b] → C, z(t) = x(t) + iy(t), is piecewise smooth iff except for finitely
many points, x0(t) and y0 (t) are continuous on [a, b] and z 0 (t) = x0 (t)+iy0 (t) 6= 0 on [a, b]. The curve is closed
iff z(a) = z(b). A closed curve is simple iff the function is one-to-one on [a, b), i.e. it has no self-intersection.
Henceforth all curves will be assumed piecewise smooth and have finite lengths.

Definitions. If C is a smooth curve given by z: [a, b] → C and f is continuous on the curve C, then we
Z Z b Z Z b
define f(z) dz = f(z(t))z 0 (t) dt and |f(z)| |dz| = |f(z(t))| |z 0 (t)| dt.
C a C a

1
Examples. (1) Let C be the unit circle given by z(t) = eit(0 ≤ t ≤ 2π). Now f(z) = is continuous on C.
Z Z 2π Z   Z 2π z
1 1 1
So dz = it
(ie ) dt = 2πi and   |dz| = dt = 2π.
eit  
C z 0 C z 0
Z Z 1
2
(2) Let C be the curve z(t) = 1 + it (0 ≤ t ≤ 1) and f(z) = z . Then 2
f(z) dz = (1 + it2 )2(2it dt) =
Z 1 1 C 0
4i  i
(2it − 4t3 − 8it5 ) dt = (it2 − t4 − t6 ) = −1 − .
0 3 0 3

Properties of Contour Integrals.

(1) If C1, C2 given by z1 : [a, b] → C, z2: [c, d] → C are smoothly equivalent (in the sense that z2 = z1 ◦ λ,
where λ: [c, d] → [a, b] is one-to-one with continuous derivative and λ(c) = a, λ(d) = b), then by the
change of variable s = λ(t), we have
Z Z t=d Z s=b Z
f(z) dz = f(z2 (t))z20 (t) dt = f(z1 (s))z10 (s) ds = f(z) dz.
C2 t=c s=a C1

Z Z
(2) f(z) dz = − f(z) dz, where −C is given by z̃: [a, b] → C, z̃(t) = z(a + b − t).
−C C
Z Z Z Z Z
(3) [f(z) + g(z)] dz = f(z) dz + g(z) dz, αf(z) dz = α f(z) dz.
C C C C C
Z
(4) Fundamental Theorem of Calculus. f 0 (z) dz = f(z(b))−f(z(a)), where C is given by z: [a, b] → C.
C

28
 Closed Curve Theorem.Z If C is
Z a closed curve and f has an antiderivative (say F (z))Z in a region
0
containing C, then f(z) dz = F (z) dz = F (z(b)) − F (z(a)) = 0. (In particular, f(z) dz =
Z C C C1

f(z) dz for curves C1 , C2 that have the same initial and terminal points because C1 followed by −C2
C2
is a closed curve.)
Z
Corollary. If C is a closed curve, then P (z) dz = 0 for every polynomial P (z).
C

(5) M-L Inequality. If |f(z)| ≤ M for every z on C and C has length L, then
Z  Z  Z
 Z Z b
  b b
 f(z) dz  =  
f(z(t))z (t) dt ≤0
|f(z(t))||z 0 (t)| dt = |f(z)||dz| ≤ M |z 0(t)| dt = M L.
   
C a a C a

Z Z
(6) If fn(z) converges uniformly to f(z) on C, then lim fn (z) dz = f(z) dz.
n→∞ C C

(Proof. Let L be the length of C. For any ε > 0, because fn (z) converges uniformly to f(z) on C,
there is N such that n ≥ N ⇒ |fn(z) − f(z)| < ε for all z on C. Then for n ≥ N ,
Z Z  Z 
   
 fn (z) dz − f(z) dz  =  (fn (z) − f(z)) dz  ≤ εL.)
 

C C C

Exercises
Z
1
1. Find dz, where C is a smooth curve from 1 to −1 not passing through the origin.
C z2
Z
f 0 (z)
2. Suppose f(z) is holomorphic and |f(z) − 1| < 1 in a region G. Show that dz = 0 for every
C f(z)
closed curve C in G, assuming f 0 is continuous.
Z
z
3. (a) Find dz, where C is the simple closed curve that goes from −2 to −1 along the real axis, then
C z
goes in the clockwise direction on the unit circle to 1, then goes from 1 to 2 along the real axis and
finally goes back to −2 in the counterclockwise direction on the circle |z| = 2.
Z
(b) Find |z − 1| |dz|, where the unit circle |z| = 1 is given the counterclockwise orientation.
|z|=1

Z
z
4. Evaluate the integral dz, where C is the curve shown below.
Z Z Z C Zz Z
[Hint: = + + + ]
C C1 C2 C3 C4

2i
C4

C2 i
C1 C3

-2 -1 0 1 2

5. If Re a ≤ 0 and Re b ≤ 0, show that |ea − eb | ≤ |a − b|.

29
6. Suppose f is holomorphic on B(0, r) and Re f 0 (z) > 0 for all z ∈ B(0, r). Show that f is one-to-one.
[Hint: Express f(a) − f(b) as an integral.]
Z  Z
  2π
 
7. If f is a continuous real-valued function and |f(z)| ≤ 1, show that  f(z) dz  ≤ | sin t| dt = 4.
 |z|=1  0
[Hint: Let I be the value of the integral, then I = |I|eiθ .]

30
10. Cauchy Theory

Cauchy theory deals with the integration of holomorphic functions and its consequences as developed
by Augustine Cauchy.
First, we deal with the local Cauchy theory. The word “local” refers to the situation in a neighborhood
of a point. We begin with a fundamental property that leads to the whole development.

Definition.
Z (Nonstandard Terminology!) A continuous function f has the rectangle property in a region Ω
iff f(z) dz = 0 for every rectangle Γ such that Γ and its inside are contained in Ω and its edges are parallel
Γ
to the real or imaginary axes. Such a rectangle will be referred to as a “rectangle” in Ω.

Rectangle Theorem (or Goursat’s Theorem). If f is holomorphic on a region Ω, then f has the
rectangle property in Ω.

Integral Theorem. If f has the rectangle property in a disk B(c, R) (0 < R ≤ ∞), then f has an
antiderivative there.
Z
Cauchy’s Theorem on Disks. If f is holomorphic on a disk B(c, R), then f(z) dz = 0 for every closed
C
curve C in B(c, R) (by the rectangle, integral and closed curve theorems.)

Γ
Proof of the Rectangle Theorem. Let Γ be Za “rectangle” in Ω. Let s
be the length of the larger side of Γ. Suppose f(z) dz = I. Divide the
Γ
interior of Γ into four congruent subrectangular regions with boundaries Γ1 ,
Ω s Z 4 Z
X
Γ2 , Γ3 , Γ4 . Since f(z) dz = f(z) dz, one of the integral must satisfy
Γ j=1 Γj
Z 
  |I|
 
 f(z) dz  ≥ . Let us define Γ(1) = Γj .
 Γj  4
Repeat this subdivision to the interior ofZΓ(1) to obtain (2)
 a Γ and so on. We obtain a sequence of
  |I|
“rectangles” Γ(1), Γ(2), Γ(3), . . . in Ω such that  f(z) dz  ≥ k . The rectangular regions bounded by the
Γ(k) 4
(k)
Γ ’s “shrink” to a point z0 in Ω.
f(z) − f(z0 )
Since f is holomorphic at z0, so g(z) = − f 0 (z0 ) → 0 as z → z0 . So given ε > 0, there is
z − z0 √
(k) s 2s
δ > 0 such that |z − z0| ≤ δ ⇒ |g(z) − 0| ≤ ε. Observe that the sides of Γ are ≤ k , so |z − z0 | ≤ k
√ 2 2
(k) 2s
for all z in Γ . Choose k large so that k < δ, then
 2 
Z  Z  Z
  
|I|      
≤  f(z) dz = (f(z ) + f 0
(z )(z − z ) +g(z)(z − z )) dz  
= g(z)(z − z ) dz .
4k    Γ(k) |
0
{z
0 0
}
0

0 
Γ(k)   Γ(k)
polynomial
√ ! √
ε 2s  s  4 2s2 ε √
≤ k
4 k = k
⇒ |I| ≤ 4 2s2 ε.
2 2 4

31
 Z
Since ε is arbitrary, f(z) dz = I = 0. QED.
Γ

Proof of the Integral Theorem.

Z Without loss of generality, let c = 0. For w
b
in the disk, define F (w) = f(z) dz, where Ca,b denotes the curve from a to
C0,w Z
Re b + i Im a then to b. By the rectangle property, f(z) dz = 0 for the boundary
a Re b + i Im a Γ
Γ of every rectangle inside the disk.
We will show F 0 = f. Fix w in the disk. For any ε > 0, since f is continuous at w, there is δ > 0 such that
|z − w| ≤ δ ⇒ |f(z) − f(w)| ≤ ε.
R
For |h| ≤ δ, length of Cw,w+h ≤ 2|h|. Observe that f(w) = h1 Cw,w+h f(w) dz. So,
w+h
   Z 

 F (w + h) − F (w)  1  1
w  − f(w) =  (f(z) − f(w)) dz  ≤ ε · 2|h| = 2ε.
 h  h Cw,w+h  |h|

O F (w + h) − F (w)
Therefore, lim = f(w). QED.
h→0 h

Occasionally, we come across some continuous functions that are “almost” holomorphic in their domains
(with exceptions only on some “thin” subsets), it is remarkable that such functions still have the rectangle
property on their domains, as the following theorem asserts.

Extension Theorem. If f is continuous on a region D and holomorphic on D \ L, where L is a (horizontal)

line segment (or in the extreme case, a point), then f has the rectangle property on D.
Proof. Let Γ be a “rectangle” in D. There are three cases to consider.
(1) If Γ and its
Z interior do not contain any point of L, then Γ is a “rectangle”
Γ
in D \ L and f(z) dz = 0 by the holomorphicity of f.
Γ Γ Γ

L (2) If Γ has a (horizontal) edge e containing points of L, let Γε be the rectangle

that has the same edges as Γ except the edge e isZ replaced by an edge eε , which
is ε unit from L on the same side as Γ. By (1), f(z) dz = 0. Now as ε → 0,
Γε
the
Z (uniform)Zcontinuity of f on e implies
Z x0 Z
x0
eε
L f(z) dz = f(x + i(y0 + ε)) dx → f(x + iy0 ) dx = f(z) dz.
y0
e
eε Z x1 Z x1 Z e

Since 0 = f(z) dz → f(z) dz as ε → 0, we get f(z) dz = 0.

x0 x1 Γε Γ Γ
Z
Γ
(3) If the interior or a vertical edge of Γ contains a point of L, then f(z) dz =
Z Z Γ
upper L
f(z) dz + f(z) dz = 0, where we used the (horizontal) line through
lower Γupper Γlower
L to form two new “rectangles” and applied (2). QED.

Now we turn to the (global) Cauchy theory, which deals with the integration of holomorphic functions
on (closed) curves in (simply connected) domains. The curves need not be restricted to lie inside disks.
Z
Definition. A continuous function f defined on a region D has the polygon property iff f(z) dz = 0 for
Γ
every right-angled polygon Γ such that Γ and its inside are contained in D.

32
Polygon Theorem. Every holomorphic function on a simply connected region has the polygon property.
Proof. Subdivide the inside of every right-angled polygons into rectangular regions. Then apply the rectangle
theorem. QED.

Integral Theorem. Every funcion f having the polygon property on a simply connected region D has an
antiderivative F on D.
Z
Proof. Fix z0 ∈ D. Define F (z) = f(z) dz, where C is a polygonal line from z0 to z consisting of
C
horizontal or vertical segments.
Z (If C̃ is another
Z such polygonal line, then C and C̃ form finitely many
right-angled polygons. Hence f(z) dz − f(z) dz = 0 and F is well-defined.) That F 0 = f follows by
C C̃
the same argument as in the first integral theorem. QED.

Cauchy’s Theorem. If f is holomorphic on a simply connectedZ region D and C is a closed curve in D,

then (by the polygon, integral and closed curve theorems,) f(z) dz = 0.
C

Cauchy’s Theorem for Homotopic Curves. If C1, C2 are two curves Z with the
same initial and terminal points in a simply connected region D, then f(z) dz =
b1 Z C1

f(z) dz for every holomorphic function f on D. (This follows because C1 and −C2
b0 C2
form a closed curve, so we may apply Cauchy’s theorem above.)

As an appication of the (global) Cauchy theorems, we will present a theorem that deals with a global
question. For a holomorphic function without roots in a domain, locally (i.e. near every point) we can take
the logarithm of the function. It is of interest to know if there is a continuous logarithm of the function on
the whole domain.

Logarithm Theorem. Let f be holomorphic on a simply Z connected region D without any root in D. Fix
f 0 (w)
z0 ∈ D and choose a choice of log f(z0 ), then G(z) = dw + log f(z0 ) (where C is any curve in D
C f(w)
from z0 to z) defines a holomorphic branch of log f(z0 ) on D (i.e. eG(z) = f(z) for all z ∈ D). In particular,
if f is entire without any root, then f = eG for some entire function G.
f0 f0
Proof. Since is holomorphic on D, the proof of the integral theorem implies G0 ≡ . Consider the
f f
function A(z) = e−G(z) f(z). We have A0 (z) = e−G(z) (−G0(z)f(z)+f 0 (z)) ≡ 0. Since A(z0 ) = e−G(z0 ) f(z0 ) =
e− log f (z0 ) f(z0 ) = 1 and D is connected, A ≡ 1. Therefore eG ≡ f. QED.
Z z
dw
Remarks. (1) If D is simply connected and not containing 0, then log z = + log z0 is a holomorphic
z0 w
logarithm on D. Again because of simply connectedness, the path of integration can be any curve in D from
z0 to z.
(2) For α ∈ C, f(z)α = eα log f (z) wherever log f is defined.
 
z+1
Example. We will show that there are continuous (in fact, holomorphic) branches of log and
r z−1
z+1
on C \[−1, 1]. Note that C \[−1, 1] is not a simply connected region.
z−1
1+w
However, the region D = C \ ((−∞, −1] ∪ [1, +∞)) is simply connected. The function is holomorphic
1−w  
1+w
on D and has no roots there. By the logarithm theorem, there is a holomorphic branch G(w) = log
1−w

33
   r
1 z+1 1 z+1
on D. Let R(z) = , then F (z) = G(R(z)) = log and e 2 F (z) = are holomorphic on
z z−1 z−1
R−1(D) = C \[−1, 1].

Exercises

1. Let C1, . . . , Cn be simple closed curves inside a simple closed curve C0 , all with the counterclock-
wise orientation. Suppose no Ci is inside a Cj for positive i, j (i 6= j). Let S be the points in-
side C0 , but outside all Cj , j = 1, . . . , n. If h is holomorphic on C0, C1, . . ., Cn and on S, show that
Z Xn Z
h(z) dz = h(z) dz. [Hint: Draw a sketch and introduce cross-cuts to connect C0 with Cj .]
C0 j=1 Cj

C0
...
C1 C2
Cn

Z
1
2. Find dz, where C is a counterclockwise regular n-gon (n > 2) with center at the origin.
C z

34
11. Power Series Representation

Below we will show that holomorphic functions have power series representations locally. First we intro-
duce the concept of the winding number of a closed curve and deduce an integral formula for a holomorphic
function. Then the power series represntation follows by applying the formula for summing geometric series
to the integrand in the integral formula.
Z
1 dz
Definition. Suppose C is a closed curve and a 6∈ C. n(C, a) = is called the winding number
2πi C z − a
of C around a.

Example. Let C: z(t) = a + Reint (0 ≤ t ≤ 2π). This is the circle of radius R centered at a that “winds”
around a |n| times (counterclockwise if n > 0, clockwise if n < 0.) We have
Z Z 2π
1 dz 1 1
n(C, a) = = Rineint dt = n.
2πi C z − a 2πi 0 Reint

Theorem. For any closed curve C and a 6∈ C, n(C, a) is an integer.

Proof. Let C be given by z: [0, 1] → C. Since C is closed, z(0) = z(1). For 0 ≤ s ≤ 1, define
Z Z s
dz z 0(t)
F (s) = = dt.
z−a 0 z(t) − a
C |[0,s]

z 0(s) d
Then F (0) = 0 and F 0(s) = . This implies ((z(s)−a)e−F (s) ) = (z 0 (s)−(z(s)−a)F 0 (s))e−F (s) ≡ 0.
z(s) − a ds
So (z(s)−a)e−F (s) is constant. Then (z(1)−a)e−F (1) = (z(0)−a)e−F (0) = z(1)−a, which implies e−F (1) = 1.
1
Therefore F (1) = 2πik, where k is an integer, and n(C, a) = F (1) = k. QED.
2πi
z 0 (s)
Remark. Integrating both sides of F 0(s) = , then using F (0) = 0, we get
z(s) − a
     
z(s) − a  z(s) − a 
F (s) = log = ln   + i arg z(s) − a .
z (0)= z (1) z(0) − a z(0) − a  z(0) − a
a Therefore, if arg is continuous, then F (1) = F (1) − F (0) = i∆C arg(z − a) =
2πi×(number of times C “winds” around a), where ∆C arg(z − a) = arg(z(1) − a) −
C arg(z(0) − a) denotes the change of argument about a as C is traversed.

Cauchy’s Integral Formula. If f isZholomorphic on B(c, R), 0 < R ≤ ∞ and C is a closed curve in
 1 f(z)
B(c, R) \ a , then f(a) = dz.
2πi n(C, a) C z − a
(
f(z) − f(a)
Proof. The function g(z) = if z 6= a is holomorphic on B(c, R) \ {a} and continuous on
z−a
0
f (a) if z = a Z Z
f(z) − f(a)
B(c, R). By the extension, integral and closed curve theorems, dz = g(z) dz = 0. There-
Z Z C z−a C
f(z) f(a)
fore, dz = dz = f(a)2πi n(C, a). QED.
C z − a C z −a

35
Power Series Representation. If f is holomorphic on the disk B(a, R), 0 < R ≤ ∞,
X∞
f (n) (a) w
then f(w) = (w − a)n for all w in B(a, R).
n=0
n! a a+r a+R

Proof. For w in B(a, R), let r be such that |w − a| < r < R. Then the circle |z − a| = r, z(t) = a +!reit
 
w − a
winds about a and w once. For z on this circle,   = |w − a| < 1 and 1 = 1 1
=
z−a  r z−w z − a 1 − w−a z−a
∞  n
1 X w−a
. By Cauchy’s integral formula,
z − a n=0 z − a
Z Z X∞ X∞  Z 
1 f(z) 1 (w − a)n f(z) 1 f(z)
f(w) = dz = n+1
dz = dz (w − a)n.
2πi z−w 2πi n=0
(z − a) n=0
2πi (z − a)n+1
|z−a|=r |z−a|=r |z−a|=r
| {z }
βn

(The last equality follows from uniform convergence for all z on the circle by Weierstrass M-test
X∞   ∞
 (w − a)n  X |w − a|n M
 f(z) ≤ M= < ∞,
 (z − a)n+1  r n+1 r − |w − a|
n=0 n=0

∞
X
where M = max |f(z)|.) The power series representation f(w) = βn (w − a)n is valid for all w ∈
|z−a|=r
n=0
f (n) (a)
B(a, r). Therefore, by Taylor’s theorem for power series, βn = . Finally, we let r → R.
n!
QED.

Examples of Power Series of Common Functions.

X∞
zn X∞
(−1)n z 2n X∞
(−1)n z 2n+1
z
e = (all z), cos z = (all z), sin z = (all z),
n=0
n! n=0
(2n)! n=0
(2n + 1)!
ez + e−z X∞
z 2n ez − e−z X∞
z 2n+1
cosh z = = (all z), sinh z = = (all z),
2 n=0
(2n)! 2 n=0
(2n + 1)!
X∞ X∞
(−1)n−1(z − 1)n α α(α − 1) . . . (α − n + 1) n
Log z = (for |z − 1| < 1), (1 + z) = 1 + z (for any com-
n=1
n n=1
n!
plex α, |z| < 1.)

Corollary 1. Holomorphic functions are infinitely differentiable (because power series are).
This is an amazing fact that is certainly not true for differentiable functions of real variables!
(
f(z) − f(a) 00 000
Corollary 2. g(z) = if z 6= a can be represented by f 0 (a)+ f (a) (z−a)+ f (a) (z−a)2 +. . .
z−a 2! 3!
f 0 (a) if z = a
near a, hence it is holomorphic at a.

Corollary 3. If f(z) is holomorphic on Ω and has roots z1 , z2, . . . , zk , then there is a holomorphic function
gk (z) on Ω such that f(z) ≡ (z − z1 )(z − z2 ) . . . (z − zk )gk (z).

 f(z) − f(z1 )
if z 6= z1
Proof. For n = 1, g1 (z) = is holomorphic on Ω. The inductive step follows by
 0 z − z1
f (z1 ) if z = z1
observing that gn(z) = (z − zn+1 )gn+1 (z) and getting gn+1 as in case n = 1. QED.

36
 Exercises

1. The following curve C divides the plane into 4 regions. For each region, state the winding number of C
around points in that regions. (Give answers by inspection, no computation is needed.)

4
1 2 3

C
49
z Log(z + 1)
2. Find lim .
z→0 (cos z 25) − 1
Z Z Z
sin z sin z sin z
3. Find dz, dz and dz, where the circles are oriented counterclockwisely.
z z2 z2 − 1
|z|=1 |z|=1 |z|=2

4. If f is holomorphic on the open unit disk D and |z| + |f(z)| ≤ 1 for all z ∈ D, then show that f ≡ 0.

5. Suppose f is holomorphic on D = {z : |z| < 1} and f is an even √ function (i.e. f(z) = f(−z)). Show
that there is a holomorphic function g on D such that g(x) = f( x) for all positive real number x < 1.
Z 1
sin zt
6. Prove that f(z) = dt is an entire function by obtaining a power series expansion for f.
0 t

7. Suppose f is holomorphic on the closed unit disk {z : |z| ≤ 1} and |a| < 1, show that (1 − |a|2)|f(a)| ≤
Z 2π
1 1 − az
|f(eiθ )| dθ. [Hint: Consider integrating f(z) .]
2π 0 z−a

v y f
8. Let f(z) = u(z)+iv(z) (or f(x, y) = (u(x, y), v(x, y))) be a one-to-
one holomorphic function from the open unit disk D = {z: |z| < 1}
onto a domain G with finite area. x u
 ∂u ∂u 
 
 
def  
(a) Show that Jf (x, y) =  ∂x ∂y  = |f 0 (z)|2 . D G
 ∂v ∂v 
 
∂x ∂y Z
(b) For distinct nonnegative integers m, n, show z m z n dA = 0 (orthogonality relation), where dA =
D
r dr dθ = dx dy is the area differential.
∞
X ∞
X
(c) Show that if f(z) = cnz n is the power series for f in D, then area of G = π n|cn|2.
n=0 n=1

∞  (n) 
X  f (0) 2
9. Suppose f is holomorphic on the open unit disk {z : |z| < 1} such that  
 n!  < ∞. Prove that
n=0
Z 2π X∞  (n) 2
1  f (0) 
lim |f(reiθ )|2 dθ =  
r→1− 2π 0
 n!  .
n=0

X
∞ X
∞
10. Suppose f(z) = anz n and g(z) = bn z n converges for all z ∈ B(0, 1). Let |z0 | < r < 1.
n=0 n=0
∞
X z0n
(a) Show that bnf(w) n+1 converges uniformly on the circle {w : |w| = r}.
n=0
w

37
 Z  z  dw X∞
1 0
(b) Show that f(w)g = anbn z0n.
2πi |w|=r w w n=0

 1/n
X  X
 M n M

11. Prove lim  aj  = max |aj |. [Hint: Consider the power series with an
j as coefficients.]
n→∞   j=1,···,M
j=1 j=1

38
12. Consequences of Cauchy Theory

Cauchy Integral Formula for Derivatives.

Z If f is holomorphic on B(a, R), then for n = 0, 1, 2, . . . and
(n) n! f(z)
0 < r < R, we have f (a) = dz (since both sides equal n!βn as shown in the proof of
2πi (z − a)n+1
|z−a|=r
the power series representation).

Notice this formula said something amazing, namely the derivative at a point can be given by an integral!

We now give an example which illustrates the fact that the Cauchy integral formula for derivatives is
also true for closed curves in a region, provided winding number factors are included. This will be proved
later.
Z b0
f(z)
Example. Let f be entire, find 20
dz, where C is the curve a
C (z − a)
.
Z Z Z b0 b0
Solution. = + , where C1 is the curve a and C2 is the curve a
C C1 C2
.
b0 b0
f(z)
Since is holomorphic on the slit planes a a , which are simply connected,
(z − a)20
b1 b1
Z Z Z Z
by Cauchy’s theorem for homotopic curves, = . Similarly, = .
C1 C2
|z−a|=ε |z−a|=ε
Z Z (19)
f(z) f(z) f (a)
So, dz = 2 dz = 2πi n(C, a) .
C (z − a)20 (z − a)20 19!
|z−a|=ε

When combined with the M -L inequality, the Cauchy integral formula for derivatives gives estimates
on the size of derivatives or coefficients of power series.

Liouville’s Theorem. A bounded entire function is constant. (More generally, if f(z) is entire and if there
are constants A and B and α ≥ 0 such that |f(z)| ≤ A + B|z|α when |z| is large (i.e. |z| > r for some r > 0),
then f(z) is a polynomial of degree k ≤ α.)
X∞
f (n) (0) n
Proof. Since f(z) is holomorphic on B(0, ∞), f(z) = z for all z (in B(0, ∞)). We will show
n=0
n!
that f (n) (0) = 0 for n > α. This implies f(z) is a polynomial of degree at most k ≤ α.
Let n > α and R > r. By the Cauchy integral formula for derivatives and M -L inequality, as R → ∞,
 
   
   n! Z f(z)  n! A + BRα n!(A + BRα )
 (n)   
f (0) =  dz  ≤ 2πR = → 0.
 2πi z n+1  2π Rn+1 Rn
 |z|=R 

39
(We can take limit as R → ∞ because f(z) is defined in the whole plane, hence the circle |z| = R would
always be in the domain of f(z), so Cauchy integral formula for derivatives can be applied!) Therefore,
f (n) = 0 and f(z) = f(0) + f 0 (0)z + . . . + f (k) (0)z k with k ≤ α. QED.
In an advanced course, a theorem called the Little Picard Theorem is usually proved, which asserts that
an entire function whose range misses (i.e. not includes) two complex numbers must be constant. This is a
big improvement over Liouville’s theorem, since bounded functions are functions whose ranges are bounded
sets, hence miss many complex numbers. An example of an entire function whose range misses only one
value is ez . Its range is C \{0}, hence misses only 0.

L’ Hôpital’s Rule. If f(z) and g(z) are holomorphic in Ω and for some a ∈ Ω, f(a) = f 0 (a) = . . . =
f (n−1) (a) = 0, g(a) = g0 (a) = . . . = g(n−1)(a) = 0 and either f (n) (a) 6= 0 or g(n) (a) 6= 0 for some a ∈ Ω, then
f(z) f (n) (a)
lim = (n) .
z→a g(z) g (a)
∞
X f (k) (a) f (n) (a)
Proof. Near a, the power series of f(z) is (z − a)k = (z − a)n + F (z)(z − a)n+1 .
k! n!
k=n+1
Similarly, the power series of g(z) near a has a similar form. Then

f(z) f (n) (a) + n!F (z)(z − a) f (n) (a)

lim = lim (n) = (n) .
z→a g(z) z→a g (a) + n!G(z)(z − a) g (a)
QED.

Definition. If a holomorphic function f(z) has a root a such that f(a) = f 0 (a) = . . . = f (n−1) (a) = 0 but
f (n) (a)
f (n) (a) 6= 0 , it is a root of order (or multiplicity) n. (Equivalently, this means that f(z) = (z − a)n +
n!
f (n+1) (a)
(z − a)n+1 + · · · near a with f (n) (a) 6= 0 or f(z) = (z − a)n g(z) with g(a) 6= 0.) A root of order
(n + 1)!
one, two or three is called a simple, double or triple root, respectively.

(sin z 2)(cos z − 1)
Example. Find lim 4 .
z→0 ez − 1
3
w w2 w4 w2
When w is near 0, sin w = w − + . . . , cos w − 1 = − + − . . . and ew − 1 = w + + . . ..
3! 2! 4! 2!
6 2 4 4
z z z z
(sin z 2 )(cos z − 1) (z 2 − + . . . )(− + −... ) − +... 1
So lim = lim 3! 2! 4! = lim 24 = − . (Incidentally,
z 4 8
z→0 e −1 z→0 z z→0 z + . . . 2
z4 + +...
2!
from the power series, we can also see that 0 is a double root for sin z 2 , cos z − 1 and a root of order 4 for
4
ez − 1.)

Fundamental Theorem of Algebra. Every nonconstant polynomial P (z) with complex coefficients has
a root in C.
1
Proof. Suppose P (z) has no root, then f(z) = is entire. Let P (z) = anz n + an−1z n−1 + · · · + a0 with
P (z)
n ≥ 1, an 6= 0. As z → ∞, eventually |z| > 1, then
  a 
 0 
|P (z)| ≥ |anz n | − |an−1z n−1| − · · · − |a0| = |z|n−1 |anz| − |an−1| − · · · −  n−1 
z
≥ |z|n−1(|anz| − |an−1| − · · · − |a0|) → ∞.
1
So, lim P (z) = ∞ and lim = 0. Then |f(z)| ≤ 1 when |z| is large. By Liouville’s theorem, f(z) is a
z→∞ z→∞P (z)
1
polynomial of degree 0, i.e. constant. Then P (z) = is constant, a contradiction. QED.
f(z)

40
Identity Theorem. Suppose f is holomorphic on a region D and the roots of f has a limit point in D (i.e.
there are an ∈ D such that f(an ) = 0 and an → a ∈ D), then f ≡ 0.

Uniqueness Theorem. Suppose g and h are holomorphic on a region D and {z: g(z) = h(z)} has a limit
point in D, then g ≡ h (by considering f = g − h).

Proof of the Identity Theorem. Let A = {z ∈ D: z is a limit of roots of f(z)} and B = D \ A. Then
A ∪ B = D, A ∩ B = ∅ and A 6= ∅ by hypothesis.
X∞
f (k) (a)
If a ∈ A, then a = lim an with f(an ) = 0. The power series of f(z) at a, namely (z − a)k ,
n→∞ k!
k=0
has roots an with limit at the center a, so f(z) = 0 in a disk around a. Then every point in this disk belongs
to A. Hence, this disk is a neighborhood of a in A. So A must be open.
If b ∈ B, then there is a disk around b such that f(z) 6= 0 in the disk (except possibly at b only). Then
all points of this disk cannot be in A. Hence they must be in B. Then this disk is a neighborhood of b in B.
So B must be open.
Since D is connected, B = ∅ and A = D. Therefore f ≡ 0. QED.

Examples. (1) The functions g(z) = sin 2z and h(z) = 2 sin z cos z are entire. Since g(x) = h(x) for every
1
real number x and the real numbers have limit points (e.g. → 0 as n → ∞), by the uniqueness theorem,
n
g ≡ h. Therefore sin 2z = 2 sin z cos z for all complex number z.

1 1
(2) The function f(z) = sin ( ) is holomorphic on D = C \{0}. Its roots are (n ∈ Z). As n → ∞,
z nπ
1
→ 0, but we cannot conclude f(z) ≡ 0 because 0 is not in D.
nπ

Mean Value Theorem. If f is holomorphic on a region D containing B(a, r), then (by the Cauchy integral
Z Z 2π
1 f(z) 1
formula) f(a) = dz = f(a + reiθ ) dθ, which is the average of f on the circle C(a, r).
2πi z −a 2π 0
|z−a|=r

Maximum Modulus Theorem. If f is a nonconstant holomorphic function on a region D and B(a, r)

is contained in D, then |f(a)| < |f(w)| for some w in B(a, r). (In addition, if D is bounded and f is also
continuous on the boundary ∂D, then the maximum of |f(z)| can only (and must) occur on ∂D.)
Proof. Let 0 < r0 < r and M = max |f(a + r0 eiθ )|. Since |f(a + r0 eiθ )| is a continuous function on [0, 2π],
0≤θ≤2π
M = |f(a + r0 eiθ0 )|, 0 ≤ θ0 ≤ 2π. Observe that
Z  Z 2π Z 2π
1  2π 0 iθ

≤ 1 0 iθ 1
|f(a)| = f(a + r e ) dθ |f(a + r e )| dθ ≤ M dθ = M.
2π  0  2π
0 2π 0

If |f(a)| < M , then |f(a)| < |f(w)| for w = a+r0eiθ0 and we are done. If |f(a)| = M , then |f(a+r0 eiθ )| ≡ M
because |f(a + r0 eiθ )| is continuous in θ. If this is true for all r0 < r, then f ≡ f(a) on B(a, r) by the Cauchy-
Riemann equations and hence on D by the uniqueness theorem, contradicting the hypothesis. Otherwise
|f(a + r0 eiθ )| 6≡ |f(a)| for some r0 < r, then some w = a + r0eiθ0 will satisfy |f(a)| < |f(w)|. QED.

Minimum Modulus Theorem. If f is a nonconstant holomorphic function on a region D and B(a, r) is

contained in D, then |f(a)| > |f(w)| for some w in B(a, r) unless f(a) = 0. (In addition, if D is bounded
and f is also continuous on the boundary ∂D, then either f has a root in D or the minimum of |f(z)| can
only occur on ∂D.)
Proof. If f(a) 6= 0, then f(z) 6= 0 in some subdisk of B(a, r). Apply the maximum modulus theorem to
1
on the subdisk. QED.
f(z)

41
Open Mapping Theorem. If f is a nonconstant holomorphic function on an open set U , then f(U ) is an
open set. (In particular, f(U ) has no boundary points.)

1
Proof. Let w ∈ U . (If each of the circles Ck = C(w, ) has a point wk such that f(wk ) = f(w), then since
k
wk → w ∈ U , the uniqueness theorem will imply f(z) ≡ f(w).) Now, since f is nonconstant, so there must
1 1
be a circle C = C(w, ) such that f(z) 6= f(w) for all z on C. Let m = min |f(z) − f(w)|. We will show
k 2 z∈C
B(f(w), m) ⊂ f(U ). (This will imply every point in f(U ) has a neighborhood in f(U ), hence f(U ) is open.)

| f ( z) - p | >
_ m Let p ∈ B(f(w), m) and consider the minimum modulus of f(z) − p on
w
B(w, r). For z ∈ C, |f(z) − p| ≥ |f(z) − f(w)| − |f(w) − p| ≥ 2m − m = m.
However, |f(w) − p| < m. So the minimum modulus of f(z) − p is not in C, hence
D
C f(z) − p has a root inside C. Then p ∈ f(U ). So B(f(w), m) ⊂ f(U ). QED.

In particular, the open mapping theorem implies the range of a holomorphic function cannot be an arc
or a curve or any closed set. We saw this before by using the Cauchy-Riemann equations.

Schwarz’s Lemma. Let D be the open unit disk. If f: D → D is holomorphic and f(0) = 0, then |f(z)| ≤ |z|
for any z ∈ D \ {0} and |f 0 (0)| ≤ 1. Equality holds in either case only for f(z) ≡ eiθ z, θ real.
(
f(z)
Proof. Define g(z) = if 0 < |z| < 1 , then g is holomorphic on D. On the circle |z| = r < 1,
0
z
f (0) if z = 0
|f(z)| 1 1
|g(z)| = ≤ . Maximum modulus theorem implies |g(z)| ≤ for |z| ≤ r. Let r → 1− , we get
|z| r r
|g(z)| ≤ 1 for |z| < 1, which implies |f(z)| ≤ |z| and |f 0 (0)| ≤ 1.
If |f(z)| = |z| for some z ∈ D or |f 0 (0)| = 1, then |g(z)| = 1 for some z ∈ D. The maximum modulus
theorem implies g ≡ constant (≡ eiθ ), then f(z) ≡ eiθ z. QED.

Remarks. If we recall that every proper simply connected domain is conformally equivalent to the open unit
disk D by the Riemann mapping theorem, then we see that Schwarz’s lemma is not just about holomorphic
mappings from D to D, but it can be viewed as a statement about mappings from simply connected domains
to simply connected domains.

This is also true of many other theorems in complex analysis with settings on D. For theorems that are
true for all simply connected domains, often all we need to prove is the case for the open unit disk, then the
general case usually follows by the Riemann mapping theorem and conformal mappings.

Example. If f is holomorphic on the right half plane such that |f(z)| ≤ 1 and f(1) = 0, what is the
maximum of |f(2)| as all such functions f are considered? Which ones attain the maximum?

f
1−z
Solution. Let T (z) = , which maps the unit disk D onto the
1+z
right half plane and let g(z) = f(T (z)), then g is holomorphic on D,
1
|g(z)| ≤ 1 and g(0) = 0. By Schwarz’s lemma, |f(T (z))| = |g(z)| ≤ |z|.
1 1 1
(We want T (z) = 2, so take z = − .) Then |f(2)| = |f(T (− ))| ≤ .
3 3 3
T
1 1 1
So max |f(2)| = . For this to happen, we must have |g(− )| = ,
3 3 3
1−w
then f(T (z)) = g(z) = eiθ z. Therefore, f(w) = eiθ T −1 (w) = eiθ .
1+w

42
 Exercises

1. If the range of an entire function lies in the right half plane Re w > 0, show that the function is a
constant function. [Hint: Compose with a Möbius mapping.]

2. Suppose a polynomial is bounded by 1 on the open unit disk. Show that all of its coefficients are
bounded by 1.

3. Show
 that if f is holomorphic on {z : |z| ≤ 1}, then there must be some positive integer k such that
1 1
f 6= .
k k+1

4. Suppose f is holomorphic on the annulus {z : 1 ≤ |z| ≤ 2}, |f(z)| ≤ 1 for |z| = 1 and |f(z)| ≤ 4 for
|z| = 2. Prove that |f(z)| ≤ |z|2 throughout the annulus.

5. Let D be the open unit disk. If f : D → D is holomorphic with at least two fixed points (i.e. points w
such that f(w) = w), show that f(z) ≡ z. [Hint: By composing with a suitable Möbius mapping, one
of the fixed points may be moved to the origin.]

6. Let f be an entire function which is real on the real axis and imaginary on the imaginary axis, show
that f is an odd function, i.e. f(z) = −f(−z).

7. Suppose f is a nonconstant holomorphic function on the closed annulus A = {z : 1 ≤ |z| ≤ 2}. If f

sends the boundary circles of A into the unit circle, show that f must have a root in A.

8. Use the uniqueness theorem to prove the identity sin(w + z) = sin w cos z + cos w sin z for all complex
numbers w and z.
Z
dz
9. Find 4 2
, where the unit circle is given the counterclockwise orientation.
|z|=1 z + 4z

Z 2π
10. Find cos(cos θ) cosh(sin θ) dθ. [Hint: Consider cos eiθ .]
0

11. Suppose f and g are holomorphic on the closed unit disk. Show that |f(z)| + |g(z)| takes its maximum
on the boundary. [Hint: Consider f(z)eiα + g(z)eiβ for appropriate α and β.]

1
12. Suppose f is entire and |f 0(z)| ≤ |z| for all z . Show that f(z) = a + bz 2 with |b| ≤ .
2
  
 1 
13. Find the maximum of f , where f is holomorphic on D = {z : |z| < 2}, f(1) = 0 and |f(z)| ≤ 10
2 
for z ∈ D.
 
1 2
14. Find all holomorphic function(s) f defined on the open unit disk D satisfying f = and f(z) =
2 3
(2 − f(z))f(2z) for all z ∈ D.

15. Let H = {z : Re z > 0}. Suppose f : H → H is holomorphic and f(1) = 1. Show that |1 − f(2)| ≤
1
|1 + f(2)|.
3

16. If f is entire and Re f 0 (z) > 0 for all complex numbers z, prove that f is a polynomial of degree 1.

43
17. Given the polynomial P (z) = z n + an−1z n−1 + · · · + a0 . Prove that max |P (z)| ≥ 1.
|z|=1

18. If f is an entire function mapping the unit circle into the unit circle (i.e. |f(z)| = 1 for |z| = 1), then
f(z) = eiθ z n for some real θ and some positive integer n. [Hint:  In the
 unit disk, f has finitely many
 z − αj 

roots α1 , . . ., αn, repeated according to multiplicities. Recall   = 1 for |z| = 1. Use the modulus
1 − αj z 
Yn
z − αj
theorems to show f(z) = eiθ first.]
j=1
1 − αj z

19. Let f and g be holomorphic on a domain U . If fg is holomorphic on U , show that either f ≡ 0 or g is

a constant function.

20. Let f be holomorphic on the open unit disk D. Show that there is a sequence {zn} in D with |zn| → 1
such that {f(zn )} is a bounded sequence. [Hint: Consider the roots of f.]

21. Let w and z be in the open unit disk D. If f : D → D is holomorphic and f(w) = z, prove that
1 − |z|2
|f 0 (w)| ≤ .
1 − |w|2

22. (Study’s Theorem) Let G be a convex domain and f be a one-to-one holomorphic function from the
open unit disk onto G. Prove that f(B(0, r)) is a convex domain for 0 < r < 1. (Recall a set S is convex
if w, z ∈ S implies tw + (1 − t)z ∈ S for allt ∈ [0, 1].)

44
13. Harmonic Functions and Conjugates

Definitions. A real-valued function u(x, y) having continuous second order partial derivatives on a region
∂2u ∂2u
D is harmonic on D iff it satisfies the Laplace equation + 2 ≡ 0 on D. If v is harmonic on D and
∂x2 ∂y
u + iv is holomorphic on D, then we say v is a harmonic conjugate of u on D.

Theorem. If f = u + iv is holomorphic on D, then u (and similarly, v) is harmonic on D.

       
∂2u ∂ ∂u ∂ ∂v ∂ ∂v ∂ ∂u
Proof. By the Cauchy-Riemann equations, we get = = = = − =
∂x2 ∂x ∂x ∂x ∂y ∂y ∂x ∂y ∂y
∂2u ∂2v ∂2v
− 2 . Similarly, = − . QED.
∂y ∂x2 ∂y2

Remark.
p In general, the converse of the theorem is false, which can be seen from the example u(x, y) =
ln x2 + y2 on C \{0}. If u = Re f there, then locally, the Cauchy-Riemann equations imply (f(z)−log z)0 =
∂ ∂
(u(z) − ln |z|) − i (u(z) − ln |z|) = 0. So the only possibility is f(z) = log z + constant, which is not
∂x ∂y
continuous on C \{0}. However, for simply connected domains, the converse of the theorem is true.

Theorem. If u is harmonic on a simply connected region D, then u is the real part of a holomorphic function
f on D.
∂u ∂u
Proof. Define g(z) = (z) − i (z), then g is holomorphic on D because u has continuous second order
∂x ∂y        
∂ ∂u ∂ ∂u ∂ ∂u ∂ ∂u
partial derivatives and the Cauchy-Riemann equations = − , − =−
∂x ∂x ∂y ∂y ∂x ∂y ∂y ∂x
0
are satisfied on D. By the integral theorem, g has an antiderivative f, i.e. f (z) = g(z). Now

∂ ∂u ∂u ∂ ∂u ∂u
(Re f − u) = Re fx − = Re f 0 − = 0 and (Re f − u) = Re fy − = Re(if 0 ) − = 0.
∂x ∂x ∂x ∂y ∂y ∂y

So Re f − u ≡ constant, i.e. u ≡ Re(f + constant) on D. QED.

Below we shall give some rules for recovering a holomorphic function f = u + iv from u (and v). Since
no proofs of these rules will be offered, proper checking should be made to ensure the functions obtained
have the correct real and imaginary parts.
∂u ∂u
Rule A. If u is harmonic on D containing an interval of the real axis, then f 0 (z) = (z, 0) − i (z, 0) and
∂x ∂y
f is obtained by integrating f 0 .
 
z + z0 z − z0
Rule B. If u is harmonic on B(z0 , r), z0 = x0 + iy0 , then take f(z) = 2u , − u(x0, y0).
2 2i
Rule C. If u, v are harmonic conjugate on B(0, r), then take f(z) = u(z, 0) + iv(z, 0).

∂u ∂v ∂u ∂v
Examples. (1) u(x, y) = x2 − y2 , (x, y) = 2x = − (x, y), (x, y) = −2y = (x, y), which imply
∂x ∂y ∂y ∂x
v(x, y) = 2xy + constant.
Rule A: f 0 (z) = 2z − i0 ⇒ f(z) = z 2 .

45
 z z   2 
z z2
Rule B: Take z0 = 0, f(z) = 2u , =2 + = z 2.
2 2i 4 4
Rule C: Take v(x, y) = 2xy. Then f(z) = (z 2 − 0) + i0 = z 2 .
∂u ∂v ∂u ∂v
(2) u(x, y) = e−y sin x, (x, y) = e−y cos x = − (x, y), (x, y) = −e−y sin x = (x, y), which imply
∂x ∂y ∂y ∂x
−y
v(x, y) = −e cos x + constant.

Rule A: f 0 (z) = cos z + i sin z = eiz ⇒ f(z) = −ieiz .

z eiz/2 − e−iz/2
Rule B: Take z0 = 0, f(z) = 2e−z/2i sin = e−z/2i = −ieiz + i.
2 i
Rule C: Take v(x, y) = −e−y cos x. Then f(z) = sin z + i(− cos z) = −ieiz .

x ∂u y 2 − x2 ∂v ∂u −2xy ∂v
(3) u(x, y) = , (x, y) = = − (x, y), (x, y) = 2 = (x, y), which imply
x2 + y2 ∂x (x2 + y2 )2 ∂y ∂y (x + y2 )2 ∂x
y
v(x, y) = − 2 + constant.
x + y2
z2 1 1
Rule A: f 0 (z) = − = − 2 ⇒ f(z) = .
z4 z z
(z + 1)/2 1
Rule B: Take z0 = 1, f(z) = 2 2 2
−1 = .
(z + 1) /4 + (z − 1) / − 4 z
y 1 1
Rule C: Although u(0, 0) is undefined, we will try and see. Take v(x, y) = − . Then f(z) = + 0 = .
x2 + y 2 z z

Mean-Value Theorem for Harmonic Functions. If u is harmonic on B(z0 , R), then for 0 ≤ r < R,
 Z 2π  Z 2π Z 2π
1 iθ 1 iθ 1
u(z0 ) = Re f(z0 ) = Re f(z0 + re )dθ = Re f(z0 + re )dθ = u(z0 + reiθ )dθ.
2π 0 2π 0 2π 0

Maximum/Minimum Principle for Harmonic Functions. If u is a nonconstant harmonic function on

an open set D, then u has no maximum or minimum on D. (If D is bounded and u is also continuous on D
and its boundary ∂D, then u attains maximum and minimum values on ∂D only.)

Proof. Suppose u has a maximum or minimum at p ∈ D. Let D0 = B(p, ε) be a disk in D. Then u has a
maximum or minimum at p on D0 . Now u = Re f on D0 for some nonconstant holomorphic function f on D0 .
By the open mapping theorem, f(D0 ) is open in C. Hence Re f(D0 ) = {Re f(z): z ∈ D0 } = {u(z): z ∈ D0 }
is open in R. Then u cannot have a maximum or minimum on D0 ⊂ D, a contradiction. QED.

Corollary. If u1, u2 are harmonic on a bounded open set D, continuous on D and its boundary ∂D, u1 = u2
on ∂D, then (considering u = u1 − u2, we get) u1 ≡ u2 on D. So, the function u 7→ u|∂D is one-to-one.

Poisson Integral Formula. If u is harmonic on B(0, 1) and continuous on B(0, 1), then for z = reiθ ∈
Z 2π
1 1 − r2
B(0, 1), u(z) = u(eit) dt. This formula gives an inverse to the function u 7→
2π 0 1 − 2r cos (θ − t) + r2
u|∂B(0,1).

Proof. Since B(0, 1) is simply connected, u = Re f for some holomorphic function f on B(0, 1). For
r < R < 1, the Cauchy integral formula gives
Z Z Z
1 f(w) 1 f(w) 1 1 1
f(z) = dw − dw= ( − )f(w) dw.
2πi w−z 2πi R2 2πi w−z R2
|w|=R |w|=R w − |w|=R w−
| {z z } z
=0

46
For w = Reit and z = reiθ , we have dw = iReit dt and

R2 iθ
1 1 1 1 (r − )e
− = − = r
w−z R2 Re − reiθ
it R2 iθ R2 iθ
w− Reit − e (Reit − reiθ )(Reit − e )
z r r
(R2 − r2 ) R2 − r2 1
= = 2 .
(Rei(t−θ) − r)(Rei(θ−t) − r)Reit R − 2Rr cos(θ − t) + r2 Reit
Z 2π
1 R2 − r2
So, f(z) = f(Reit ) dt. Taking the real part of both sides and letting
2π 0 R2 − 2Rr cos (θ − t) + r2
Z 2π
1 1 − r2
R → 1− , we get u(z) = u(eit ) dt. QED.
2π 0 1 − 2r cos (θ − t) + r2

Remark. In an advanced course, it can be shown that if u is continuous on C(0, 1), then u(z), defined by
the Poisson integral above, is harmonic on B(0, 1) and continuous on B(0, 1).

Exercises

1. If f = u + iv is holomorphic, prove that u + v and uv are harmonic.

2. If u is harmonic, prove that all the partial derivatives of u are harmonic, but u2 is not harmonic, unless
u is constant.

3. Find holomorphic functions whose real parts are

2
−y 2
(a) x3y − xy3 ; (b) ex cos(2xy);
y 1 − x2 − y 2
(c) arctan ; (d) .
x (1 − x)2 + y2

4. Is the parenthetical part of the maximum/minimum principle true if D is unbounded? [Hint: Consider
u(x, y) = ±y on {z : Im z ≥ 0}.]
Z 2π Z 2π
dt 3ecos t cos(sin t)
5. Using the Poisson integral formula, find and √ dt.
0 5 − 4 cos t 0 5 − 4 cos(t − 2)
 
1 ∂ ∂u 1 ∂2u
6. Show that in polar coordinates, Laplace’s equation is r + 2 2 = 0.
r ∂r ∂r r ∂θ

7. (a) Suppose g is holomorphic on the closed unit disk, g(0) = 1 and Re g(z) > 0 for |z| < 1. Use the
1 + |z|
Poisson integral formula to show that Re g(z) ≤ for |z| < 1.
1 − |z|
1+|z|
(b) Suppose f : {z : |z| ≤ 1} → {z : 0 < |z| < 1} is holomorphic. Show that |f(z)| ≤ |f(0)| 1−|z| for
|z| < 1.

47
14. Morera's Theorem

The rectangle theorem asserts that if f is holomorphic, then f has the rectangle property. Below we shall
prove the converse for continuous functions. Thus, holomorphicity is equivalent to the rectangle property.

Morera’s Theorem. If a continuous function f has the rectangle property on an open set U , then f is
holomorphic on U .
Proof. For a ∈ U , take a disk B(a, r) in U . Since f has the rectangle property, the integral theorem implies
f has an antiderivative F on B(a, r). That is, F is differentiable on B(a, r) and F 0 = f. Then F is infinitely
differentiable, so f is differentiable on B(a, r). Since a is arbitrary, f is holomorphic on U . QED.
Z Z  zt  Z
∞
ezt ∞  e  ∞
Example. Define F (z) = dt for U = {z: Re z < 0}. (Observe that   dt ≤ ext dt
t+1 t + 1
0 0 0
1 1
=− = for x = Re z < 0. So, F is defined on U .) Now, let Γ be a “rectangle” in U . We have
x |x|
Z Z ∞ Z ∞ Z
ezt ezt
dt dz = dz dt = 0
Γ 0 t+1 0 Γ t+1

because for a fixed positive t, ezt is holomorphic on U . (The interchange of integrals is valid because
Z Z  zt  Z
∞  e  1
  dt dz ≤ dz < ∞,
t + 1
Γ 0 Γ |x|

since x = Re z is bounded away from 0 on Γ.) Therefore, F is holomorphic on U by Morera’s theorem.

As an application of Morera’s theorem, we will consider limits of holomorphic functions. In general, the
limit of an sequence of continuous functions may not be continuous (e.g. fn (x) = xn , 0 ≤ x ≤ 1 has the limit
f(x) = 0 if 0 ≤ x < 1 , which is not continuous). So the limit of a sequence of holomorphic functions
1 if x = 1
may not be holomorphic. However, if the convergence to the limit is uniform, then it is holomorphic. In
that case, the derivatives will also converge uniformly to the derivative of the limit.

Weierstrass’ Theorem. If fn is a sequence of holomorphic functions on a region D and fn converges

uniformly to f on every closed disk B(a, r) = {z: |z − a| ≤ r} in D, then f is holomorphic on D. (Fur-
(k)
thermore, for any positive integer k, fn will also converge uniformly to f (k) on every closed disk in D, i.e.
k k
d d
lim fn = k lim fn uniformly on D.)
n→∞ dz k dz n→∞
Proof. For every a in D, take a closed disk B(a, r) in D. Since the fn ’s areZcontinuous and
Z the convergence
is uniform, f is continuous on B(a, r). For a “rectangle” Γ in B(a, r), f(z) dz = lim fn (z) dz =
Z Γ Γ n→∞

lim fn (z) dz = 0. By Morera’s theorem, f is holomorphic at a for any a ∈ D.

n→∞ Γ
(For the parenthetical statement, let k be a positive integer, B(a, r) ⊂ D and ε > 0 be given. Then
there is R > r such that B(a, r) ⊂ B(a, R) ⊂ D. Fix a ρ such that 0 < ρ < R − r. Since fn converges
uniformly to f on B(a, R), there is N such that n ≥ N ⇒ |fn(w) − f(w)| < ε for all w ∈ B(a, R). Now for

48
any arbitrary z ∈ B(a, r), C(z, ρ) ⊂ B(a, R). By the Cauchy integral formula for derivatives and the M -L
inequality,
   
   k! Z fn (w) − f(w)  k! 1 k!ε
 (k) (k)   
fn (z) − f (z) =  dw  ≤ ε 2πρ = k .
 2πi |w−z|=ρ (w − z) k+1  2π ρ k+1 ρ

Since z is arbitrary in B(a, r), uniform convergence is proved.) QED.

The next type of applications of Morera’s theorem concerns extending holomorphic functions across line
segments.

Schwarz Reflection Principle. Let D be a symmetric region with respect to the real axis, Dupper = {z ∈
D : Im z > 0} and Dlower = {z ∈ D : Im z < 0}. If f is continuous on Dupper ∪ (D ∩ R), holomorphic
on Dupper and real-valued on D ∩ R, then f can be extended to a holomorphic function on D by defining
f(z) = f(z) for z ∈ Dlower .
Proof. The hypothesis that f is real-valued on D ∩ R is equivalent to f(x) = f(x)
for all x ∈ D ∩ R. This makes the extension continuous on all of D. On Dlower , we
can apply the definition of derivative to get
D upper
f(z + h) − f(z) f(z + h) − f(z)
D lower
R
f 0 (z) = lim = lim = f 0 (z)
h→0 h h→0 h
for all z ∈ Dlower . So, f is holomorphic on D \ R . Then the extension theorem and
Morera’s theorem imply f is holomorphic on D.
QED.

Exercises

Z 1
sin zt
1. Use Morera’s Theorem to show that f(z) = dt is an entire function.
0 t

2. Prove that if f is continuous on the closed unit disk {z : |z| ≤ 1}, holomorphic on the open unit disk
{z : |z| < 1} and real-valued on the unit circle {z : |z| = 1}, then f is a constant function.

3. Let f be an entire function which is real on the real axis.

(a) Use Schwarz reflection principle to prove that f(z) = f(z) for every complex number z.
(b) If f is also imaginary on the imaginary axis, prove that f is an odd function by considering the
power series of f at the origin.

49
15. Isolated Singularities

In this section, we will consider a two-sided infinite series representation near a singular point in the
domain of a holomorphic function.

X
∞ X
−1 X
∞ X
−1 X
∞
Definition. We say wk converges to L iff wk , wk converge and wk + wk = L.
k=−∞ k=−∞ k=0 k=−∞ k=0

Laurent Series Representation. If f is holomorphic on the annulus A = {z: 0 ≤ R1 < |z − a| < R2 ≤ ∞},
∞
X Z
k 1 f(w)
then for all z ∈ A, f(z) = ak (z − a) , where ak = dw for any r between R1 and
2πi (w − a)k+1
k=−∞
|w−a|=r
R2. (Also the convergence is absolute on A, uniform on any smaller annulus {z: R1 < R01 ≤ |z−a| ≤ R02 < R2}
and the coefficients ak are unique for A .) The series is called the Laurent series of f(z) on A.

Proof. By a change of variable, we may assume a = 0. Let R1 < r1 < r2 < R2 .

Also, let C1 , C2 be the counterclockwise
 circles |z| = r1, |z| = r2. For r1 < |z0 | <
z0
 f(z) − f(z0 )
if z 6= z0 , z ∈ A
R1 r2 , the function g(z) = is holomorphic on A.
r1 r 2  0 z − z0
R2 f (z0 ) z = z0
Taking the line Zsegments [r1, r2] and [−r2, −r1 ] as cross-cuts, we get by
Cauchy’s theorem, g(w) dw = 0. Using the definition of g(z) and grouping
C2 −C1
simply terms, we get
connected
Z Z Z Z 
f(w) f(w) dw dw
dw− dw = f(z0 ) − = 2πif(z0 ).
C2 w − z0 C1 w − w0 C2 w − z0 C1 w − z0
upper half
  X
z0  z0 2
∞
1 1 1 z0k
For w ∈ C2, |z0| < |w| = r2 , = z = 1+ + +... = .
w − z0 w(1 − )
0 w w w wk+1
k=0
w !
 2 X−1
1 −1 1 w w z0k
For w ∈ C1, |w| = r1 < |z0|, = w =− 1+ + +... = − .
w − z0 z0 (1 − ) z0 z0 z0 wk+1
k=−∞
z0
z   
 0 w
Both series converge uniformly in w because   < 1 on C2 and   < 1 on C1. So, for R1 < r < R2,
w z0
X∞ Z X−1 Z X∞  Z 
f(w)z0k f(w)z0k f(w)
2πif(z0 ) = k+1
dw + k+1
dw = k+1
dw z0k .
C2
w C1
w w
k=0 k=−∞ k=−∞
| {z } | {z } |w|=r
I II
Z Z Z
(Observe that = = by using cross-cuts as above.) Letting r1 → R1 and r2 → R2, we get the
C2 C1
|w|=r
result. (For the parenthetical statement, we observe that the series I and II are power series in z0 and
z0−1 , respectively. Hence, the absolute and uniform convergence properties and the uniqueness of coefficients
follow from the corresponding properties for power series.) QED.

50
 An important point of the theorem is that the Laurent series coefficients for a holomorphic function
X∞
on an annulus are unique. (To elaborate on this, suppose f(z) = bk (z − a)k uniformly on the same
k=−∞ !
Z X
∞ Z
1 f(w) 1 k−n−1
annulus, then dw = bk (w − a) dw = bn. Therefore, the
2πi |w−a|=r (w − a)n+1 2πi |w−a|=r
k=−∞
coefficients bk ’s are the same as the coefficients ak ’s in the Laurent series representation above.) So in
computing Laurent series, we do not have to compute the coefficients by integrals. Instead, we can apply
other methods (such as the formula for summing geometric series or power series of common functions,)
depending on the given functions and annuli.

1
Example. Find the Laurent series of f(z) = for A1 = {z: 0 < |z| < 1}, A2 = {z: 1 < |z| < ∞}
z 2(1 − z)
and A3 = {z: 0 < |z − 1| < 1}.
1 1 1 1
Solution. On A1, = 2 (1 + z + z 2 + . . .) = 2 + + 1 + z + z 2 + . . ..
z 2 (1 − z) z z z
   
1 1 1 1 1 1 1 1 1 1 1 1
On A2 ,   < 1, 2 =− 3 1 = − 3 (1 + + 2 + 3 + . . .) = . . . − 6 − 5 − 4 − 3 .
z z (1 − z) z 1− z z z z z z z z z
X
∞
On A3 , the series is of the form ak (z−1)k . We make use of the fact |z−1| < 1 and write z = 1+(z−1), so
k=−∞
  X∞
1 1 1 1 d 1 d
= · . For w = z −1, |w| < 1, =− =− (−w)n =
z 2 (1 − z) (1 + (z − 1))2 (1 − z) (1 + w)2 dw 1+w n=0
dw
X∞
(−1)n+1 nwn−1, so
n=1

1 1 1
=− [1 − 2(z − 1) + 3(z − 1)2 − 4(z − 1)3 + . . .] = − + 2 − 3(z − 1) + 4(z − 1)2 − . . . .
z 2 (1 − z) (z − 1) z −1

Definitions. If f(z) is holomorphic in a deleted neighborhood of z0 (i.e. on some B(z0 , ε) \ {z0 }), then f(z)
is said to have an isolated singularity at z0 . The Laurent expansion of f(z) at z0 is the Laurent series of f(z)
on A = {z: 0 < |z − z0 | < ε}. If it is of the form
(i) a0 + a1(z − z0 ) + a2 (z − z0 )2 + . . ., we say z0 is a removable singularity of f(z) (in this case, we can
define f(z0 ) = a0 and f(z) becomes holomorphic at z0 because the Laurent series is the same as the
power series of f(z) in B(z0 , ε));
z3 z5
sin z sin z z− + − ...
(Example: f(z) = has a removable singularity at 0. For 0 < |z| < ε, = 3! 5!
z z z
z2 3 z4
= 1 + 0z − + 0z + − . . .. So defining f(0) = 1 will make f(z) holomorphic at 0.)
3! 5!
a−k a−k+1 a−1
(ii) k
+ k−1
+ ...+ + a0 + a1 (z − z0) + . . . with a−k 6= 0, we say z0 is a pole
(z − z0 ) (z − z0 ) (z − z0 )
of order k for f(z) (in this case, F (z) = (z − z0 )k f(z) is holomorphic at z0 , F (z0 ) = a−k 6= 0 and
F (z)
lim f(z) = lim = ∞);
z→z0 z→z0 (z − z0 )k

1
(Example: f(z) = has a pole (of order 1) at 0. For 0 < |z| < ε,
sin z
1 1 1 z2 7 4 1 1 7 3
= 2 4 = (1 + + z + . . .) = + 0 + z + 0z 2 + z + . . . .)
sin z z z z 6 360 z 6 360
z(1 − + − . . .)
3! 5!
51
 a−k a−k+1 a−1
(iii) . . . + + +...+ + a0 + a1 (z − z0 ) + . . ., we say z0 is an essential singular-
(z − z0 )k (z − z0 )k−1 z − z0
ity of f(z).
1
(Example: f(z) = sin has an essential singularity at 0. For 0 < |z| < ε,
z
1 1 1 1 1 1 1 0 1 0 1
sin = − + −... = ...+ + − 3 − 2 + + 0 + 0z + . . . .)
z z 3! z 3 5! z 5 120z 5 z 4 6z z z

X
∞ X
−1
Definition. If the Laurent series of f(z) on 0 < |z − z0 | < ε is ak (z − z0 )k , then ak (z − z0 )k is
k=−∞ k=−∞
∞
X
k
the principal part of f at z0 and ak (z − z0 ) is the holomorphic part of f at z0 .
k=0

Theorem. Let z0 be an isolated singularity of f(z).

(i) (Riemann’s principle.) If lim (z − z0 )f(z) = 0 (e.g. |f(z)| ≤ M near z0 ), then z0 is a removable
z→z0
singularity.
(ii) If there is a positive integer k such that lim (z − z0 )k f(z) 6= 0, but lim (z − z0)k+1 f(z) = 0, then z0 is
z→z0 z→z0
a pole of order k.

(z − z0 )f(z)
if z 6= z0,
Proof. (i) The function g(z) = is holomorphic near z0 and continuous at z0.
0 if z = z0,
g(z) X∞
g(k+1) (z0 )
By the extension theorem, g is holomorphic at z0 . So, near z0 , f(z) = = (z − z0 )k .
z − z0 (k + 1)!
k=0
Therefore, z0 is a removable singularity of f.

(ii) By (i), lim (z − z0 )[(z − z0 )k f(z)] = 0 implies (z − z0 )k f(z) has a removable singularity at z0 . So,
z→z0
X
∞
near z0 , (z − z0 )k f(z) = cn(z − z0 )n . Now c0 = lim (z − z0 )k f(z) 6= 0. Therefore, near z0 , f(z) =
z→z0
n=0
c0 c1
+ + · · ·, i.e. f has a pole of order k at z0 . QED.
(z − z0 )k (z − z0 )k−1

Casorati-Weierstrass Theorem. If z0 is an essential singularity of f(z), then the range set f(D∗ ) =
{f(z): 0 < |z − z0 | < ε} is dense in C for every ε > 0. (Here “dense” means in every open disk in C,
regardless of size and location, we can find a point of f(D∗ ).)
Proof. Suppose f(D∗ ) is not dense in C for some ε > 0. Then f(D∗ ) misses some disk B(a, r), i.e.
1 1
|f(z) − a| ≥ r for 0 < |z − z0 | < ε. It follows that g(z) = is bounded (by ) near z0 . By
f(z) − a r
Riemann’s principle, z0 is a removable singularity of g(z), i.e. g can be defined at z0 so as to be holomorphic
1
there. Now f(z) = a + . If g(z0 ) 6= 0, then lim (z − z0 )f(z) = 0, which forces f to have a removable
g(z) z→z0
singularity at z0 , a contradiction. If g(z0 ) = 0, then g(z) = (z − z0 )k h(z) for some positive integer k and
1
some holomorphic function h(z) near z0 with h(z0 ) 6= 0. It follows that lim (z − z0 )k f(z) = 6= 0 and
z→z0 h(z0 )
lim (z − z0 )k+1 f(z) = 0. Then f has a pole of order k at z0 , a contradiction. QED.
z→z0

In an advanced course, a theorem called the Great Picard Theorem is usually proved, which asserts that
f(D∗ ) can miss at most one complex number and in fact, the equation f(z) = c has infinitely many solutions
in D∗ , except perhaps for one complex value c. (For example, 0 is an essential singularity of e1/z and in every
neighborhood of 0, the range of e1/z is C \{0}.) This is stronger than the Casorati-Weierstrass theorem.

52
Summary. Let z0 be an isolated singularity of f. Then
(i) z0 is a removable singularity of f ⇐⇒ lim f(z) ∈ C ⇐⇒ f(z) is bounded near z0 (i.e. |f(z)| ≤ M
z→z 0
near z0 for some M ) ⇐⇒ lim (z − z0 )f(z) = 0;
z→z0
(ii) z0 is a pole of f ⇐⇒ lim f(z) = ∞; (also, z0 is a pole of order k ⇐⇒ lim (z − z0)k f(z) 6= 0 and
z→z0 z→z0
lim (z − z0 )k+1 f(z) = 0;)
z→z0
(iii) z0 is an essential singularity of f ⇐⇒ f(D∗ ) = {f(z): 0 < |z − z0 | < ε} is dense in C for any
ε > 0 ⇐⇒ lim f(z) doesn’t exist.
z→z0

(Reasons.
X
∞
(i) If z0 is a removable singularity, then near z0 , f(z) = ak (z − z0 )k , so lim f(z) = a0 ∈ C.
z→z0
k=0
If lim f(z) = A ∈ C, then for each ε > 0, there is δ > 0 such that 0 < |z − z0 | < ε ⇒ |f(z) − A| < δ.
z→z0
So |f(z)| < A + δ in 0 < |z − z0 | < ε.
If |f(z)| ≤ M near z0 , then |(z − z0)f(z)| ≤ M |z − z0 | → 0 as z → z0 .
If lim (z − z0 )f(z) = 0, then z0 is a removable singularity by Riemann’s Principle.
z→z0
(ii) If z0 is a pole (of order k), then from the definition, we have lim f(z) = ∞ (and lim (z − z0 )k f(z) 6= 0,
z→z0 z→z0
lim (z − z0 )k+1 f(z) = 0).
z→z0
If lim f(z) = ∞, then z0 is not a removable singularity by (i) above and it is not an essential singularity
z→z0
by Casorati-Weierstrass Theorem (because the range of f misses small values near z0). (The converse
part of the parenthetical statement was proved in a theorem earlier.)
(iii) If z0 is an essential singularity, then f(D∗ ) is dense in C for any ε > 0 by Casorati-Weierstrass Theorem.
If f(D∗ ) is dense in C, then lim f(z) cannot exist.
z→z0
If lim f(z) doesn’t exist, then by (i) and (ii), z0 is not a removable singularity or a pole, hence it must
z→z0
be an essential singularity.)

Exercises

1. Identify the isolated singularities of the following functions and classify each as a removable singularity,
a pole (and its order) or an essential singularity:

1
(a) ; (b) cot z;
z4 + z2
2
e1/z z2 − 1
(c) ; (d) .
z−1 sin πz
1
2. Find the Laurent series of on
z2 − 4
(a) 0 < |z − 2| < 4; (b) 2 < |z| < ∞.
Z
1 1
3. Find sin dz for positive r 6= . As usual, the circle |z| = r is given the counterclockwise
|z|=r z nπ
orientation.

p 1
4. Suppose f is holomorphic on C \{0} and satisfies |f(z)| ≤ |z| + p . Prove that f is a constant
|z|
function.

53
 5. Find the roots of sinh z. What is the largest r for which there exist c0, c1 , c2 , . . . ∈ C such that
∞
(z 2 + π2)(ez − 1) X
= cnz n for |z| < r?
sinh z n=0

6. Let G be a region and f : G → C be continuous. If f 2 is holomorphic on G, show that f is holomorphic

on G. [Hint: First show f is holomorphic at z such that f(z) 6= 0. Then consider the singularity type
of the roots of f.]

7. If f has a pole at 0, show that ef cannot have a pole at 0.

8. Let f be holomorphic on {z : R < |z| < ∞}. We say ∞ is a removable singularity, a pole of order k or
an essential singularity
  of f(z) if and only if 0 is a removable singularity, a pole of order k or essential
1
singularity of f , respectively.
z
(a) Prove that an entire function with a pole at ∞ is a polynomial.
(b) Prove that a holomorphic function on C ∪{∞} except for isolated poles must be a rational function.

9. Prove that the image of the plane under  a nonconstant

 entire mapping f is dense in the plane. [Hint:
1
If f is not a polynomial, then consider f .]
z

10. Can the positive integers {1, 2, 3, . . .} be partitioned into a finite number of sets S1 , S2 , . . ., Sk , each of
which is an arithmetic progression, i.e.

S1 = {a1 , a1 + d1 , a1 + 2d1, . . .}

S2 = {a2 , a2 + d2 , a2 + 2d2, . . .}
···
Sk = {ak , ak + dk , ak + 2dk , . . .},
and
X such that there are no equal common differences (i.e. di 6= dj for i 6= j)? [Hint: Consider the series
z n for j = 1, 2, . . ., k.] (This exercise is due to D. J. Newman.)
n∈Sj

54
16. Residues and Roots

X
∞
Definition. Let ak (z − z0 )k be the Laurent series of f(z) on 0 < |z − z0 | < ε, then the residue of f(z)
k=−∞
at z0 is Res f(z) = Res (f, z0 ) = a−1 .
z=z0
X
∞ Z
k
Observe that if f(z) = ak (z − z0 ) for 0 < |z − z0| < ε, then for 0 < r < ε, f(z) dz
k=−∞ |z−z0 |=r
∞
X Z
(z − z0 )k+1
= ak (z − z0 )k dz = 2πia−1 because (z − z0 )k has antiderivative for k =
6 −1. This
k+1
k=−∞
|z−z0 |=r
is the reason the word “residue” is used (because a−1 is the only coefficient that remains).

Definition. A function f is meromorphic on a region D iff f is holomorphic on D except for (isolated) poles.

Residue Theorem. Suppose f is meromorphic on a simply connected region D with poles at z1 , . . . , zn.
Let Γ be a closed curve on D not passing through z1 , . . . , zn, then
Z X
n
f(z) dz = 2πi n(Γ, zj ) Res f(z).
Γ z=zj
j=1

  X
−1
1
Proof. Let Pj = ak,j (z − zj )k be the principal part of f at zj , where mj is the order of the
z − zj
k=−mj
X
n  
1
pole at zj . Then the function g(z) = f(z) − Pj is holomorphic on D. (At zj , the singularity
z − zj
j=1
is cancelled.) By Cauchy’s theorem,

Z Z n Z
X X
−1 Z X
n
0= g(z) dz = f(z) dz − ak,j (z − zj )k dz = f(z) dz − 2πi a−1,j n(Γ, zj ).
Γ Γ j=1 Γ k=−m Γ j=1
j

QED.

Remark. The residue theorem is true even if there are essential singularities. The proof for that amounts
to replacing Γ by a sequence of small circles about the singularities.

Cauchy Integral Formula for Derivatives. If f is holomorphic Z on a simply connected region D and a is
n! f(w)
a point not on a closed curve Γ in D, then f (n) (a) = dw.
2πi n(Γ, a) Γ (w − a)n+1
f (n) (a)
Proof. Near a, f(z) = f(a) + f 0 (a)(z − a) + . . . + (z − a)n + . . .. By the residue theorem,
n!
Z
f(w) f(z) f (n) (a)
n+1
dw = 2πi n(Γ, a)Res = 2πi n(Γ, a) .
Γ (w − a) z=a (z − a)n+1 n!
QED.

55
Argument Principle. If f is meromorphic on a simply connected region Z D and Γ is a simple closed curve
1 f 0 (z)
in D not passing through the roots zj nor the poles pk of f, then dz = n − m, where n and m
2πi Γ f(z)
are the number of roots and poles (counting multiplicities) of f inside Γ, respectively.
Z 0 Z
z1 f 1 f (z) 1 dw
p [Note: If we let w = f(z), then n − m = dz =
z2 1
2πi Γ f(z) 2πi f ◦Γ w
z 3 p2 ∆Γ arg f(z)
0 = n(f ◦ Γ, 0) = , where ∆Γ arg f(z) is the change of argument in
z4 2π
Γ f Γ f(z) along Γ. This is the reason for the name Argument Principle.]

f0
Proof. The isolated singularities of are at the zj ’s and pk ’s. If the order of zj as a root of f is nj , then
f
f 0 (z) nj
f(z) = anj (z − zj )nj + . . . near zj and = + . . . near zj . If the order of pk as a pole of f is mk ,
f(z) z − zj
a−mk f 0 (z) −mk
then f(z) = m
+ . . . near pk and = + . . .. By the residue theorem,
(z − pk ) k f(z) z − pk
Z 0 X
1 f (z) f 0 (z) X f 0 (z) X X
dz = Res + Res = nj + (−mk ) = n − m.
2πi Γ f(z) j
z=z j f(z) z=p k f(z)
j
k k

QED.

Rouché’s Theorem (Estermann-Glicksberg Form). If f and g are holomorphic inside and on a simple
closed curve Γ and |f(z) + g(z)| < |f(z)| + |g(z)| for all z ∈ Γ, then f and g have the same number of roots
inside Γ. (Note the inequality on Γ implies f and g have no roots on Γ.)
f(z)
Proof. For z ∈ Γ, |f(z) + g(z)| < |f(z)| + |g(z)| implies is not zero or a positive real number (oth-
g(z)
f(z) f(z)
erwise |f(z) + g(z)| = |f(z)| + |g(z)|). Then log can be defined with 0 < arg < 2π. Therefore
g(z) g(z)
Z  0 Z 0 Z 0
f(z) f (z) g (z)
0= log dz = dz − dz and the result follows from the argument principle.
Γ g(z) Γ f(z) Γ g(z)
QED.

Remark. To show f and g have the same number of roots inside Γ, it is sufficient to show |f(z)+g(z)| < |f(z)|
or |f(z) + g(z)| < |g(z)|.

Examples. (1) Find the number of roots of f(z) = 17z 5 + z 2 − 1 in each quadrant and on the real or
imaginary axes.
r
iR 0 4 2
γ1 Solution. Because f (z) = 85z + 2z, f(0) = −1, f(− 3 ) < 0, there is one positive
85
γ2 2 5
real root. Because f(iy) = −y − 1 + 17y i 6= 0 for y real, there is no imaginary root.
We now apply the argument principle to the right half plane. Consider Γ to be
the boundary of the half disk of radius R (R huge) in the right half plane. On γ1 ,
π π
z = Reiθ , − ≤ θ ≤ , because R is huge, f(z) ≈ 17z 5 = 17R5ei5θ . This implies
2 2
-iR π π
∆γ1 arg f(z) ≈ 5( − (− )) = 5π.
f ( iR) 2 2
On γ2 , z = iy, where y decreases from R to −R, f(iy) = −y2 − 1 +17y5 i, f(iR) =
| {z }
negative
5
17R
−R2 − 1 + 17R5 i, f(−iR) = −R2 − 1 −17R5 i. Since lim = +∞, these im-
- 1 = f (0) | {z } | {z } | {z } | {z } R→+∞ R2 + 1
negative positive negative negative
3π π
f (-iR ) ply ∆γ2 arg f(z) = arg f(−iR) − arg f(iR) ≈ − = π.
2 2

56
 6π
So ∆Γ arg f(z) = 5π + π = 6π and there are = 3 roots inside Γ when R is huge (one of which is real
2π
as shown above). Therefore there is one positive real root and two pairs of complex roots in conjugate (one
complex root in each quadrant).

(2) Show that all the roots of f(z) = z 5 + 3z + 1 are in the disk |z| < 2.
Solution. Let g(z) = −z 5 , which has 5 roots (counting multiplicities) in the disk |z| < 2. Let Γ be the circle
|z| = 2, then for z ∈ Γ, |f(z) + g(z)| = |3z + 1| ≤ 3|z| + 1 = 7 < |g(z)| = |z|5 = 25 = 32. So by Rouché’s
theorem, f has 5 roots inside |z| = 2.

(3) What is the smallest positive integer r such that f(z) = z 5 + 48z + 64 has a root in the disk |z| < r?
Solution. For |z| = 1, let g(z) = −64, then |f(z) + g(z)| = |z 5 + 48z| ≤ |z|5 + 48|z| = 49 < 64 = |g(z)|, so
by Rouché’s theorem, f has no roots inside |z| = 1.
For |z| = 2, let g(z) = −48z, then |f(z) + g(z)| = |z 5 + 64| ≤ |z|5 + 64 = 96 = 48|z| = |g(z)| ≤
|f(z)| + |g(z)|. For equality to hold throughout, we must have z 5 = 32 and f(z) = 0. Then 32 + 48z + 64 = 0
implies z = −2, which contradicts z 5 = 32. So one of the inequality is strict, i.e. on |z| = 2, |f(z) + g(z)| <
|f(z)| + |g(z)|. By Rouché’s theorem, f has one root inside |z| = 2. Therefore r = 2.

Yn
z − aj
(4) If a1, . . . , an, b ∈ D = {z: |z| < 1} and f(z) = , then the equation f(z) = b has exactly n
j=1
1 − aj z
solutions (counting multiplicities) in D.
n 
Y

 z − aj 
Solution. Let g(z) = b − f(z), then for |z| = 1, |f(z) + g(z)| = |b| < 1 =  
 1 − aj z  = |f(z)|. By Rouché’s
j=1
theorem, g has n roots in D, since f has n roots a1 . . . , an in D. So f(z) = b has n solutions in D.

1
(5) Let w ∈ C \[−2, 2] and k a positive integer. Define f(z) = z k + k . Show that f takes the value w
z
exactly k times in D = {z: |z| < 1} (i.e. f(z) = w has exactly k solutions in D).
Solution. Let g(z) = f(z) − w, we want to show g has exactly k roots in D. Let Γ be the unit circle
z(θ) = eiθ , 0 ≤ θ ≤ 2π, g(eiθ ) = eikθ + e−ikθ − w = 2 cos kθ − w.
Then g ◦ Γ is the line segment joining −2 − w to 2 − w. Since w 6∈ [−2, 2], g ◦ Γ does
w not wind around 0, so by the argument principle,
   
-2 - w 2- w number of roots number of poles
0 = n(g ◦ Γ, 0) = − .
of g(z) in D of g(z) in D

Since 0 is the only pole of f(z) (hence also of g(z)) of order k, we have the number of roots of g(z) in D =
number of poles of g(z) in D = k.

Local Mapping Theorem. Let f be a nonconstant holomorphic function on Ω and z0 be a root of f of

order m on Ω. Then for ε > 0 small, there is a δ > 0 such that the equation f(z) = w with 0 < |w| < δ has
exactly m distinct roots (each of order 1) in B(z0 , ε), i.e. near z0 , f is locally a m-to-1 mapping.

Proof. Since f is nonconstant, by the identity theorem, there cannot be any sequence of roots of f (or f 0 )
converging to z0 . So we may pick ε > 0 small so that f(z) 6= 0 and f 0 (z) 6= 0 for z ∈ B(z0 , ε) \ {z0 } ⊂ Ω. Let
δ = min |f(z)|, then δ > 0. Suppose 0 < |w| < δ, then for |z − z0 | = ε, we have |f(z) + (w − f(z)) | =
|z−z0 |=ε
|w| < δ ≤ |f(z)| + |w − f(z)|. By Rouché’s theorem, f and w − f must have the same number of roots in
B(z0 , ε), namely m. Since f 0 (z) 6= 0 for z ∈ B(z0 , ε), the roots of f(z) = w are distinct and of order 1.
QED.

57
Inverse Mapping Theorem. If f is univalent (i.e. one-to-one and holomorphic) on an open set U , then
f −1 : f(U ) → U is holomorphic.

Proof. (Since f is one-to-one, f is nonconstant. Hence by the open mapping theorem, f(U ) is open.) For an
arbitrary a ∈ U , we will show that f −1 is continuous at f(a). Given ε > 0, by the open mapping theorem,
f(B(a, ε)) is open. Since f(a) ∈ f(B(a, ε)), there is an open disk B(f(a), δ) ⊂ f(B(a, ε)) for some δ > 0.
Now w = f(f −1 (w)). So,

|w − f(a)| < δ ⇒ w ∈ B(f(a), δ) ⊂ f(B(a, ε)) ⇒ f −1 (w) ∈ B(a, ε) ⇒ |f −1 (w) − a| < ε.

Next we will show f 0 (z) 6= 0 for all z ∈ U. Otherwise, suppose f 0 (c) = 0. Then g(z) = f(z) − f(c)
has a root of order m ≥ 2 at c. By the local mapping theorem, there are distinct a and b near c such that
g(a) = g(b). Then f(a) = f(b), contradicting the fact f is one-to-one.
Therefore, by the inverse rule, f −1 is holomorphic on f(U ). QED.

z−a
Earlier we proved that for |a| < 1, θ real, f(z) = eiθ is a one-to-one function from the open unit
1 − az
disk D onto D. Here we will prove the converse.

z−a
Theorem. If f is a one-to-one holomorphic function from D onto D, then f(z) = eiθ for some real θ
1 − az
and a ∈ D.
z +a
Proof. Let a ∈ D such that f(a) = 0. Let T (z) = , which maps D onto D, the unit circle onto the
1 + az
−1
unit circle and T (0) = a. The function g = f ◦ T and g are one-to-one holomorphic functions of D onto
D such that g(0) = 0 = g−1 (0). By Schwarz’s lemma, |g(z)| ≤ |z| and |g−1 (z)| ≤ |z| ( ⇐⇒ |z| ≤ |g(z)|).
So the equality |g(z)| = |z| holds for all z ∈ D. Then f ◦ T (z) = g(z) ≡ eiθ z. Setting w = T (z), we get
w−a
f(w) = eiθ T −1 (w) = eiθ . QED.
1 − az

Hurwitz’s Theorem. Let {fn} be a sequence of holomorphic functions on Ω converging uniformly on each
closed disk in Ω to the (holomorphic, by Weierstrass’ theorem) function f. If fn has no roots in Ω for all n,
then either f has no roots in Ω or f ≡ 0. If fn is one-to-one on Ω for all n, then either f is one-to-one on Ω
or f is a constant function.

Proof. For the first assertion, suppose f 6≡ 0 and f has a root at w ∈ Ω. Then by the identity theorem,
there is a circle C = C(w, r) such that f(z) 6= 0 for z ∈ C and B(w, r) ⊂ Ω. Let ε = min |f(z)|, then ε > 0.
z∈C
Since fn converges uniformly on B(w, r) to f, there is N such that n ≥ N ⇒ |fn(z) − f(z)| < ε for all
z ∈ B(w, r). Then for z ∈ C, |fn(z) − f(z)| < ε ≤ |f(z)| ≤ |fn(z)| + |f(z)|. By Rouché’s theorem, it follows
that fn must have a root inside C, which is a contradiction.
For the second assertion, suppose f(a) = f(b) for distinct a, b ∈ Ω and f is not the constant function
f(b). Then by the uniqueness theorem, there is a closed disk D centered at a in Ω such that f(z) 6= f(b) for
all z ∈ D \ {a}. In particular, b 6∈ D. Since fn converges uniformly on D to f, it follows that fn (z) − fn (b)
converges uniformly on D to f(z) − f(b). By the first assertion, since fn (z) − fn (b) has no roots in D,
f(z) − f(b) cannot have any root in D, which is a contradiction to f(a) = f(b). QED.

Exercises
Z
dz
1. Find all possible values of I = , where C is a curve with initial point 0 and final point 1 that
C 1 + z2
1
does not meet the poles of .
1 + z2

58
 2. Is there a holomorphic function f on the closed unit disk which sends the unit circle with the counter-
clockwise orientation into the unit circle with the clockwise orientation?

3. Determine the number of roots 17z 5 + z 2 + 1 on the real and imasginary axis and in each quadrant.

4. Show that if α and β 6= 0 are real, the equation z 2n + α2z 2n−1 + β 2 = 0 has n − 1 roots with positive
real parts if n is odd, and n roots with positive real parts if n is even.

5. If a > e, show that the equation ez = az n has n solutions inside the unit circle.

6. Suppose f is holomorphic and one-to-one

Z on B(0, r) = {z : |z − a| ≤ r}. If f has a root in B(a, r), show
1 wf 0 (w)
that the root is given by dw.
2πi f(w)
|w−a|=r

7. Suppose f is entire and f(z) is real if and only if z is real. Use the argument principle to show that f
can have at most one root. [Hint: Let Γ be a large circle |z| = R, what is n(f ◦ Γ, 0)?]

8. If f is holomorphic on and inside a simple closed curve Γ and f is one-to-one on Γ, prove that f is
one-to-one inside Γ. [Hint: Is f ◦ Γ a simple closed curve? For w 6∈ f ◦ Γ, let g(z) = f(z) − w, what is
n(g ◦ Γ, 0)?]

9. Let P (z) be a polynomial

Z of degree at least 2.
dz
(a) Show that = 0 when all the roots of P are inside the circle |z| = r.
|z|=r P (z)
Xn
1
(b) Suppose P has n distinct roots z1 , . . . , zn. Show that 0 (z )
= 0.
j=1
P j

10. Use Rouché’s theorem to give another proof of the fundamental theorem of algebra.

X
∞ X
∞
11. Let f(z) = z + an z n . Suppose n|an| ≤ 1.
n=2 n=2
(a) Prove that f is holomorphic on the open unit disk D, i.e. the power series converges for every
z ∈ D.
(b) Prove that f is one-to-one on D. [Hint: Use Rouché’s theorem to show that g(z) = f(z) − f(z0 )
has exactly one solution in D for each fixed z0 ∈ D.]

59
17. Applications of Residue Theory

a−k a−1
If f has a pole of order k at z0 , then the Laurent expansion of f(z) is +. . .+ +a0 +a1 (z−
(z − z0 )k (z − z0 )
z0 ) + . . .. So the function ϕ(z) = (z − z0 )k f(z) = a−k + . . .+ a−1 (z − z0 )k−1 + a0 (z − z0 )k + . . . is holomorphic
ϕ(k−1)(z0 ) ϕ(k−1)(z) 1 dk−1
at z0 . Then Res(f, z0 ) = a−1 = = lim = lim [(z − z0 )k f(z)]. If f is
(k − 1)! z→z0 (k − 1)! z→z0 (k − 1)! dz k−1
holomorphic at z0 , Res(f, z0 ) = 0 (because an = 0 for n < 0).

Theorem. If g, h are holomorphic near z0 , g has a root of order k at z0 (i.e. g(z0 ) = . . . = g(k−1)(z0 ) =
g(z)
0, g(k)(z0 ) 6= 0) and h a root of order l at z0 (i.e. h(z0 ) = . . . = h(l−1) (z0 ) = 0, h(l) (z0 ) 6= 0), then has
    h(z)
removable singularity k≥l
a at z0 in case .
pole (of order l − k) k<l
∞
X g(n) (z0 )
(z − z0)n
g(z) n! g(k)(z0 ) l!
n=k
Proof. Near z0 , = ∞ (n) = (z − z0 )k−l + . . .. If k − l ≥ 0, then z0 is a
h(z) X h (z0 ) k! h (l) (z )
0
(z − z0 )n
n!
n=l
removable singularity. If k − l < 0, then it is a pole of order l − k. QED.
g
Special Cases: (1) If f = , z0 is a root of order k for g, a root of order k + 1 for h, then z0 is a simple
h
g(k) (z0 )
pole (i.e. pole of order 1) and Res(f, z0 ) = (k + 1) (k+1) .
h (z0 )
g
(2) If f = , g(z0 ) 6= 0, h(z0 ) = h0(z0 ) = 0, h00 (z0 ) 6= 0, then z0 is a double pole (i.e. pole of order 2)
h
g0 (z0 ) 2 g(z0 )h000(z0 ) g
and Res(f, z0 ) = 2 00 − . If f = , g(z0 ) = 0, g0 (z0 ) 6= 0, h(z0 ) = h0 (z0 ) = h00(z0 ) = 0,
h (z0 ) 3 (h00 (z0 ))2 h
g00 (z0 ) 2 g0 (z0 )h(4) (z0 )
h000(z0 ) 6= 0, then z0 is a double pole and Res(f, z0 ) = 3 000 − .
h (z0 ) 3 (h000(z0 ))2
z
Examples. (1) has a pole at πi. It is a simple pole. So,
1 + ez
z z πi
Res( , πi) = lim (z − πi) = πi = −πi.
1 + ez z→πi 1 + ez e
1
(2) has a pole at 0. It is a double pole. So,
z sin z
 0
1 1 1 sin z − z cos z z
Res( , 0) = lim z2 = lim = lim = 0.
z sin z z→0 1! z sin z z→0 sin2 z z→0 2 cos z

1 1 1 1 1 1
Alternatively, near 0, = 1 3 = 2 1 4 = 2 + + · · · . So, Res( , 0) = 0.
z sin z z(z − 6 z + · · ·) z − 6z + · · · z 6 z sin z

cot z cos z
(3) = has a pole at 0. It is a double pole. Let g(z) = cos z, h(z) = z(z − 1) sin z, then
z(z − 1) z(z − 1) sin z
cot z 0 2 1×6
g(0) = 1, g0 (0) = 0, h00(0) = −2, and h000(0) = 6. So, Res( , 0) = 2 − = −1.
z(z − 1) −2 3 (−2)2

60
 Alternatively, near 0,
2 2
cos z 1 − z2 + · · · 1 − z2 + · · · 1 1
= z 3
= 2 3
= − 2 − + ···.
z(z − 1) sin z 2
(−z + z ) z − 6 + · · · −z + z + · · · z z
cot z
So, Res( , 0) = −1.
z(z − 1)
1
(4) z cos has an essential singularity at −1, since near −1,
z +1
 
1 1 1
z cos = [(z + 1) − 1] 1 − 2
+ · · · = (z + 1) − 1 − + · · ·.
z+1 2(z + 1) 2(z + 1)
1 1
So Res(z cos , −1) = − .
z+1 2

Recall the residue theorem implies that if f is holomorphic on and inside a simple closed curve C except
Z n
X
at poles a1, . . . , an, then f(z) dz = 2πi Res(f, aj ). (In view of the examples above, notice this formula
C j=1
said something amazing, namely the integral on a closed curve can be computed by doing some derivatives!)
Now let us compute some nonelementary integrals.

Some Common Types of Integrals.

Z 2π Z  k 
z − z −k z m + z −m dz
Type I. F (sin kθ, cos mθ) dθ = F , , where z = eiθ , dz = ieiθ dθ.
0 2i 2 iz
|z|=1

Z 2π Z Z
dθ 1 dz dz 1
Example. = −1
 =  = 2πi Res
0 5 + 4 sin θ z − z iz (2z + i)(z + 2i) z=− i (2z + i)(z + 2i)
2
|z|=1 5 + 4 |z|=1
2i
2 i
= π, where we observed that only − is inside |z| = 1 and it is a simple pole.
3 2

Z ∞
P (x)
Type II. dx, where P , Q are polynomials and Q has no real roots.
−∞ Q(x)
Z ∞ Z 0 Z b Z ∞
Remark. The improper integral is defined (in the Riemann sense) to be lim + lim . If
−∞ a→−∞ a b→+∞ 0 −∞
Z R Z ∞
exists, it agrees with the limit lim (which is called the principal value of and is denoted by P.V.
R→+∞ −R
Z ∞ Z ∞ Z ∞ −∞

). If doesn’t exist, sometimes P.V. may exist.

−∞ −∞ −∞

Z ∞ Z 0 Z b Z ∞
Example. x dx = lim x dx + lim x dx doesn’t exist, but P.V. x dx = 0.
−∞ a→−∞ a b→+∞ 0 −∞
Z ∞ Z ±∞
P (x) 1
If deg Q(x) > deg P (x) + 1, then dx exists (because near ±∞, it is like
n
dx, n > 1).
−∞ Q(x) c xZ ∞
P (x)
In this case, we often compute its principal value instead. If deg Q(x) ≤ deg P (x) + 1, then dx
Z ∞ −∞ Q(x)
P (x)
doesn’t exist, but P.V. dx may exist.
−∞ Q(x)

61
 Z Z !
∞ R
x2 − 1
Example. Find dx = lim f(x) dx .
−∞ (x2 + 1)(x2 − 2x + 2) R→∞ −R

Factoring the denominator, we get (x2 + 1)(x2 − 2x + 2) = (x − i)(x + i)(x − 1 − i)(x − 1 + i).

CR

i 1 +i
Step 1 . Consider the contour where R is large.
-R 0 R

-i 1 -i

By the residue
Z theorem,  
f(z) dz = 2πi Res f(z) + Res f(z)
z=i z=1+i
CR +[−R,R]
 
z2 − 1 z2 − 1 π
= 2πi lim 2
+ lim 2
= .
z→i (z + i)(z − 2z + 2) z→1+i (z + 1)(z − 1 − i) 5
Z 
  2
R +1
Step 2 . Now z ∈ CR ⇒ |z| = R,  f(z) dz  ≤ M L = 2 − 1)(R2 − 2R − 2)
πR → 0 as R → +∞.
CR (R
Z ∞ Z R  Z Z 
π
Step 3 . So, f(x) dx = lim f(x) dx = lim f(z) dz − f(z) dz = .
−∞ R→∞ −R R→∞ CR 5
CR +[−R,R]

Z ∞    Z ∞
P (x) cos ax Re P (x) iax
Type III. dx = e dx, where P , Q are polynomials and Q has no
−∞ Q(x) sin ax Im −∞ Q(x)
real roots.
Z ∞  
P (x) cos ax
If deg Q(x) > deg P (x) + 1, then dx exists (and equals its principal value). If
sin ax
Z ∞  Q(x) 
−∞
P (x) cos ax
deg Q(x) ≤ deg P (x) + 1, then dx or its principal value may exist.
−∞ Q(x) sin ax

The following is a useful inequality when dealing with Type III integrals.
Z π Z π Z π
2 2 2θ π
Jordan’s Inequality. For M > 0, e−M sin θ dθ = 2 e−M sin θ dθ ≤ 2 e−M π dθ < (because on
0 0 0 M
π 2θ 2θ
[0, ], sin θ ≥ as can be seen from the graphs of y = sin θ and y = ).
2 π π
Z ∞  Z ∞ 
x sin 9x zei9z x sin 9x
Example. Find 2
dx = Im 2
dz . (Note 2 is even. If the principal value of
−∞ x + 4 z +4 x +4
Z 0 Z ∞−∞ Z ∞
1
the integral exists, then = P.V. = and the integral (in the Riemann sense) will exist.)
−∞ 2 −∞ 0

CR

Step 1 . Consider the contour 2i

where R is large.

-R 0 R
-2 i
Z
zei9z zei9z zei9z
By the residue theorem, 2
dz = 2πi Res 2 = 2πi lim = e−18πi.
z +4 z=2i z + 4 z→2i z + 2i
CR +[−R,R]

62
Step 2 . As R → ∞,
Z  Z   Z π
 zei9z   z   i9z 
 dz  ≤   e  |dz| ≤ R e−9R sin θ R dθ ≤ 2
R
R
π
→ 0,
 2
z +4   2  2
CR CR z + 4 R −4 0 R − 4 9R

where we have applied Jordan’s inequality to get the last inequality.

Step 3 . Therefore,
Z ∞   Z Z 
x sin 9x zei9z zei9z
P.V. dx = Im lim dz − dz = e−18 π.
−∞ x2 + 4 R→∞ z2 + 4 CR z2 + 4
CR +[−R,R]

We extract from step 2 a useful fact.

Z
Jordan’s Lemma. If lim ( max |h(z)|) = 0, then lim h(z)eiaz dz = 0 for a > 0.
R→∞ z∈CR R→∞ CR

Proof. As R → ∞,
Z  Z Z
    π
π
 h(z)eiaz dz  ≤ |h(z)| eiaz  |dz| ≤ max |h(z)| e−aR sin θ R dθ ≤ max |h(z)| → 0.
 z∈CR a z∈CR
CR CR 0

QED.

 
 f(z) 
Big O-Little o Notations. We say f(z) = O(g(z)) as z → a if there is M such that    ≤ M as z → a
g(z) 
f(z)
(i.e. in a neighborhood of a). We say f(z) = o(g(z)) as z → a if → 0 as z → a.
g(z)
Z ∞ Z ∞
Type IV. xαf(x) dx (−1 < α < 0) or (ln x)k f(x) dx (k = 1, 2, 3 . . .), where f is meromorphic on C
0 0
1
and continuous at 0. (In the first integral, we require f(z) = o( ) as z → ∞ and in the second integral,
|z|α+1
1
we require f(z) = O( ) as z → ∞. These conditions ensure the integrals will exist.)
|z|2
Z ∞ Z R
dx 1 1
Example. Find √ 2
= lim x− 2 2 dx.
0 x(x + 1) R→∞
r→0+ r x +1

CR

i
-C r γ1
Step 1 . Consider the contour R where r is small and R is large.
r θ
−γ 2
-i

Here γ1 (x) = x, γ2(x) = xei(2π−θ) , r ≤ x ≤ R. Inside and on the contour, we have 0 ≤ arg z < 2π.
Z  
− 12 1 z −1/2 z −1/2
By the residue theorem, z dz = 2πi Res 2 + Res 2
z2 + 1 z=i z + 1 z=−i z + 1
CR −γ2 −Cr +γ1
 
z −1/2 z −1/2 √
= 2πi lim + lim = π(i−1/2 − (−i)−1/2 ) = π(e−iπ/4 − e−3iπ/4 ) = π 2.
z→i z + i z→−i z − i

63
 Z Z R Z Z R Z R
z −1/2 x−1/2 z −1/2 x−1/2e−iπ x−1/2
Step 2 . We have dz = dx, lim dz = dx = − dx,
γ1 z2 + 1 r x2 + 1 θ→0+ γ2 z2 + 1 r x2 + 1 r x2 + 1
Z 
 z −1/2  R−1/2
 dz ≤ ML = 2 (2π − θ)R → 0 as R → +∞ and
 2 
CR z + 1 R −1
Z 
 z −1/2  −1/2
 dz  ≤ ML = r (2π − θ)r → 0 as r → 0+ .
 z2 + 1  1 − r2
Cr

√ Z Z ∞
−1/2 1 x−1/2
Step 3 . We have π 2 = lim z dz = 2 dx.
r,θ→0+ ,R→+∞ z2 + 1 0 x2 + 1
CR −γ2 −Cr +γ1
Z ∞
√
dx π 2
Therefore, √ = .
0 x(x2 + 1) 2

Miscellaneous Examples
Z ∞  Z ∞ ix 
sin x e sin x
Example. Find dx = Im dx . (Note is even.)
−∞ x −∞ x x

CR
Step 1 . Consider the contour where r is small and R is large.
-C r

-R -r r R
Z Z −r Z Z R  iz
e
By the residue theorem, + − + dz = 0 (since there is no singularity inside the con-
CR −R Cr r z
tour).
  Z
1 1 eiz
Step 2 . Since lim max   = lim = 0, Jordan’s lemma implies lim dz = 0. Now
R→+∞ z∈CR z R→+∞ R R→+∞ CR z
 
(iz)2
Z Z 1 + iz + +... Z   Z
eiz 2 1
dz = dz = + g(z) dz = πi + g(z) dz,
Cr z Cr z Cr z Cr

where
Z g(z) has
 a removable singularity (hence holomorphic) at 0, so it is bounded near 0 (say by M ). Then,
 
 g(z) dz  ≤ M L = M πr → 0 as r → 0+ .

Cr

Step 3 . Therefore,

Z Z −r Z Z R  iz Z ∞ Z ∞ Z ∞
e eix sin x eix
0= lim + − + dz = dx − πi ⇒ dx = Im dx = π.
r→0+ ,R→+∞ CR −R Cr r z −∞ x −∞ x −∞ x

We extract from step 2 the following useful rule for dealing with arcs around simple poles.

Rule.Z If f(z) has an isolated simple pole at z = c and Cr is the arc z(θ) = c + reiθ , α ≤ θ ≤ β, then
lim+ f(z) dz = i(β − α) Res f(z).
r→0 Cr z=c

64
 a−1 X
∞
a−1
Cr Proof. The Laurent expansion of f at c is + ak (z − c)k = + g(z) where
z −c z−c
k=0
g(z) is holomorphic at c. So
β Z Z   Z β Z
α a−1 1 iθ
f(z) dz = + g(z) dz = a−1 iθ
ire dθ + g(z) dz
c Cr Cr z −c Z α re Cr

= i(β − α)Res f(z) + g(z) dz.

z=c
Z Cr 
 
Now g is bounded near c (say by M ), then  g(z) dz  ≤ M L = M (β − α)r → 0 as r → 0+ . QED.
Cr

Z ∞
eax
Example. Find P.V. dx, where 0 < a < 1.
−∞ 1 + ex
eaz
Step 1 . Consider the function f(z) = , its poles are at nπi (n odd integer).
1 + ez
Consider the contour on the left, where R is large.
−γ 2 3π i γ1 (t) = R + it, 0 ≤ t ≤ 2π,
γ2 (t) = t + 2πi, −R ≤ t ≤ R,
−γ 3 πi γ1 γ3 (t) = −R + it, 0 ≤ t ≤ 2π.

-R R
−π i
Z R Z 
(z − πi)eaz
By the residue theorem, + f(z) dz = 2πi Res f(z) = 2πi lim = −2πieπai .
−R z=πi z→πi 1 + ez
γ1 −γ2 −γ3
Z  Z 2π a(R+it) 
 eaz   e i dt  eaR
Step 2 . We have   dz  =  ≤ M L = 2π → 0 as R → +∞ (because a < 1),
z   1 + eR+it  eR − 1
γ1 1 + e 0
Z Z R a(t+2πi) Z R
eaz e 2πai eat
z
dz = t+2πi
dt = e t
dt and
γ2 1 + e −R 1 + e −R 1 + e
Z  Z 2π a(−R+it) 
 eaz   e i dt  e−aR
 dz = ≤ ML = 2π → 0 as R → +∞ (because 0 < a).
 z   −R+it  1 − e−R
γ3 1 + e 0 1+e
Z R Z  Z ∞
eax
Step 3 . So, −2πieπai = lim + f(z) dz = (1 − e2πai ) x
dx.
R→+∞ −R −∞ 1 + e
γ1 −γ2 −γ3
Z ∞  
eax −2πieπai 2i π
Therefore, x
dx = 2πai
= π πai −πai
= .
−∞ 1 + e 1−e e −e sin πa

The following example illustrates how the residue theorem can produce interesting series representations
of some common functions.
cos z
Example. The poles of cot z = are at kπ, k any integer. For a fixed z 6= kπ, let n be large so that
sin z
1 1
z is inside the rectangle Γn with vertices ±(n + )π ± i(n + )π. Now for w = (n + 12 )π + iy on the right
 iw   −y 2 2
 e + e−iw   e − ey 

edge of Γn , | cot w| =  iw  
=  −y  ≤ 1 and for w = x + i(n + 12 )π on the top edge of Γn,
 2iw  e − e−iw  e + ey 
e + 1  1 + e−(2n+1)π 1 + e−π
| cot w| =  2iw  ≤ −(2n+1)π
≤ = C. Since | cot w| = | cot(−w)| and 1 ≤ C, so | cot w| ≤ C
e −1 1−e 1 − e−π
on Γn. By the M -L inequality, as n → ∞,
Z 
 z cot w dw  |z|C
 ≤
(8n + 4)π → 0.
 w(w − z)  (n + )π (n + 1 )π − |z|
1
Γn 2 2

65
By the residue theorem,
Z !
1 z cot w dw z cot w X
n
z cot w
0 = lim = lim Res + Res
n→∞ 2πi Γn w(w − z) n→∞ w=z w(w − z) w=kπ w(w − z)
k=−n

∞
1 X z 1 X 2z
= cot z − + = cot z − − .
z kπ(kπ − z) z z − k2π2
2
k6=0 k=1

∞
X ∞ 
X 
1 2z 1 1 1
Therefore, cot z = + = + + for z 6= kπ, k any integer.
z z 2 − k2 π2 z z − kπ z + kπ
k=1 k=1

Remarks. The last example can be modified to yield formulas for summing series. To be more precise, let
f be meromorphic on C with finitely many poles zp , p = 1, 2, . . ., k and lim zf(z) = 0. Suppose Cn is the
z→∞
1 1
counterclockwise square with vertices ±(n + ) ± i(n + ), then by the residue theorem,
2 2
 
Z
 X
n X k

πf(z) cot πz dz = 2πi 
 f(j) + Res (πf(z) cot πz, zp ) .

Cn j=−n p=1
j6=zp

1 1 √ 1
Since lim zf(z) = 0 and n + ≤ |z| ≤ (n + ) 2 for all z ∈ Cn , we have lim (n + ) max |f(z)| = 0. By
z→∞ 2 2 n→∞ 2 z∈Cn
the M -L inequality, as n → ∞,
Z   
  1 + e−π
 πf(z) cot πz dz  ≤ π max |f(z)| (8n + 4) → 0.
  z∈Cn 1 − e−π
Cn

It follows that
X
∞ X
k
f(j) = − Res (πf(z) cot πz, zp ) .
j=−∞ p=1
j6=zp

1
For instance, if we set f(z) = , then
z2
X∞ ∞  
1 1 X 1 1 π cot πz π2
= = − Res , 0 = .
j=1
j2 2 j=−∞ j 2 2 z2 6
j6=0

If we replace cot z by csc z in the above argument, then we get a similar formula

X
∞ X
k
(−1)j f(j) = − Res (πf(z) csc πz, zp ) .
j=−∞ p=1
j6=zp

Exercises

Z π Z 2π
2 dθ π 1 (2n)!
1. Show that = √ and (2 cos θ)2n dθ = for every positive integer n.
0 1 + sin2 θ 2 2 2π 0 n!n!

66
 Z ∞
dx
2. Find , where n ≥ 2 is a positive integer. [Hint: This can be done following the example for
0 1 + xn
Type II integrals. Alternatively, the contour below can be considered.]
Re 2πi/n

0 R

Z ∞ Z ∞ Z ∞
sin2 x sin3 x sin2 x
3. Find dx, dx and dx. [Hint: 4 sin3 x = Im(3eix − e3ix).]
−∞ x2 −∞ x3 −∞ 1 + x2

Z ∞ Z ∞
4. Find cos x2 dx and sin x2 dx. [ Hint: Use the contour . ]
π/4
0 0 0 R

Z ∞
ln x
5. Find dx. [ Hint: Use the contour . ]
0 x2 + 1
-R -r r R
Z ∞ √ Z ∞
2 π 2
6. Given that e−x dx = . Find e−x cos 2x dx.
0 2 0
-R+i R+i
2
[Hint: Use the contour and f(z) = e−z . ]
-R R

Z
1
7. Suppose f is holomorphic on the annulus {z : r < |z| < ∞}. Define Res(f, ∞) = − f(z) dz.
2πi |z|=R>r
oo
(Here, the minus sign appears because the counterclockwise orientation around ∞ cor-
responds to the clockwise orientation around 0 as can be seen from stereographic projec-
X∞
tion.) Equivalently, if f(z) = ak z k on {z : r < |z| < ∞}, then Res(f, ∞) = −a−1.
0 k=−∞
X
n
(a) If f is meromorphic on C with isolated poles at a1 , . . . , an, show that Res(f, aj )+Res(f, ∞) = 0.
j=1
That is, the sum of all residuesin C ∪{∞}
 is 0.
1 1
(b) Show that Res(f(z), ∞) = Res − 2 f ,0 .
Z z z
dz
(c) Find

1 .
|z|=1 sin z

X∞
1 X (−1)j+1
∞ X∞
1
8. Find 4
, 2
and .
j=1
j j=1
j j=1
1 + j2

67
18. Infinite Products

Suppose a polynomial P (z) has roots at 0 (of order m), z1 , z2 , . . ., zn . Then

Y
n n 
Y 
z
P (z) = Az m (z − zk ) = Bz m 1− ,
zk
k=1 k=1

P (z)
where B = A × (−z1 ) × · · · × (−zn ) and can be determined from P , e.g. B = lim .
zm z→0

Now sin z is entire and has simple roots at kπ, k any integer. If we think of (the power series of) an
entire function as an “infinite” polynomial, then we may conjecture that
∞  ∞  
Y z  z  Y z2
sin z = z 1− 1+ =z 1− 2 2 .
kπ kπ k π
k=1 k=1

π
This formula is true for all complex numbers and was first discovered by Euler. Taking z = and
2
transposing terms, we get

Y
∞     
π 4k2 2×2 4×4 6×6 8×8
= = ···,
2 4k2 − 1 1×3 3×5 5×7 7×9
k=1

which is known as Wallis’ Formula. Also we have

∞   " ∞
! #
z3 Y z2 X 1 2
z− + · · · = sin z = z 1− 2 2 = z 1− z + ··· .
6 k π k π2
2
k=1 k=1

X
∞
1 π2
3
By the uniqueness of the coefficients of z , we deduce that = .
k2 6
k=1

Now to prove Euler’s formula for sin z, we begin with the definition of infinite products.

Y
∞ Y
n
Definitions. For a sequence {ak } of nonzero complex numbers, we define ak = lim ak . If the limit
n→∞
k=1 k=1
∞
Y ∞
Y
is a nonzero number, we say ak converges. Otherwise, we say ak diverges. For sequences with finitely
k=1 k=1
∞
Y ∞
Y
many of the ak ’s being 0 and ak converges, we also say ak converges to 0.
k=1,ak 6=0 k=1

∞ 
Y  n
Y
1 k−1 1
Examples. (1) 1− = lim = lim diverges to 0.
k n→∞ k n→∞ n
k=2 k=2

Y
∞   Y
n
1 k+1
(2) 1+ = lim = lim (n + 1) diverges to ∞.
k n→∞ k n→∞
k=1 k=1

68
 Y∞   Yn   
1 (k − 1)(k + 1) n+1 1 1 1 1
(3) 1 − 2 = lim = lim = . (Note that 1 − 2 = 1 + 1− . If
k n→∞ k2 n→∞ 2n 2 k k k
k=2 k=2
1
we consider the factor 1 − as a “weight”, then we see that an infinite product can be made to converge
k
by putting proper weights on the terms.)

Y
∞
Term Test. If ak converges, then
k=1

 
Y
k  Y
k−1 Y
∞  Y
∞
lim ak = lim  ap ap  = ap ap = 1.
k→∞ k→∞
p=1,ap 6=0 p=1,ap 6=0 p=1,ap 6=0 p=1,ap 6=0

Remark. For convergent products, the terms are close to 1. Since the logarithm of a finite product is the
∞
Y ∞
X
sum of the logarithms, we see easily that ak converges if and only if Log ak converges.
k=1 k=1,ak 6=0
1
In the lemma below, it will be shown that for |z| ≤ , | Log(1−z)| ≤ 2|z|. So, by the absolute convergence
∞
2 ∞
X Y
test and the comparison test, |1 − ak | converges implies ak converges.
k=1 k=1

Definition. An elementary factor is one of the following entire functions

z2 zp
z+ 2 + ···+ p
E0 (z) = 1 − z and Ep (z) = (1 − z)e for p = 1, 2, 3, . . ..

1
Lemma. If |z| ≤ , then | Log Ep(z)| ≤ 2|z|p+1.
2
1
Proof. For |z| ≤ , let C be the line segment from 0 to z. Then
2
Z Z X
∞ ∞
X
dw z k+1
Log(1 − z) = − =− wk dw = − and
C 1−w C k=0 k+1
k=0

 
X 
 ∞ z k+1  X ∞
|z|k+1 |z|p+1 X 1
∞
| Log Ep (z)| =  ≤
 ≤ k−p
≤ 2|z|p+1.
k=p k + 1  k=p k + 1 p + 1 2
k=p

QED.

Theorem. Let {zk } be a sequence of nonzero complex numbers converging to ∞ and {mk } a sequence of
X∞  mk
R R 1
nonnegative integers such that for any fixed R > 0, converges (e.g. mk = k because <
|zk | |zk | 2
k=1
for k large). Then the product

Y
∞   ∞ 
Y  z + z2
+···+ z mk −1
z z zk 2zk2 m −1
(mk −1)zk k
Emk −1 = 1− e
zk zk
k=1 k=1

converges to an entire function with roots zk (repeated according to multiplicities.)

69
  
z  1
Proof. Fix R > 0. Let N be such that k > N ⇒ 2R < |zk |. Then for |z| ≤ R and k > N , we have   < .
zk 2
By the lemma, 
X∞   X∞  mk
 
Log Emk −1 z  ≤ 2 R
< ∞.
 zk  |zk |
k=N +1 k=N +1

X
∞  
z
By Weierstrass’ M -test (and Weierstrass’ theorem), Log Emk −1 converges uniformly (to a holo-
zk
k=N +1
morphic function) on |z| ≤ R. Exponentiating this function and multiplying the first N terms, we see that
Y∞   Y∞  
z z
Emk −1 is holomorphic on |z| < R. Since R is arbitrary, Emk −1 must be entire.
zk zk
k=1 k=1
  Y∞  
w w
Clearly, the zk ’s are roots. For w 6= zk , Emk −1 6= 0 and the convergence of Emk −1
zk zk
k=1
Y∞   Y∞  
w w
implies lim Emk −1 = 1. So for K large, Emk −1 is close to 1; in particular, it is
K→∞ zk zk
k=K+1 k=K+1
Y∞   Y K   Y ∞  
w w w
nonzero. Then Emk −1 = Emk −1 6= 0. Therefore, there cannot be any other
zk zk zk
k=1 k=1 k=K+1
roots. QED.

Weierstrass Factorization Theorem. Let f be an entire function with roots 0 (of order m ≥ 0), z1 , z2, . . .
converging to ∞. Suppose there is a sequence {mk } of nonnegative integers as in the previous theorem, then
there is an entire function g such that
∞ 
Y  z + ···+ z mk −1
g(z) m z z m −1
(mk −1)zk k
f(z) ≡ e z 1− e k .
zk
k=1

Y
∞  
z
Proof. By the previous theorem, h(z) = z m Emk −1 is entire and has the same roots (with the
zk
k=1
f
same multiplicities) as f(z). Then is an entire function without any roots. By the logarithm theorem,
h
f
= eg for some entire function g. Therefore, f = eg h. QED.
h
X  R 2
Finally, Euler’s formula for sin z follows easily. Since < ∞ for all R > 0, we may take mk = 2
kπ
k6=0
for all k. By the Weierstrass factorization theorem,
Y z  z/kπ
∞ 
Y z  z 
sin z = eg(z) z 1− e = eg(z) z 1− 1+ .
kπ kπ kπ
k6=0 k=1

Taking the logarithmic derivative, we get

1 X
∞
d 2z
cot z = (log sin z) = g0(z) + + .
dz z z 2 − k2 π2
k=1

sin z
From the last example of last section, we see that g must be constant. Now eg = lim = 1. Therefore,
z→0 z
∞ 
Y 
z2
sin z = z 1− 2 2 .
k π
k=1

70
 Exercises

1. Prove that
∞ 
Y  Y∞
1 n3 − 1 2
(a) 1+ = 2; (b) 3
= .
n=1
n(n + 2) n=2
n +1 3

∞
Y α sin α 2 2 2 π
2. Prove that cos = and √ p √ r q ··· = .
n=1
2n α 2 2+ 2 √ 2
2+ 2+ 2

Y∞   X (−1)k−1 n!
n−1
(−1)n−1
3. Find 1+ , where an = .
n=2
an k!
k=1

∞
Y n 1
4. If |z| < 1, prove that (1 + z 2 ) = .
n=0
1−z

∞
X ∞
Y
5. Suppose |zn |2 < ∞. Is it necessary that cos zn must converge?
n=1 n=1

∞
X ∞
Y (−1)k
6. Show that (1 − ak ) converges does not imply ak converges by considering ak = 1 + √ .
k=1 k=1
k

71
 Suggested Readings and References

There are a large number of textbooks on complex analysis. Some good texts for reference alongside
this set of notes are:
[1] J. Bak and D. J. Newman, Complex Analysis, Springer-Verlag, 1982;
[2] R. P. Boas, Invitation to Complex Analysis, Random House, 1987;
[3] J. E. Marsden and M. J. Hoffman, Basic Complex Analysis, 2nd ed.,Freeman, 1987;
[4] Yu. V. Sidorov, M. V. Fedoryuk and M. I. Shabunin, Lectures on the Theory of Functions of a Complex
Variable, Mir, 1985.

For students who like to have a simple glance at applications in science and engineering, we refer to
[5] S. Fisher, Complex Variables, 2nd ed., Brooks/Cole, 1990.

For mature students who like to read more about the subject, there are the classics:
[6] L. Ahlfors, Complex Analysis, 3rd ed., Mc Graw-Hill, 1979;
[7] S. Saks and A. Zygmund, Analytic Functions, 3rd ed., Elsevier, 1971;
[8] E. C. Titchmarsh, The Theory of Functions, 2nd ed., Oxford, 1939.

For students who like to do more problems, we recommend

[9] K. Knopp, Problem Book in the Theory of Functions, vol. I & II, Dover, 1948,1952;
[10] J. G. Krzyz, Problems in Complex Variable Theory, Elsevier, 1971;
[11] L. I. Volkovyskii, G. L. Lunts and I. G. Aramanovich, A Collection of Problems on Complex Analysis,
Dover, 1991.

In passing, we will like to mention that there are many well-written historical accounts of the subject
appeared in the periodical The Mathematical Intelligencer, published by Springer-Verlag. On the other hand,
for those who prefer to read a book on the development of the subject with some history, the following is
highly recommended:
[12] R. Remmert, Theory of Complex Functions, Springer-Verlag, 1991.

Among the references above, [1] and [4] are the best to read along with this set of notes. One of the
many outstanding features of [4] is its large collection of examples, from simple to advanced. This reference
also have excellent discussions on conformal mappings, multiple-valued analytic functions and asymptotic
methods. On the other hand, [1] is a combination of beauty and elegance. There are interesting materials in
[1] which do not appear anywhere else in the literature. Also, in [1], there are many ingenious ideas useful
in different parts of mathematics and clever proofs of important theorems. For [2] and [3], the students
will find interesting techniques, useful formulas and detailed discussions on specific materials. For example,
the discussions on analytic continuation in [2] is very, very well presented, informative and worthy of any
student’s attention.

As for exercises, the problems in [9], [10] and [11] are very good for practice. Some are quite challenging
and most will serve to strengthen the students’ understanding.

For students who like to continue pursuing the subject, [8] should be a good place to start, then the
second half of [6] and most of [7] would provide a solid foundation. Finally, those who plan to eventually
specialize in analysis, we suggest the standard graduate texts:
[13] J. B. Conway, Functions of One Complex Variable, 2nd ed., Springer-Verlag, 1978;
[14] W. Rudin, Real and Complex Analysis, 3rd ed., Mc Graw-Hill, 1987.

72
 Index

absolute convergence test 2 domain of convergence 12

analytic 10 double root 36
annulus 46
argument principle 52 elementary factor 65
entire 10
big O notation 59 essential singularity 48, 49
bilinear transformation 21 Euler’s equation 17
boundary 4 exponential function 17, 18
bounded set 4 extended complex plane 9
branch 18 extension theorem 29

Casaroti-Weierstrass theorem 48 fixed point 21, 39

Cauchy integral formula 32, 35, 51 fundamental theorem
for derivatives 35, 51 of algebra 36
Cauchy sequence 1 of calculus 26
Cauchy-Riemann equations 15, 16
polar form 16 Goursat’s theorem 28
Cauchy-Schwarz inequality 3 great Picard theorem 48
Cauchy’s theorem 28, 30
for homotopic curves 30 harmonic 41
on disks 28 conjugate 41
chordal metric 10 holomorphic 10
closed curve 26 holomorphic part 48
closed curve theorem 27 Hurwitz’s theorem 54
closed disk 4
closed set 4 identity theorem 13, 37
closure 4 for power series 13
compact 4 improper integral 57
conformal mapping 20 principal value 57
conformally equivalent 20 Riemann sense 57
connected 4 infinite product 64
continuous 7, 9 integral function 10
contour integral 26 integral of contour 26
converge 1, 2, 46, 64 integral theorem 28, 30
infinite product 64 interior 4
sequence 1 inverse mapping theorem 54
series 2, 46 inverse of function 17
converge uniformly 8 inverse rule 18
cross ratio 22 isolated singulaity 47

differentiable 10 Jordan’s inequality 58

disconnected 4 Jordan’s lemma 59
disk of convergence 12
diverge 1, 2, 64 Laplace equation 41, 43
infinite prouct 64 polar form 43
sequence 1 Laurent expansion 47
series 2 Laurent series 46
domain 4 Laurent series representation 46

73
left side 23 radius of convergence 12
l’ Hôpital’s rule 36 rational function 7
limit 7, 9 “rectangle” 28
limit superior 11 rectangle property 28
linear fractional transformation 21 rectangle theorem 28
Liouville’s theorem 35 region 4
little o notation 59 regular 10
little Picard theorem 36 removable singularity 47, 49
local mapping theorem 53 residue 51, 63
logarithm function 17 residue theorem 51
logarithm theorem 30 Riemann mapping 24
logarithm function 17 Riemann mapping theorem 23
principal branch 17 Riemann sense 57
Riemann sphere 9
maximum modulus theorem 37 Riemann’s principle 48
maximum/minimum principle 42 right side 23
mean value theorem 37, 42 right-angled polygon 5
for harmonic functions 42 root 36
meromorphic 51 root test 11
minimum modulus theorem 37 Rouché’s theorem 52
M -L inequality 27
Möbius transformation 21 schlicht 10
Morera’s theorem 44 Schwarz reflection principle 45
multiplicity 36 Schwarz’s lemma 38
simple closed curve 26
neighborhood 4 simple root 36
simply connected 5
open disk 4 stereographic projection 9
open mapping theorem 38 Study’s theorem 40
open set 4 symmetric points 22
order 36, 47, 49 symmetry principle 23
orientation 23
orientation principle 23 Taylor’s theorem 13
term test 2, 65
Picard theorem 36, 48 for series 2
great 48 for infinite product 65
little 36 trigonometric functions 17
piecewise smooth 26 triple root 36
Poisson integral formula 42
polar representation 1 uniform convergence 7,8
pole 47, 49 uniqueness theorem 13, 37
order 47, 49 for power series 13
polygon property 30 univalent 10
polygon theorem 30 upper limit 11
polygonally connected 4
power series 11 Wallis’ formula 64
power series representation 33 Weierstrass factorization theorem 66
principal branch 17 Weierstrass M -test 7
principal part 48 Weierstrass’s theorem 44
principal value 18, 57 winding number 32
of improper integral 57

74