2.

Functions

2.1 Basic notions

1. Write the following functions f (z) in the forms f (z) = u(x, y) + iv(x, y) under Carte-
sian coordinates with u(x, y) = Re( f (z)) and v(x, y) = Im( f (z)):
(a) f (z) = z3 + z + 1
(b) f (z) = z3 − z;
1
(c) f (z) = ;
i−z
(d) f (z) = exp(z2 ).

Solution. (a)

f (z) = (x + iy)3 + (x + iy) + 1

= (x + iy)(x2 − y2 + 2ixy) + x + iy + 1
= x3 − xy2 + 2ix2 y + ix2 y − iy3 − 2xy2 + x + iy + 1
= x3 − 3xy2 + x + 1 + i(3x2 y − y3 + y).

(b)

f (z) = z3 − z = (x + yi)3 − (x + yi)

= (x3 + 3x2 yi − 3xy2 − y3 i) − (x + yi)
= (x3 − 3xy2 − x) + i(3x2 y − y3 − y),
22 Chapter 2. Functions

(c)
1 1
f (z) = =
i − z −x + (1 − y)i
−x − (1 − y)i
= 2
x + (1 − y)2
x 1−y
=− 2 2
−i 2
x + (1 − y) x + (1 − y)2
(d)
f (z) = exp(z2 ) = exp((x + yi)2 )
= exp((x2 − y2 ) + 2xyi)
2 −y2
= ex (cos(2xy) + i sin(2xy))
x2 −y2 2 −y2
=e cos(2xy) − iex sin(2xy)


2. Suppose that f (z) = x2 − y2 − 2y + i(2x − 2xy), where z = x + iy. Use the expressions
z+z z−z
x= and y =
2 2i
to write f (z) in terms of z and simplify the result.

Solution. We have
f (z) = x2 − y2 − 2y + i(2x − 2xy)
= x2 − y2 + i2x − i2xy − 2y
= (x − iy)2 + i(2x + 2iy)
= z2 + 2iz.

3. Suppose p(z) is a polynomial with real coefficients. Prove that
(a) p(z) = p(z);
(b) p(z) = 0 if and only if p(z) = 0;
(c) the roots of p(z) = 0 appear in conjugate pairs, i.e., if z0 is a root of p(z) = 0,
so is z0 .

Proof. Let p(z) = a0 + a1 z + ... + an zn for a0 , a1 , ..., an ∈ R. Then

p(z) = a0 + a1 z + · · · + an zn
= a0 + a1 z + · · · + an zn
= a0 + (a1 )z + · · · + (an )zn
= a0 + a1 z + · · · + an zn = p(z).
If p(z) = 0, then p(z) = 0 and hence p(z) = p(z) = 0; on the other hand, if p(z) = 0,
then p(z) = p(z) = 0 and hence p(z) = 0.
By the above, p(z0 ) = 0 if and only if p(z0 ) = 0. Therefore, z0 is a root of p(z) = 0
if and only if z0 is. 
2.1 Basic notions 23

4. Let
z
T (z) = .
z+1
Find the inverse image of the disk |z| < 1/2 under T and sketch it.

Solution. Let D = {|z| < 1/2}. The inverse image of D under T is

1
T −1 (D) = {z ∈ C : T (z) ∈ D} = {|T (z)| < }
  2
z 1
= z: < = {2|z| < |z + 1|}.
z+1 2

Let z = x + yi. Then

2|z| < |z + 1| ⇔ 4(x2 + y2 ) < (x + 1)2 + y2

⇔ 3x2 − 2x + 3y2 < 1
1 2
 
4
⇔ x− + y2 <
3 9
1 2
⇔ z− <
3 3

So
 
−1 1 2
T (D) = z : z − <
3 3

is the disk with centre at 1/3 and radius 2/3. 

5. Sketch the following sets in the complex plane C and determine whether they are
open, closed, or neither; bounded; connected. Briefly state your reason.
(a) |z + 3| < 1;
(b) | Im(z)| ≥ 1;
(c) 1 ≤ |z + 3| < 2.

Solution. (a) Since {|z + 3| < 1} = {(x + 3)2 + y2 − 1 < 0} and f (x, y) = (x + 3)2 +
y2 − 1 is a continuous function on R2 , the set is open. It is not closed since the only
sets that are both open and closed in C are 0/ and C.
Since

|z| = |z + 3 − 3| ≤ |z + 3| + | − 3| = |z + 3| + 3 < 4

for all |z + 3| < 1, {|z + 3| < 1} ⊂ {|z| < 4} and hence it is bounded.
It is connected since it is a convex set. 

Solution. (b) We have

{| Im(z)| ≥ 1} = {|y| ≥ 1} = {y ≥ 1} ∪ {y ≤ −1}.

24 Chapter 2. Functions

Since f (x, y) = y is continuous on R2 , both {y ≥ 1} and {y ≤ −1} are closed and

hence {| Im(z)| ≥ 1} is closed. It is not open since the only sets that are both open
and closed in C are 0/ and C.
Since zn = n + 2i ∈ {| Im(z)| ≥ 1} for all n ∈ Z and
p
lim |zn | = lim n2 + 4 = ∞,
n→∞ n→∞
the set is unbounded.
The set is not connected. Otherwise, let p = 2i and q = −2i. There is a polygonal
path
p0 p1 ∪ p1 p2 ∪ ... ∪ pn−1 pn
with p0 = p, pn = q and pk ∈ {| Im(z)| ≥ 1} for all 0 ≤ k ≤ n.
Let 0 ≤ m ≤ n be the largest integer such that pm ∈ {y ≥ 1}. Then pm+1 ∈ {y ≤ −1}.
So Im(pm ) ≥ 1 > 0 and Im(pm+1 ) ≤ −1 < 0. It follows that there is a point p ∈
pm pm+1 such that Im(p) = 0. This is a contradiction since pm pm+1 ⊂ {| Im(z)| ≥ 1}
but p 6∈ {| Im(z)| ≥ 1}. Therefore the set is not connected. 
Solution. (c) Since −2 ∈ {1 ≤ |z + 3| < 2} and {|z + 2| < r} 6⊂ {1 ≤ |z + 3| < 2}
for all r > 0, {1 ≤ |z + 3| < 2} is not open. Similarly, −1 is a point lying on its
complement
{1 ≤ |z + 3| < 2}c = {|z + 3| ≥ 2} ∪ {|z + 3| < 1}
and {|z + 1| < r} 6⊂ {1 ≤ |z + 3| < 2}c for all r > 0. Hence {1 ≤ |z + 3| < 2}c is not
open and {1 ≤ |z + 3| < 2} is not closed. In summary, {1 ≤ |z + 3| < 2} is neither
open nor closed.
Since
|z| = |z + 3 − 3| ≤ |z + 3| + | − 3| < 5
for all |z + 3| < 2, {1 ≤ |z + 3| < 2} ⊂ {|z| < 5} and hence it is bounded.
The set is connected. To see this, we let p1 = −3/2, p2 = −3 + 3i/2, p3 = −9/2
and p4 = −3 − 3i/2. All these points lie on the circle {|z + 3| = 3/2} and hence lie
in {1 ≤ |z + 3| < 2}.
It is easy to check that for every point p ∈ {1 ≤ |z + 3| < 2}, ppk ⊂ {1 ≤ |z + 3| < 2}
for at least one pk ∈ {p1 , p2 , p3 , p4 }. So the set is connected. 

6. Show that
| sin z|2 = (sin x)2 + (sinh y)2
for all complex numbers z = x + yi.
Proof.
| sin(z)|2 = | sin(x + yi)|2 = | sin(x) cos(yi) + cos(x) sin(yi)|2
= | sin(x) cosh(y) − i cos(x) sinh(y)|2
= sin2 x cosh2 y + cos2 x sinh2 y
= sin2 x(1 + sinh2 y) + cos2 x sinh2 y
= sin2 x + (cos2 x + sin2 x) sinh2 y = (sin x)2 + (sinh y)2 .

2.1 Basic notions 25

7. Show that

| cos(z)|2 = (cos x)2 + (sinh y)2

for all z ∈ C, where x = Re(z) and y = Im(z).

Proof.

| cos(z)|2 = | cos(x + yi)|2 = | cos(x) cos(yi) − sin(x) sin(yi)|2

= | cos(x) cosh(y) − i sin(x) sinh(y)|2
= cos2 x cosh2 y + sin2 x sinh2 y
= cos2 x(1 + sinh2 y) + sin2 x sinh2 y
= cos2 x + (cos2 x + sin2 x) sinh2 y = (cos x)2 + (sinh y)2

8. Show that
tan z1 + tan z2
tan(z1 + z2 ) =
1 − (tan z1 )(tan z2 )

for all complex numbers z1 and z2 satisfying z1 , z2 , z1 + z2 6= nπ + π/2 for any integer
n.

Proof. Since

i(e−iz1 − eiz1 ) i(e−iz2 − eiz2 )

tan z1 + tan z2 = + iz
eiz1 + e−iz1 e 2 + e−iz2
−iz iz iz −iz −iz
(e 1 − e 1 )(e 2 + e 2 ) + (e 2 − eiz2 )(eiz1 + e−iz1 )
=i
(eiz1 + e−iz1 )(eiz2 + e−iz2 )
ei(z1 +z2 ) − e−i(z1 +z2 )
= −2i
(eiz1 + e−iz1 )(eiz2 + e−iz2 )

and
i(e−iz1 − eiz1 )
  −iz
i(e 2 − eiz2 )
 
1 − (tan z1 )(tan z2 ) = 1 −
eiz1 + e−iz1 eiz2 + e−iz2
(e−iz1 + eiz1 )(e−iz2 + eiz2 ) + (e−iz1 − e−iz1 )(e−iz2 − eiz2 )
=
(e−iz1 + eiz1 )(e−iz2 + eiz2 )
ei(z1 +z2 ) + e−i(z1 +z2 )
=2 ,
(eiz1 + e−iz1 )(eiz2 + e−iz2 )

we have

tan z1 + tan z2 ei(z1 +z2 ) − e−i(z1 +z2 )

= −i i(z +z ) = tan(z1 + z2 ).
1 − (tan z1 )(tan z2 ) e 1 2 + e−i(z1 +z2 )
26 Chapter 2. Functions

Alternatively, we can argue as follows if we assume that the identity holds for z1 and
z2 real. Let
tan z1 + tan z2
F(z1 , z2 ) = tan(z1 + z2 ) − .
1 − (tan z1 )(tan z2 )
We assume that F(z1 , z2 ) = 0 for all z1 , z2 ∈ R with z1 , z2 , z1 + z2 6= nπ + π/2.
Fixing z1 ∈ R, we let f (z) = F(z1 , z). Then f (z) is analytic in its domain
C\({nπ + π/2} ∪ {nπ + π/2 − z1 }).
And we know that f (z) = 0 for z real. Therefore, by the uniqueness of analytic
functions, f (z) ≡ 0 in its domain. So F(z1 , z2 ) = 0 for all z1 ∈ R and z2 ∈ C in its
domain.
Fixing z2 ∈ C, we let g(z) = F(z, z2 ). Then g(z) is analytic in its domain
C\({nπ + π/2} ∪ {nπ + π/2 − z2 }).
And we have proved that g(z) = 0 for z real. Therefore, by the uniqueness of analytic
functions, g(z) ≡ 0 in its domain. Hence F(z1 , z2 ) = 0 for all z1 ∈ C and z2 ∈ C in
its domain. 

9. Find all the complex roots of the equation cos z = 3.

Solution. Since cos z = (eiz + e−iz )/2, it comes down to solve the equation eiz +
e−iz = 6, i.e.,
w + w−1 = 6 ⇔ w2 − 6w + 1 = 0
√
if we let w = eiz . The roots of w2 − 6w + 1 = 0 are w = 3 ± 2 2. Therefore, the
solutions for cos z = 3 are
√ √ √
iz = log(3 ± 2 2) ⇔ z = −i(ln(3 ± 2 2) + 2nπi) = 2nπ − i ln(3 ± 2 2)
for n integers. 
π 
10. Calculate sin +i .
4
Solution.
π  1
sin + i = (ei(π/4+i) − e−i(π/4+i) )
4 2i
1
= (e−1 eπi/4 − ee−πi/4 )
2i
1  −1 π π π π 
= e (cos + i sin ) − e(cos − i sin )
2i
√  4 4 4 4
 √  
2 1 2 1
= e+ + e− i
4 e 4 e

2.1 Basic notions 27
π 
11. Compute cos +i .
3

Solution.
π  1
cos + i = (ei(π/3+i) + e−i(π/3+i) )
3 2
1 −1 πi/3
= (e e + ee−πi/3 )
2
1  −1 π π π π 
= e (cos + i sin ) + e(cos − i sin )
2 3 3 3 3
  √  
1 1 3i 1
= e+ − e−
4 e 4 e

12. Find ii and its principal value.

Solution. We have

ii = ei log i = ei(2nπi+πi/2) = e−2nπ−π/2

for n integers and its principal value given by

ii = ei Log i = ei(πi/2) = e−π/2 .

√
13. Let f (z) be the principal branch of 3 z.

(a) Find f (−i).

Solution.
√
1 1 πi πi 3 i
f (−i) = exp( Log(−i)) = exp( (− )) = exp(− ) = −
3 3 2 6 2 2


(b) Show that

f (z1 ) f (z2 ) = λ f (z1 z2 )

√ √
−1 + 3i −1 − 3i
for all z1 , z2 6= 0, where λ = 1, or .
2 2
28 Chapter 2. Functions

Proof. Since
f (z1 ) f (z2 ) 1 1 1
= exp( Log z1 + Log z2 − Log(z1 z2 ))
f (z1 z2 ) 3 3 3
1
= exp( (Log z1 + Log z2 − Log(z1 z2 )))
3
i 2nπi
= exp( (Arg z1 + Arg z2 − Arg(z1 z2 ))) = exp( )
3 3
for some integer n, λ = exp(2nπi/3). Therefore,

1 √
 if n = 3k
λ= −1+ 3i
2 if n = 3k + 1
 −1−√3i

2 if n = 3k + 2
where k ∈ Z. 

14. Let f (z) be the principal branch of z−i .

(a) Find f (i).
Solution.
f (i) = i−i = exp(−i Log(i)) = exp(−i(πi/2)) = eπ/2 .

(b) Show that
f (z1 ) f (z2 ) = λ f (z1 z2 )
for all z1 , z2 6= 0, where λ = 1, e2π or e−2π .
Proof. Since
f (z1 ) f (z2 )
= exp(−i Log z1 − i Log z2 + i Log(z1 z2 ))
f (z1 z2 )
= exp(−i(Log z1 + Log z2 − Log(z1 z2 )))
= exp(−i(i Arg z1 + i Arg z2 − i Arg(z1 z2 )))
= exp(Arg z1 + Arg z2 − Arg(z1 z2 ))
= exp(2nπ)
for some integer n, λ = exp(2nπ). And since
−π < Arg(z1 ) ≤ π, −π < Arg(z2 ) ≤ π
and
−π < Arg(z1 z2 ) ≤ π,
we conclude that
−3π < Arg z1 + Arg z2 − Arg(z1 z2 ) < 3π
and hence −3 < 2n < 3. So n = −1, 0 or 1 and λ = e−2π , 1 or e2π . 
2.1 Basic notions 29

15. Determine the Möbius transformation mapping 0 to 2, 2i to 0, and i to 3/2.

Solution. Consider the function
az + b
f (z) = . (2.1)
cz + d
Then
b
f (0) = 2 then =2 (2.2)
d
f (−2) = 0 then −2ia + b = 0 (2.3)
3 ai + b 3
f (i) = then = . (2.4)
2 ci + d 2
From 2.2 we get d 6= 0. Notice also that we can take d to be any nonzero complex
number.
Let us take d = i. Thus, from expression 2.2, we have b = 2i. From 2.3, a = 1, and
from 2.4 we have c = 1.
Hence the transformation we are looking for is
z + 2i
f (z) = .
z+i
Remark: Other formulations are possible for different choices of the constant d,
e.g.
iz − 2
f (z) = .
iz − 1

16. Let T be a mapping from C to C. A fixed point of T is a point z satisfying T (z) = z.
(a) Show that any Möbius transformation, apart from the identity, can have at most
2 fixed points in C.
(b) Give examples of Möbius transformations having (i) 2; (ii) 1 and (iii) no fixed
points in C.

Proof. (a) Let the Möbius transformation be

az + b
T (z) = .
cz + d
If T (z) = z, then
az + b
= z,
cz + d
cz2 + dz = az + b,
cz2 + (d − a)z − b = 0. (2.5)
Now, if c 6= 0, then the expression 2.5 is a quadratic equation with one or two
solutions. That is
−(d − a) + [(d − a)2 + 4bc]1/2
.
2c
30 Chapter 2. Functions

If c = 0, expression 2.5 becomes

(d − a)z = b.

This equation has one solution b/(d − a), if d 6= a. It has no solution, if d = a and
b 6= 0. And, finally, it has infinitely many solutions, if d = a and b = 0.
Notice that in the last case we have d = a and b = c = 0. Since ad − bc 6= 0, then
a = d 6= 0. Thus we have
az
T (z) = = z.
d
That is the identity, which is excluded.
(b) (i) From part (a), if c 6= 0 and (d − a)2 + 4bc 6= 0, we will have 2 fixed points.
Hence, the function
1
T (z) =
z
is an example with two fixed points: −1 and 1.
(ii) Now, a function with one fixed point, is the following
z
T (z) = .
2
(iii) Finally, with no fixed points, we have

T (z) = z + 1.

17. For z ∈ C, show that:

(a) sin z = sin z;
(b) cosh z = cosh z.

Proof. (a) Using the definition of sin z, we have

eiz − e−iz e−iz − eiz

 −iz
e − eiz e − e−iz
  iz 
sin z = = = =
2i −2i −2i 2i
= sin z

(b) Recall that for z = x + iy, we have that

ez = ex−iy = ex e−iy = ex eiy = ez .

Thus
ez − e−z ez − e−z
 
cosh z = = = cosh z
2 2

2.1 Basic notions 31

18. Find all solutions z ∈ C of of the following (express your answers in the form x + iy):
(a) log z = 4i;
(b) zi = i.

Solution. (a) We have that exp (log z) = z. Thus

z = exp (log z) = exp (4i) = cos 4 + i sin 4

Notice that, if z = exp (4i), then we have

log (exp (4i)) = 4i + 2nπi (n ∈ Z)

In particular, for n = 0, we have that log [exp (4i)] = 4i.

(b) Method one: We know that zi = exp (i log z). Thus

exp (i log z) = i

Since i = exp (i (π/2 + 2nπ)), with n ∈ Z, then we have

 π 
exp i + 2nπ = exp [i(ln |z| + i arg(z))]
2
= exp [− arg(z) + i ln |z|]
= exp [− arg(z)] · exp [i ln |z|]

Thus
π
arg(z) = 2kπ (k ∈ Z) and ln |z| = + 2nπ (n ∈ Z)
2
Hence, hπ i
z = exp + 2nπ (n ∈ Z).
2
Method two: Consider the following identity

log [exp (i log z)] = log i (2.6)

Thus

log [exp (i log z)] = i log z + 2n1 πi (n1 ∈ Z)

Hence, substituting in (2.6), we obtain

i log z + 2n1 πi = log i (n1 ∈ Z)

Thus
log z = −i log i − 2n1 π (n1 ∈ Z)
From the polar form of i, we have that r = 1 and Θ = π2 . Thus
π 
log i = ln 1 + i + 2n2 π (n2 ∈ Z)
2
So π
h i π
log z = −i ln 1 + i + 2n2 π − 2n1 π = + 2π(n2 − n1 )
2 2
32 Chapter 2. Functions

with n1 , n2 ∈ Z. Since n2 − n1 ∈ Z, we have

π
log z = + 2nπ (n ∈ Z)
2
Therefore, zi = i when
hπ i
z = exp + 2nπ (n ∈ Z)
2


19. Show that  

−1 1 1+z
tanh z = log . (2.7)
2 1−z

Solution. If w = tanh−1 z, then

sinh w ew − e−w e2w − 1
z = tanh w = = w =
cosh w e + e−w e2w + 1
Thus

z(e2w + 1) = e2w − 1
z+1
e2w =
1 −z 
z+1
2w log e = log
1−z
 
1 z+1
w = log
2 1−z
Hence  
−1 1 z+1
tanh z = log .
2 1−z


20. Find all solutions of the equation tanh z = i and express them in the form x + iy.

Solution. Method one:

Applying the inverse hyperbolic function in both sides we have

z = tanh−1 z = tanh−1 (i).

Using the formula (2.7), we have

 
−1 1 1+i
tanh (i) = log
2 1−i
1
= log (i)
2
2.1 Basic notions 33

Using the definition of log for i with r = 1 and Θ = π/2 we have that
π  π 
log (i) = ln 1 + i + 2nπ = i + 2nπ (n ∈ Z).
2 2
Therefore
1 h π i π 
z = tanh−1 (i) = i + 2nπ = i + nπ
2 2 4
for n ∈ Z.

Method two:
We know that
sinh z −i sin(iz)
tanh z = = = −i tan(iz).
cosh z cos(iz)
Thus,
tanh z = i ⇐⇒ −i tan(iz) = i ⇐⇒ tan(iz) = −1.
Applying the inverse trigonometric function in both sides we have

iz = tan−1 (iz) = tan−1 (−1). (2.8)

Now, using the formula

 
−1 i i+z
tan z = log
2 i−z
we obtain
 
−1 i i−1
tan (−1) = log
2 1+i
i
= log (i)
2
i h π i
= i + 2nπ (n ∈ Z)
2 2
π 
= − + nπ (n ∈ Z).
4
Thus, substituting in (2.8), we have
π 
iz = − + nπ (n ∈ Z).
4
Hence π 
z=i + nπ (n ∈ Z).
4

34 Chapter 2. Functions

21. Let Ω1 and Ω2 be nonempty, closed sets in C.

(a) Show that the set Ω1 ∪ Ω2 is closed.
(b) If instead Ω2 is nonempty and open:
(i) could Ω1 ∪ Ω2 still be closed?
(ii) Need it be closed?
Give proofs or examples/counterexamples.

Proof. (a) We need to prove that

∂ (Ω1 ∪ Ω2 ) ⊆ Ω1 ∪ Ω2 .

Let z ∈ ∂ (Ω1 ∪ Ω2 ). Thus, for every ε > 0, we have that

Bε (z) ∩ (Ω1 ∪ Ω2 ) 6= 0/ (2.9)

and

Bε (z) ∩ (Ω1 ∪ Ω2 )c = Bε (z) ∩ (Ωc1 ∩ Ωc2 ) 6= 0.

/ (2.10)

From expression (2.9),

Bε (z) ∩ Ω1 6= 0/ or Bε (z) ∩ Ω2 6= 0,
/ or both.

But, from expression (2.10), we have

Bε (z) ∩ Ωc1 6= 0/ and Bε (z) ∩ Ωc2 6= 0.

/

Hence
Bε (z) ∩ Ω1 6= 0/ and Bε (z) ∩ Ωc1 6= 0,
/
or
Bε (z) ∩ Ω2 6= 0/ and Bε (z) ∩ Ωc2 6= 0,
/
or both. This means that z ∈ ∂ Ω1 or z ∈ ∂ Ω2 (or both). That is

z ∈ ∂ Ω1 ∪ ∂ Ω2 .

Since Ω1 and Ω2 are closed, we have that z ∈ ∂ Ω1 ∪ ∂ Ω2 ⊆ Ω1 ∪ Ω2 . Therefore

Ω1 ∪ Ω2 is closed.
(b) (i) The answer is ’yes’. For example, Ω1 = B1 (0), Ω2 = B1 (0).
(ii) The answer is ’no’. For example, Ω1 = B1 (0), Ω2 = B2 (0). In this case, Ω1 ∪
Ω2 = Ω2 is open. 
 2.2 Limits, Continuity and Differentiation 35

2.2 Limits, Continuity and Differentiation

1. Compute the following limits if they exist:
iz3 + 1
(a) lim 2 ;
z→−i z + 1
4 + z2
(b) lim .
z→∞ (z − 1)2
Im(z)
(c) lim .
z→0 z
Solution. (a)

iz3 + 1 i(z3 + i3 )
lim = lim
z→−i z2 + 1 z→−i z2 + 1
i(z + i)(z2 − iz + i2 )
= lim
z→−i (z + i)(z − i)
i(z − iz + i2 )
2
= lim
z→−i z−i
limz→−i (z2 − iz + i2 )
=i
limz→−i (z − i)
limz→−i z2 − i limz→−i z + limz→−i i2
=i
limz→−i z − limz→−i i
i((−i) − i(−i) + i2 ) 3
2
= =
−i − i 2
(b)

4 + z2 4 + z−2
lim = lim
z→∞ (z − 1)2 z→0 (z−1 − 1)2
4z2 + 1 limz→0 (4z2 + 1)
= lim =
z→0 (1 − z)2 limz→0 (1 − z)2
4(limz→0 z)2 + limz→0 1
= =1
(limz→0 1 − limz→0 z)2

(c) Since

Im(z) y
lim = lim = −i
Re(z)=0,z→0 z y→0 yi

and
Im(z) 0
lim = lim = 0
Im(z)=0,z→0 z x→0 x

the limit does not exist.

36 Chapter 2. Functions

2. Show the following limits:

4z5
(a) lim 5 = 4;
z→∞ z − 42z
z4
(b) lim 2 = ∞;
z→∞ z + 42z
(az + b)3 a3
(c) lim = , if c 6= 0.
z→∞ (cz + d)3 c3
Solution. (a)
1 5


4z5 4 z 4
lim 5 = lim = lim = 4.
z→∞ z − 42z 1 5 1 z→0 1 − 42z4
z→0



z − 42 z

(b) Notice that

−1
z4 1/z4

lim ⇐⇒ lim
z→∞ z2 + 42z z→0 1/z2 + 42/z
 −1
1
⇐⇒ lim
z→0 z2 + 42z3

lim z2 + 42z3 = 0.


⇐⇒
z→0

z4
Therefore, lim = ∞.
z→∞ z2 + 42z
(c)
(az + b)3 (a/z + b)3 (a + bz)3 a3
lim = lim = lim = 3.
z→∞ (cz + d)3 z→0 (c/z + d)3 z→0 (c + dz)3 c


3. Show that lim z/z does not exist.

z→0
Hint: Consider what happens to the function at points of the form x + 0i for x → 0,
x 6= 0, and then at points of the form 0 + yi for y → 0, y 6= 0.

Proof. For z = x + 0i, x 6= 0,

z x
= =1→1 as x → 0.
z x
On the other hand, for z = 0 + yi, y 6= 0,
z yi
= = −1 → −1 as y → 0.
z −yi

However, lim z/z must be independent of direction of approach. Hence limit does
z→0
not exist. 
2.2 Limits, Continuity and Differentiation 37

4. Show that if f (z) is continuous at z0 , so is | f (z)|.

Proof. Let f (z) = u(x, y) + iv(x, y). Since f (z) is continuous at z0 = x0 + y0 i, u(x, y)
and v(x, y) are continuous at (x0 , y0 ). Therefore,

(u(x, y))2 + (v(x, y))2

is continuous at (x0 , y0 ) since the sums and products of continuous functions are
continuous. It follows that
q
| f (z)| = (u(x, y))2 + (v(x, y))2

is continuous at z0 since the compositions of continuous functions are continuous.



5. Let
(
z3 /z2 if z 6= 0
f (z) =
0 if z = 0

Show that
(a) f (z) is continuous everywhere on C;
(b) the complex derivative f 0 (0) does not exist.

Proof. Since both z3 and z2 are continuous on C∗ = C\{0} and z2 6= 0, f (z) = z3 /z2
is continuous on C∗ .
At z = 0, we have

z3
lim | f (z)| = lim = lim |z| = 0
z→0 z→0 z2 z→0

and hence limz→0 f (z) = 0 = f (0). So f is also continuous at 0 and hence continuous
everywhere on C.
The complex derivative f 0 (0), if exists, is given by

f (z) − f (0) z3
lim = lim 3 .
z→0 z−0 z→0 z

Let z = x + yi. If y = 0 and x → 0, then

z3 x3
lim = lim = 1.
z=x→0 z3 x→0 x3

On the other hand, if x = 0 and y → 0, then

z3 (−yi)3
lim = lim = −1.
z=yi→0 z3 x→0 (yi)3

So the limit limz→0 z3 /z3 and hence f 0 (0) do not exist. 

38 Chapter 2. Functions

6. Show that f (z) in (2) is actually nowhere differentiable, i.e., the complex derivative
f 0 (z) does not exist for any z ∈ C.

Proof. It suffices to show that C-R equations fail at every z 6= 0:

∂ z3
   
∂ ∂ ∂
+i f (z) = +i
∂x ∂y ∂x ∂ y z2
 3
∂ z3
 
∂ z
= +i
∂ x z2 ∂ y z2
 2
2z3 3iz2 2iz3
  
3z
= − 3 +i − 2 − 3
z2 z z z
6z 2
= 2 6= 0
z
for z 6= 0. 

7. Find f 0 (z) when

(a) f (z) = z2 − 4z + 2;
(b) f (z) = (1 − z2 )4 ;
z+1 1
(c) f (z) = (z 6= − );
2z + 1 2
(d) f (z) = e1/z (z 6= 0).
Answer. (a) 2z − 4; (b) −8(1 − z2 )3 z; (c) −1/(2z + 1)2 ; (d) −e1/z /z2 .
8. Prove the following version of complex L’Hospital: Let f (z) and g(z) be two
complex functions defined on |z − z0 | < r for some r > 0. Suppose that f (z0 ) =
g(z0 ) = 0, f (z) and g(z) are differentiable at z0 and g0 (z0 ) 6= 0. Then
f (z) f 0 (z0 )
lim = 0
z→z0 g(z) g (z0 )
[Refer to: problems 1c and 6 in section 3.1; and problem 12 in section 3.2]
Proof. Since f (z) and g(z) are differentiable at z0 , we have
f (z) − f (z0 )
lim = f 0 (z0 )
z→z0 z − z0
and
g(z) − g(z0 )
lim = g0 (z0 ).
z→z0 z − z0
And since g0 (z0 ) 6= 0,
f (z) − f (z0 ) limz→z0 ( f (z) − f (z0 ))/(z − z0 ) f 0 (z0 )
lim = = 0 .
z→z0 g(z) − g(z0 ) limz→z0 (g(z) − g(z0 ))/(z − z0 ) g (z0 )
Finally, since f (z0 ) = g(z0 ) = 0,
f (z) f 0 (z0 )
lim = 0 .
z→z0 g(z) g (z0 )

 2.3 Analytic functions 39

9. Show that if f (z) satisfies the Cauchy-Riemann equations at z0 , so does ( f (z))n for
every positive integer n.

Proof. Since f (z) satisfies the Cauchy-Riemann equations at z0 ,

 
∂ ∂
+i f (z) = 0
∂x ∂y
at z0 . Therefore,
 
∂ ∂ ∂ ∂
+i ( f (z))n = ( f (z))n + i ( f (z))n
∂x ∂y ∂x ∂y
∂ ∂
= ( f (z))n−1 f (z) + i( f (z))n−1 f (z)
∂x ∂y
 
∂ ∂
= ( f (z))n−1 +i f (z) = 0
∂x ∂y
at z0 . 

2.3 Analytic functions

1. Explain why the function f (z) = 2z2 − 3 − zez + e−z is entire.

Proof. Since every polynomial is entire, 2z2 − 3 is entire; since both −z and ez are
entire, their product −zez is entire; since ez and −z are entire, their composition e−z
is entire. Finally, f (z) is the sum of 2z3 − 3, −zez and e−z and hence entire. 

2. Let f (z) be an analytic function on a connected open set D. If there are two constants
c1 and c2 ∈ C, not all zero, such that c1 f (z) + c2 f (z) = 0 for all z ∈ D, then f (z) is
a constant on D.

Proof. If c2 = 0, c1 6= 0 since c1 and c2 cannot be both zero. Then we have c1 f (z) =

0 and hence f (z) = 0 for all z ∈ D.
If c2 6= 0, f (z) = −(c1 /c2 ) f (z). And since f (z) is analytic in D, f (z) is anlaytic in
D. So both f (z) and f (z) are analytic in D. Therefore, both f (z) and f (z) satisfy
Cauchy-Riemann equations in D. Hence
 
∂ ∂
+i (u + vi) = 0
∂x ∂y
and
 
∂ ∂
+i (u − vi) = 0
∂x ∂y
in D, where f (z) = u(x, y) + iv(x, y) with u = Re( f ) and v = Im( f ). It follows that
   
∂ ∂ ∂ ∂
+i u= +i v=0
∂x ∂y ∂x ∂y
and hence ux = uy = vx = vy = 0 in D. Therefore, u and v are constants on D and
hence f (z) ≡ const. 
40 Chapter 2. Functions

3. Show that the function sin(z) is nowhere analytic on C.

Proof. Since
 
∂ ∂ ∂ ∂
+i sin(z) = sin(z) + i sin(z)
∂x ∂y ∂x ∂y
∂z ∂z
= cos(z) + i cos(z)
∂x ∂y
= cos(z) + i cos(z)(−i) = 2 cos(z)

sin(z) is not differentiable and hence not analytic at every point z satisfying cos(z) 6=
0. At every point z0 satisfying cos(z0 ) = 0, i.e., z0 = nπ + π/2, sin(z) is not differ-
entiable in |z − z0 | < r for all r > 0. Hence sin(z) is not analytic at z0 = nπ + π/2
either. In conclusion, sin(z) is nowhere analytic. 

4. Let f (z) = u(x, y) + iv(x, y) be an entire function satisfying u(x, y) ≤ x for all z =
x + yi. Show that f (z) is a polynomial of degree at most one.

Proof. Let g(z) = exp( f (z) − z). Then |g(z)| = exp(u(x, y) − x). Since u(x, y) ≤ x,
|g(z)| ≤ 1 for all z. And since g(z) is entire, g(z) must be constant by Louville’s
theorem. Therefore, g0 (z) ≡ 0. That is, ( f 0 (z) − 1) exp( f (z) − z) ≡ 0 and hence
f 0 (z) = 1 for all z. So f (z) ≡ z + c for some constant c. 

5. Show that
3 −3xy2
| exp(z3 + i) + exp(−iz2 )| ≤ ex + e2xy

where x = Re(z) and y = Im(z).

Proof. Note that |ez | = eRe(z) . Therefore,

| exp(z3 + i) + exp(−iz2 )| ≤ | exp(z3 + i)| + | exp(−iz2 )|

= exp(Re(z3 + i)) + exp(Re(−iz2 ))
= exp(Re((x3 − 3xy2 ) + (3x2 y − y3 + 1)i))
+ exp(Re(2xy − (x2 − y2 )i))
3 −3xy2
= ex + e2xy .

6. Let f (z) = u(x, y) + iv(x, y) be an entire function satisfying

v(x, y) ≥ x

for all z = x + yi, where u(x, y) = Re( f (z)) and v(x, y) = Im( f (z)). Show that f (z)
is a polynomial of degree 1.
2.3 Analytic functions 41

Proof. Let g(z) = exp(i f (z) + z). Then

|g(z)| = | exp((−v(x, y) + iu(x, y)) + (x + iy))| = exp(x − v(x, y)).

Since v(x, y) ≤ x, x − v(x, y) ≤ 0 and |g(z)| ≤ 1 for all z. And since g(z) is entire,
g(z) must be constant by Louville’s theorem. Therefore, g0 (z) ≡ 0. That is, (i f 0 (z) +
1) exp(i f (z) + z) ≡ 0 and hence f 0 (z) = i for all z. So f (z) ≡ iz + c for some constant
c. 

7. Show that the entire function cosh(z) takes every value in C infinitely many times.

Proof. For every w0 ∈ C, the quadratic equation y2 − 2w0 y + 1 = 0 has a complex

root y0 . We cannot have y0 = 0 since 02 − 2w0 · 0 + 1 6= 0. Therefore, y0 6= 0 and
there is z0 ∈ C such that ez0 = y0 . Then

ez0 + e−z0 y2 + 1 2w0 y0

cosh(z0 ) = = 0 = = w0 .
2 2y0 2y0

And since cosh(z + 2πi) = cosh(z), cosh(z0 + 2nπi) = w0 for all integers n. There-
fore, cosh(z) takes every value w0 infinitely many times. 

8. Determine which of the following functions f (z) are entire and which are not? You
must justify your answer. Also find the complex derivative f 0 (z) of f (z) if f (z) is
entire. Here z = x + yi with x = Re(z) and y = Im(z).
1
(a) f (z) =
1 + |z|2
Solution. Since u(x, y) = Re( f (z)) = (1 + x2 + y2 )−1 and v(x, y) = 0, ux =
2x(1 + x2 + y2 )−2 6= 0 = vy . Hence the Cauchy-Riemann equations fail for f (z)
and f (z) is not entire.
z
(b) f (z) = 2(3 ) (here 2z and 3z are taken to be the principle values of 2z and 3z ,
respectively, by convention)
z
Solution. Let g(z) = 2z and h(z) = 23 . Since both g(z) and h(z) are entire,
f (z) = g(h(z)) is entire and
z
f 0 (z) = g0 (h(z))h0 (z) = 23 3z (ln 2)(ln 3)

by chain rule.
(c) f (z) = (x2 − y2 ) − 2xyi
Solution. Since
 
∂ ∂
+i f = (2x − 2yi) + i(−2y − 2xi) = 4x − 4yi 6= 0,
∂x ∂y

the Cauchy-Riemann equations fail for f (z) and hence f (z) is not entire.
42 Chapter 2. Functions

(d) f (z) = (x2 − y2 ) + 2xyi

Solution. Since
 
∂ ∂
+i f = (2x − 2yi) + i(2y + 2xi) = 0,
∂x ∂y

the Cauchy-Riemann equations hold for f (z) everywhere. And since fx and fy
are continous, f (z) is analytic on C. And f 0 (z) = fx = 2x + 2yi = 2z.

9. Let CR denote the upper half of the circle |z| = R for some R > 1. Show that

eiz 1
2
≤
z +z+1 (R − 1)2

for all z lying on CR .

Proof. For z ∈ CR , |z| = R and Im(z) ≥ 0. Let z = x + yi. Since y = Im(z) ≥ 0,

|eiz | = |ei(x+yi) | = |e−y+xi | = e−y ≤ 1

for z ∈ CR . And since

√ ! √ !
2 −1 + 3i −1 − 3i
|z + z + 1| = z − z−
2 2
√ √
−1 + 3i −1 − 3i
= z− z−
2 2
√ ! √ !
−1 + 3i −1 − 3i
≥ |z| − |z| −
2 2
= (R − 1)(R − 1) = (R − 1)2 ,

we obtain

eiz 1
≤
z2 + z + 1 (R − 1)2

for z ∈ CR . 

10. Let
(
z2 /|z| if z 6= 0
f (z) =
0 if z = 0

Show that f (z) is continuous everywhere but nowhere analytic on C.

2.3 Analytic functions 43

Proof. Since both z and |z| are continuous on C, z2 /|z| is continuous on C∗ . There-
fore, f (z) is continuous on C∗ . To see that it is continuous at 0, we just have to show
that
z2
lim f (z) = lim = f (0) = 0.
z→0 z→0 |z|

This follows from

z2 |z|2
lim = lim = lim |z| = 0.
z→0 |z| z→0 |z| z→0

Therefore, f (z) is continuous everywhere on C.

To show that f (z) is nowhere analytic, it suffices to show that the Cauchy-Riemann
equations fail for f (z) on C∗ . This follows from
  2 
∂ ∂ z
+i
∂x ∂y |z|

2z xz 2  
2iz yz2

= − +i − −
|z| |z|3 |z| |z|3
(4z − x − iy)z2 3z
= = 6= 0
|z|3 |z|

for z 6= 0. Consequently, f (z) is nowhere analytic. 

11. Find where

i i+z
tan−1 (z) = Log
2 i−z
is analytic?
Solution. The branch locus of tan−1 (z) is
   
i+z w−1
z: = w ∈ (−∞, 0] = z : z = i , w ∈ (−∞, 0] .
i−z w+1

For w ∈ (−∞, 0],

w−1 2
= 1− ∈ (−∞, −1] ∪ (1, ∞)
w+1 w+1

so tan−1 (z) is analytic in

C\ {z : Re(z) = 0, Im(z) ∈ (−∞, −1] ∪ [1, ∞)} .

44 Chapter 2. Functions

12. Show that the following functions are defined on all of C, but are nowhere analytic
(here z = x + iy):
(a) z 7→ 2xy + i(x2 + y2 );
(b) z 7→ ey eix .

Proof. (a) Notice that the real and imaginary components of the function

f (z) = f (x + iy) = 2xy + i(x2 + y2 )

are well defined. Hence the function f (z) is defined on all C. If f (z) were analytic
then we will have

ux = vy ⇒ 2y = 2y (2.11)

and

uy = −vx ⇒ 2x = −2x (2.12)

Equation (2.11) is true for all z ∈ C. However, equation (2.12) is true only on
the imaginary axis. Hence, there is no point in C for which the functions f (z)
is differentiable on a neighbourhood (since the Cauchy Riemann equations are
necessary for differentiability) and therefore, f (z) is nowhere analytic.
(b) Note that the real and imaginary components of the function

f (z) = f (x + iy) = ey eix = ey cos x + iey sin x

are well defined. Hence the function f (z) is defined on all C. If f (z) were analytic
then we will have

ux = vy ⇒ −ey sin x = ey sin x ⇒ 2ey sin x = 0 ⇒ sin x = 0

and
uy = −vx ⇒ ey cos x = −ey cos x ⇒ 2ey cos x = 0 ⇒ cos x = 0.
On the one hand, we have that the roots of sin x are nπ (n = 0, ±1, ±2, . . .), but
cos(nπ) = (−1)n 6= 0. On the other hand, the roots of cos x are (2n − 1)π/2 but
sin((2n − 1)π/2) = − cos(nπ) = −(−1)n 6= 0. Consequently, the Cauchy-Riemann
equations are not satisfied anywhere. 

13. Show where the function z 7→ x3 + i(1 − y)3 is:

(a) analytic;
(b) differentiable (here z = x + iy).

Solution. If the function

f (z) = f (z + iy) = x3 + i(1 − y)3

were analytic then we will have

ux = vy ⇒ 3x2 = −3(1 − y)2 ,

 2.3 Analytic functions 45

which is only satisfied at x = 0, y = 1; and, on the other hand,

uy = −vx ⇒ 0 = 0,

which holds everywhere. Note also that the componentes of f (z), and all its first-
order partials exist everywhere.
Since the Cauchy-Riemann equations only hold at z = i, the function f (z) is only dif-
ferentiable at z = i. Hence, in particular, it is not differentiable on any neighbourhood
of any point, and therefore is nowhere analytic. 

14. Verify that the following functions are analytic on their domain of definition, and
state the derivative, (here z = x + iy = eiθ ):
(a) z 7→ ln r + iθ , domain {z : r > 0, 0 < θ < 2π};
4z + 1
(b) z 7→ 3 , domain C \ {0, 1, −1}.
z −z
Solution. (a) We have ur = 1/r, vr = 0, uθ = 0, vθ = 1. Hence u, ur , uθ , v, vr , vθ
are defined and continuous on the domain, with the Cauchy-Riemann equations
being satisfied. That is

1
rur = vθ ⇔ r· = 1, and uθ = −vr ⇔ 0 = 0.
r
The derivative is then
 
0 −iθ −iθ 1 1 1
f (z) = e (ur + ivr ) = e = iθ = ,
r re z

since z = reiθ . This is defined on the whole domain, Therefore the function f (z) is
analytic there.
(b) The functions is a rational function and hence exists and is differentiable as long
as the denominator does not vanish, which is true on the whole domain. Hence f is
analytic on the whole domain, with

(z3 − z) · 4 − (4z + 1)(3z2 − 1) −(8z3 + 3z + 1)

f 0 (z) = = .
(z3 − z)2 (z3 − z)2

2.3.1 Harmonic functions

1. Verify that the following functions u are harmonic, and in each case give a conjugate
harmonic function v (i.e. v such that u + iv is analytic).
(a) u(x, y) = 3x2 y + 2x2 − y3 − 2y2 ,
(b) y(x, y) = ln(x2 + y2 ).

Solution. (a) If u(x, y) = 3x2 y + 2x2 − y3 − 2y2 , then

ux = 6xy + 4x, uy = 3x2 − 3x2 − 4y

uxx = 6y + 4, uyy = −6y − 4
46 Chapter 2. Functions

Thus
∆u = uxx + uyy = 6y + 4 + (−6y − 4) = 0.
Hence, u is harmonic.
The harmonic conjugate of u will satisfy the Cauchy-Riemann equations and have
continuos partials of all orders. By Cauchy-Riemann equations

ux = vy , vx = −uy ,

we have that vy = 6xy + 4x. Thus

Z
v= (6xy + 4x)dy = 3xy2 + 4xy + g(x).

Thus
vx = 3y2 + 4y + g0 (x)
Since vx = −uy ,

3y2 + 4y + g0 (x) = −3x2 + 3y2 + 4y

g0 (x) = −3x2
g(x) = −x3
Therefore, the harmonic conjugate is

v(x, y) = 3xy2 + 4xy − x3 .

(b) If u(x, y) = ln(x2 − y2 ), then

2x 2y
ux = , uy =
x2 + y2 x2 + y2
2(y2 − x2 ) 2(x2 − y2 )
uxx = , uyy =
(x2 + y2 )2 (x2 + y2 )2
Thus
2(y2 − x2 ) 2(x2 − y2 ) 2
∆u = uxx + uyy = 2 2
+ 2 2
x + y2 = 0.
x +y x +y
Hence, u is harmonic.
Similarly to the previous part, by Cauchy-Riemann equations

ux = vy , vx = −uy ,
2x
we have that vy = . Thus
x2 + y2
2x y
Z
v= dy = 2 arctan + g(x)
x2 + y2 x
So we have
−2y
vx = + g0 (x)
x2 + y2
2.3 Analytic functions 47

Since vx = −uy ,
−2y −2y
+ g0 (x) =
x2 + y2 x2 + y2
g0 (x) = 0
g(x) = c c ∈ R

Hence the harmonic conjugate is

y
v(x, y) = 2 arctan
x
Notice that u is defined on C \ {0} and v is not defined if x = 0. 

2. Suppose that U solves a Neumann problem for Laplace’s equation on a domain Ω.

Show that U + c also solves this problem for any c ∈ R. Does the same result hold
for the corresponding Dirichlet problem?

Proof. Let n be the external unit normal to Ω. If U solves a Neumann problem then
U satisfies the following conditions

∆U = 0, in Ω;
dU
 = g, on ∂ Ω
dn
Considering U + c, we have

∆(U + c) = ∆U = 0.

So U + c satisfies Laplace’s equation in ∂ Ω. Furthermore,

d dU
(U + c) = ∇(U + c) · n = ∇U · n = =g
dn dn
Then U + c also satisfies the boundary condition on Ω. Therefore, U + c solves the
Neumann problem.
On the other hand, if U solves a Dirichlet problem, it satisfies the conditions
(
∆U = 0, in Ω;
U = α, on ∂ Ω

where α ∈ R. Certainly, U + c satisfies Laplace’s equation for any c ∈ R. However,

U + c satisfies the second condition only when c = 0. Hence, U + c does not solve a
Dirichlet problem. 

3. Let Λ be the domain {w| Im w < π}. Denote the two components of the boundary
of Λ by Γ1 = {w| Im w = 0} and Γ2 = {w| Im w = π}. Let C be an arbitrary real
constant.
48 Chapter 2. Functions

(a) Verify that the function

w 
T = Im +C cosh w
π
is harmonic on Λ, and satisfies the Dirichlet boundary conditions

T = 0, T = 1.
Γ1 Γ2

(b) For what values (if any) of C is T a bounded function on Λ, i.e. for what values
of C does there exist an M > 0 such that |T (w)| ≤ M for all w ∈ Λ?

Solution. (a) Let w = u + iv, then we have

w  v
T = Im +C cosh w = +C sinh u sin v.
π π
Then
1
Tu = C cosh u sin v, Tv = +C sinh u cos v
π
Tuu = C sinh u sin v, Tvv = −C sinh u sin v
Thus
∆T = Tuu + Tvv = C sinh u sin v + (−C sinh u sin v) = 0.
Hence, T is harmonic on Λ.
Notice that on Γ1 , we have that v = 0 and so T = 0. Finally, on Γ2 we have that
π
v = π, then T = + 0 = 1.
π
v
(b) If C = 0, the solution is and |T | < 1 on Λ. If C 6= 0, considering the point
π
π
w0 = µ + i, we have
2
1
T (w0 ) = +C sinh µ
2
1
1 +Ceµ +Ce−µ


=
2
For a function µ sufficiently large

|C| µ
|T (w0 )| > e ,
2
which tends to ∞ as µ → ∞. Therefore, T is unbounded on Λ. 
 3. Complex Integrals

3.1 Contour integrals

1. Evaluate the following integrals:
Z 2
2
(a) t 2 + i dt;
Z1π/4
(b) e−2it dt;
Z0∞
(c) tezt dt when Re(z) < 0.
0

Solution.
(a)

Z 2
2 Z 2
t 2 + i dt = (t 4 + 2it 2 + i2 )dt
1 1
2
t5 2it 3 26 14
= + −t = + i
5 3 1 5 3

(b)

π/4
e−2it
Z π/4
−2it
e dt = −
0 2i 0
1+i 1 i
= = −
2i 2 2
50 Chapter 3. Complex Integrals

(c)
1 ∞
Z ∞ Z
tezt dt = td(ezt )
0 z 0 
1
Z ∞
zt ∞ zt
= te 0 − e dt
z 0
 
1 zt 1 zt
= lim te − lim e − 1
z t→∞ z t→∞
1
= 2
z
where
lim tezt = lim ezt = 0
t→∞ t→∞

because
t 1
lim |tezt | = lim tet Re(z) = lim = − lim =0
t→∞ t→∞ t→∞ e−xt t→∞ xe−xt

by L’Hospital (see Problem 8 in section 2.2), and

1
lim |ezt | = lim et Re(z) = lim =0
t→∞ t→∞ t→∞ e−xt

as x = Re(z) < 0.
R
2. Find the contour integral γ zdz for
(a) γ is the triangle ABC oriented counterclockwise, where A = 0, B = 1 + i and
C = −2;
(b) γ is the circle |z − i| = 2 oriented counterclockwise.

Solution. (a)
Z Z Z Z
zdz = zdz + zdz + zdz
γ AB BC CA
Z 1
= t(1 + i)d(t(1 + i))
0
Z 1
+ (1 − t)(1 + i) − 2td((1 − t)(1 + i) − 2t)
0
Z 1
+ −2(1 − t)d(−2(1 − t))
0
Z 1 Z 1 Z 1
= 2tdt + ((2i − 4) + 10t)dt + 4(t − 1)dt
0 0 0
= 1 + (2i − 4) + 5 − 2 = 2i
(b)
Z 2π Z 2π
i + 2eit d(i + 2eit ) = 2i(−i + 2e−it )eit dt = 8πi.
0 0

3.1 Contour integrals 51

3. Compute the following contour integral

Z
zdz,
L

where L is the boundary of the triangle ABC with A = 0, B = 1 and C = i, oriented

counter-clockwise.

Solution.
Z Z Z Z
zdz = zdz + zdz + zdz
L AB BC CA
Z 1 Z 1
= tdt + (1 − t) + tid((1 − t) + ti)
0 0
Z 1
+ (1 − t)id((1 − t)i)
0
Z 1 Z 1 Z 1
= tdt + (−1 + i) ((1 − t) − ti)dt − (1 − t)dt = i
0 0 0

4. Evaluate the contour integral

Z
f (z)dz
C

using the parametric representations for C, where

z2 − 1
f (z) =
z
and the curve C is
(a) the semicircle z = 2eiθ (0 ≤ θ ≤ π);
(b) the semicircle z = 2eiθ (π ≤ θ ≤ 2π);
(c) the circle z = 2eiθ (0 ≤ θ ≤ 2π).
Solution.
(a)
Z π 2iθ
4e − 1
Z
π
f (z)dz = d(2eiθ ) = (2e2iθ − i) 0
= −πi
C 0 2eiθ

(b)
Z 2π 2iθ
4e − 1
Z
2π
f (z)dz = d(2eiθ ) = (2e2iθ − i) = −πi
C π 2eiθ π

(c) Adding (a) and (b), we have −2πi.

52 Chapter 3. Complex Integrals

5. Redo previous Problem 4 using an antiderivative of f (z).

Solution. For (a),
−2
!
z2
Z
f (z)dz = − lim Log(z) − Log(2)
C 2 2
z→−2
Im(z)>0

= −(ln 2 + πi − ln 2) = −πi.

For (b),
2
!
z2
Z
f (z)dz = − Log(2) − lim Log(z)
C 2 −2
z→−2
Im(z)<0

= −(ln 2 − (ln 2 − πi)) = −πi.

6. Let CR be the circle |z| = R (R > 1) oriented counterclockwise. Show that

Log(z2 )
 
π + ln R
Z
dz < 4π
CR z2 R
and then
Log(z2 )
Z
lim dz = 0.
R→∞ CR z2

Proof. Using expression (1.1) in Problem 5, we have

| Log(z2 )| ≤ ln |z2 | + π = 2 ln R + π

for |z| = R > 1. Therefore,

Log(z2 )
 
π + 2 ln R
Z
dz ≤ 2πR
CR z2 R2
   
π/2 + ln R π + ln R
= 4π < 4π .
R R
And since
 
π + ln R 1
lim 4π = 4π lim =0
R→∞ R R→∞ R

by L’Hospital (see Problem 8 in section 2.2),

Log(z2 )
Z
lim dz = 0.
R→∞ CR z2

3.1 Contour integrals 53

7. Without evaluating the integral, show that

dz 9π
Z
≤
C z2 + z + 1 16
where C is the arc of the circle |z| = 3 from z = 3 to z = 3i lying in the first quadrant.
Proof. Since
|z2 + z + 1| ≥ |z2 | − |z| − 1 = |z|2 − |z| − 1 = 5
for |z| = 3,
1 1
≤ .
z2 + z + 1 5
Therefore,
 
dz 6π 1 3π 9π
Z
2
≤ = < .
C z +z+1 4 5 10 16

R
8. Evaluate the integral C Re(z)dz for the following contours C from −4 to 4:
(a) The line segments from −4 to −4 − 4i to 4 − 4i to 4;
(b) the lower half of the circle with radius 4, centre 0;
(c) the upper half of the circle with radius 4, centre 0.
(d) What conclusions (if any) can you draw about the function f (z) = Re(z) from
this?

Solution. (a) Notice that the contour C consists of three contours:

i. C1 defined by z(t) = −4 − 4it, with 0 ≤ t ≤ 1, followed by
ii. C2 defined by z(t) = −4(1 − 2t) − 4i, with 0 ≤ t ≤ 1, and finally
iii. C3 defined by z(t) = 4 − 4i(1 − t), with 0 ≤ t ≤ 1
Thus Z Z Z Z Z
f (z)dz = f (z)dz = f (z)dz + f (z)dz + f (z)dz (3.1)
C C1 +C2 +C3 C1 C2 C3
Rb
f (z(t))z0 (t)dt.
R
where f (z) = Re(z). Recall also that C f (z)dz = a
Now, for C1 we have that z0 (t) = −4i. Then
Z Z 1 Z 1 1
f (z)dz = (−4)(−4i)dt = 16i dt = 16it = 16i.
C1 0 0 0

For C2 we have that z0 (t) = 8, then

Z Z 1
f (z)dz = [−4(1 − 2t)] (8)dt
C2 0
Z 1
= (64t − 32)dt
0
Z 1 Z 1
= 64 tdt − 32 dt
0 0
t2 1 1
= 64 − 32t
2 0 0
= 32 − 32 = 0
54 Chapter 3. Complex Integrals

Finally for C3 we have that z0 (t) = 4i, then

Z Z 1 Z 1 1
f (z)dz = (4)(4i)dt = 16i dt = 16it = 16i.
C3 0 0 0
Therefore, using expression 3.1, we obtain
Z
Re(z)dz = 32i
C

(b) In this case, the contour C is defined by

z(t) = 4eit = 4 cost + 4i sint,
with π ≤ t ≤ 2π. Here we have z0 (t) = 4ieit .
Thus
Z Z 2π
Re(z)dz = (4 cost)(4ieit )dt
C π
Z 2π
= 16i costeit dt
π
Z 2π Z 2π
= 16i cos2 tdt − 16 cost sintdt
π π
2π
sin2 t
Z 2π
= 8i [1 + cos(2t)]dt − 16 ·
π 2
π
  2π
sin(2t)
= 8i t + − 0 = 8π i
2
π
(c) Finally, in this case, the contour C is defined by
z(t) = −4e−it = −4 cost + 4i sint,
with 0 ≤ t ≤ π. Here we have z0 (t) = 4ie−it .
Thus
Z Z π
Re(z)dz = (−4 cost)(4ie−it )dt
C 0
Z π
= −16i coste−it dt
Z0π Z π
2
= −16i cos tdt + 16 cost sintdt
0 0
π
sin2 t
Z π
= −8i [1 + cos(2t)]dt + 16 ·
0 2
0
  π
sin(2t)
= −8i t + + 0 = −8π i
2
0
(d) We have seen that the integral along each contour has a different value.
The reason is that the function f (z) = Re(z) is not analytic on any domain containing
any of the contours discussed in parts (a), (b) and (c). In fact, this function is nowhere
analytic.

 3.2 Cauchy Integral Theorem and Cauchy Integral Formula 55

3.2 Cauchy Integral Theorem and Cauchy Integral Formula

1. Evaluate the following integrals, justifying your procedures. For c) and d) you
shouldZ also state why the integral is well defined (i.e., independent of the path taken).
2dz
(a) 2 −1
, where C is the circle with radius 1/2, centre 1, positively oriented;
C z
 
1
Z
z
(b) e + dz, where C is the lower half of the circle with radius 1, centre 0,
C z
negatively
Z oriented;
2
(c) zez dz;
ZC
(d) cosh zdz.
C

Solution. (a) Notice that

2 1 1
= − .
z2 − 1 z−1 z+1
Thus
2 1 1
Z Z Z
2
dz = dz − dz.
C z −1 C z−1 C z+1
On the one hand,
1
Z
dz = 2πi
C z−1
1 R f (z)
by Cauchy integral formula, f (z0 ) = 2πi C z−z0 dz.

On the other hand, since 1/(z + 1) is analytic on and inside C, then

1
Z
dz = 0
C z+1

by Cauchy’s Theorem. Therefore,

2
Z
dz = 2πi.
C z2 − 1

(b) Notice that the integrand f (z) = ez − 1/z is analytic on C. The function

F(z) = ez − Log z

serves as an antiderivative of f (z). Here Log z is a branch of the logarithm chosen

with the branch cut on the positive imaginary axis. That is,
 
π 5π
Log z = ln r + iθ , r > 0, < θ < .
2 2

Thus
Z   −1
1
z 1 1
e − dz = (ez − Log z) = − πi − e + 2πi = − e + πi.
C z e e
1
56 Chapter 3. Complex Integrals
2
(c) Since the integrand f (z) = zez is analytic, the integral is path independent. An
antiderivative of f (z) is
2
ez
F(z) = .
2
Thus
2 i
ez 1 1
Z
z2
ze dz = = − .
C 2 2e 2
0

(d) Since the integrand f (z) = cosh z is analytic, the integral is path independent. An
antiderivative of f (z) is
F(z) = sinh z.
Thus
2πi
Z
cosh zdz = sinh z = sinh(2πi) − sinh(πi) = 0.
C πi

2. Let CR be the circle with radius R, centre 0, positively oriented. Show that

z2 + 4z + 7
Z
lim 2 2
dz = 0.
R→∞ CR (z + 4)(z + 2z + 2)

Use this fact to prove that

z2 + 4z + 7
Z
2 2
dz = 0.
C (z + 4)(z + 2z + 2)

where C is the circle with radius 5, centre 2, positively oriented.

Solution. Recall that for a contour C of length L and a piecewise continuous f (z) on
C, if M is a nonnegative constant such that | f (z)| < M for all points z on C at which
f (z) is defined, then
Z
f (z)dz < ML.
C

Now, consider the function

z2 + 4z + 7
f (z) = .
(z2 + 4)(z2 + 2z + 2)

For |z| large, we have that

1 + 4z + z72 1
| f (z)| = ≈
1 + z42 |z2 + 2z + 2| |z|2
3.2 Cauchy Integral Theorem and Cauchy Integral Formula 57

On the circle CR we have that |z| = R. Thus, for R large we have that
2
| f (z)| <
R2
Since the length of CR is 2πR, then
2 4π
Z
f (z)dz < ML = 2
2πR =
CR R R
which tends to 0 as R → ∞.
On the other hand, notice that he singularities

z1 = 2i, z2 = −2i, z3 = −1 + i, and z4 = −1 − i

of the function
z2 + 4z + 7
f (z) =
(z2 + 4)(z2 + 2z + 2)
are inside C and since f (z) is analytic on the annulus defined by C and CR with
R > 7, then Z Z
f (z)dz = f (z)dz.
C CR
Thus Z Z
lim f (z)dz = lim f (z)dz = 0.
R→∞ C R→∞ CR
Hence Z
f (z)dz = 0.
C

3. Evaluate
sin z
Z
dz,
C (z + 1)7
where C is the circle of radius 5, centre 0, positively oriented.

Solution. Recall the extension of the Cauchy integral formula:

n! f (z)
Z
(n)
f (z0 ) = n+1
dz.
2πi C (z − z0 )

Considering the function f (z) = sin z, which is analytic on C, we have

6! sin z 6! sin z
Z Z
(6)
f (−1) = = .
2πi C (z − (−1)) 6+1 2πi C (z + 1)7

Since f (6) (z) = − sin z, then

sin z sin z 2πi 2π sin(1)
Z Z
7
= 6+1
=− sin(−1) = i
C (z + 1) C (z − (−1)) 6! 6!

58 Chapter 3. Complex Integrals

4. Let C be the boundary of the triangle with vertices at the points 0, 3i and −4 oriented
counterclockwise. Compute the contour integral
Z
(ez − z)dz.
C

z is closed and ez is entire.

R
Solution. By Cauchy Integral Theorem, C e dz = 0 since C
Therefore,
Z Z Z Z Z
z
(e − z)dz = − zdz = − zdz − zdz − zdz
C C p1 p2 p2 p3 p3 p1
Z 1 Z 1
=− (−3it)d(3it) − (−3i(1 − t) − 4t)d(3i(1 − t) − 4t)
0 0
Z 1
− (−4)(1 − t)d((−4)(1 − t))
0
9 7
= − − − 12i + 8 = −12i
2 2
where p1 = 0, p2 = 3i and p3 = −4. 

5. Compute
Z 1
zi dz
−1

where the integrand denote the principal branch

zi = exp(i Log z)

of zi and where the path of integration is any continuous curve from z = −1 to z = 1

that, except for its starting and ending points, lies below the real axis.

Solution. Note that zi+1 /(i + 1) is an anti-derivative of zi outside the branch locus
(−∞, 0]. So

zi+1 zi+1
Z 1
zi dz = − lim
−1 i + 1 1 Im(z)<0
z→−1 i + 1

1 exp((i + 1)(−πi))
= −
i+1 i+1
1 + eπ 1 + eπ
= = (1 − i)
i+1 2

3.2 Cauchy Integral Theorem and Cauchy Integral Formula 59

6. Apply Cauchy Integral Theorem to show that

Z
f (z)dz = 0
C

when C is the unit circle |z| = 1, in either direction, and when

z3
(a) f (z) = 2 ;
z + 5z + 6
(b) f (z) = etan z ;
(c) f (z) = Log(z + 3i).
R
Solution. By Cauchy Integral Theorem, |z|=1 f (z)dz = 0 if f (z) is analytic on and
inside the circle |z| = 1. Hence it is enough to show that f (z) is analytic in {|z| ≤ 1}.
(a) f (z) is analytic in {z 6= −2, −3} and hence analytic in {|z| ≤ 1}.
(b) f (z) is analytic in {z : cos z = 0} = {z = nπ + π/2, n ∈ Z}. Since |nπ + π/2| >
1 for all integers n, f (z) is analytic in {|z| ≤ 1}.
(c) Log(z) is analytic in C\(−∞, 0] and hence Log(z + 3i) is analytic in C\{z :
z = x − 3i, x ∈ (−∞, 0]}. Since |x − 3i| > 1 for all x real, f (z) is analytic in
{|z| ≤ 1}.


7. Let C1 denote the positively oriented boundary of the curve given by |x| + |y| = 2
and C2 be the positively oriented circle |z| = 4. Apply Cauchy Integral Theorem to
show that
Z Z
f (z)dz = f (z)dz
C1 C2

when
z+1
(a) f (z) = 2 ;
z +1
z+2
(b) f (z) = ;
sin(z/2)
sin(z)
(c) f (z) = 2 .
z + 6z + 5
R R
Solution. By Cauchy Integral Theorem, C1 f (z)dz = C2 f (z)dz if f (z) is analytic
on and between C1 and C2 . Hence it is enough to show that f (z) is analytic in
{|x| + |y| ≥ 2, |z| ≤ 4}.
(a) f (z) is analytic in {z 6= ±i}. Since ±i ∈ {|x| + |y| < 2}, f (z) is analytic in
{|x| + |y| ≥ 2, |z| ≤ 4}.
(b) f (z) is analytic in {z : sin(z/2) 6= 0} = {z 6= 2nπ : n ∈ Z}. Since 2nπ ∈ {|x| +
|y| < 2} for n = 0 and |2nπ| > 4 for n 6= 0 and n ∈ Z, f (z) is analytic in
{|x| + |y| ≥ 2, |z| ≤ 4}.
(c) f (z) is analytic in {z 6= −1, −5}. Since −1 ∈ {|x| + |y| < 2} for n = 0 and
| − 5| > 4, f (z) is analytic in {|x| + |y| ≥ 2, |z| ≤ 4}.

60 Chapter 3. Complex Integrals

8. Let C denote the positively oriented boundary of the square whose sides lie along
Z x = ±2 and y = ±2. Evaluate each of these integrals
the lines
zdz
(a) ;
ZC z + 1
cosh z
(b) 2
dz;
ZC z + z
tan(z/2)
(c) dz.
C z − π/2

Solution. (a) By Cauchy Integral Formula,

zdz
Z
= 2πi(−1) = −2πi.
C z+1

(b) By Cauchy Integral Theorem,

cosh z cosh z cosh z

Z Z Z
2
dz = 2
dz + 2
dz
C z +z |z|=r z + z |z+1|=r z + z

for r = 1/2. By Cauchy Integral Formula,

cosh z cosh(z)
Z
2
dz = 2πi = 2πi
|z|=r z + z z+1 z=0

and
cosh z cosh z
Z
2 +z
dz = 2πi = −2πi cosh(−1).
|z+1|=r z z z=−1

Hence
cosh z
Z
2
dz = 2πi(1 − cosh(−1)).
C z +z

(c) Note that tan(z/2) is analytic in {z 6= (2n + 1)π : n ∈ Z} and hence analytic
inside C. Therefore,

tan(z/2)
Z
dz = 2πi tan(π/4) = 2πi
C z − π/2

by Cauchy Integral Formula.



9. Find the value of the integral g(z) around the circle |z − i| = 2 oriented counterclock-
wise when
1
(a) g(z) = 2 ;
z +4
1
(b) g(z) = 2 .
z(z + 4)
3.2 Cauchy Integral Theorem and Cauchy Integral Formula 61

Solution. (a) Since −2i 6∈ {|z − i| ≤ 2} and 2i ∈ {|z − i| ≤ 2},

(z + 2i)−1
Z Z
π
g(z)dz = dz = 2πi(2i + 2i)−1 =
|z−i|=2 |z−i|=2 z − 2i 2

by Cauchy Integral Formula.

(b) By Cauchy Integral Theorem,
Z Z Z
g(z)dz = g(z)dz + g(z)dz
|z−i|=2 |z|=r |z−2i|=r

for r < 1/2. Since

1
Z
πi
g(z)dz = 2πi =
|z|=r z2 + 4 z=0 2

and
1
Z
πi
g(z)dz = 2πi =−
|z−2i|=r z(z + 2i) z=2i 4

by Cauchy Integral Formula,

Z
πi
g(z)dz =
|z−i|=2 4

10. Compute the integrals of the following functions along the curves C1 = {|z| = 1}
and C2 = {|z − 2| = 1}, both oriented counterclockwise:
1
(a) ;
2z − z2
sinh z
(b) .
(2z − z2 )2

Solution. (a)

dz (2 − z)−1
Z Z
2
= dz = 2πi(2 − 0)−1 = πi
|z|=1 2z − z |z|=1 z

(b)

sinh z (sinh z)(2 − z)−2

Z Z
2 2
dz = dz
|z|=1 (2z − z ) |z|=1 z2
πi
= 2πi((sinh z)(2 − z)−2 )0 =
z=0 2


62 Chapter 3. Complex Integrals

11. Show that if f is analytic inside and on a simple closed curve C and z0 is not on C,
then
f (m) (z) f (z)
Z Z
(n − 1)! n
dz = (m + n − 1)! m+n
dz
C (z − z0 ) C (z − z0 )

for all positive integers m and n.

Proof. If z0 lies outside C, then

f (m) (z) f (z)
Z Z
n
dz = m+n
dz = 0
C (z − z0 ) C (z − z0 )

by Cauchy Integral Theorem, since f (m) z/(z −z0 )n and f (z)/(z −z0 )m+n are analytic
on and inside C.
If z0 lies inside C, then
f (m) (z)
Z
(n − 1)! n
dz = ( f (m) (z))(n−1) z=z0
= f (m+n−1) (z0 )
C (z − z0 )

and
f (z)
Z
(m + n − 1)! m+n
dz = f (m+n−1) (z0 )
C (z − z0 )

by Cauchy Integral Formula. Therefore,

f (m) (z) f (z)
Z Z
(n − 1)! n
dz = (m + n − 1)! m+n
dz.
C (z − z0 ) C (z − z0 )

12. Let f (z) be an entire function. Show that f (z) is a constant if | f (z)| ≤ ln(|z| + 1) for
all z ∈ C.

Proof. For every z0 ∈ C, we have

1 f (z)
Z
0
f (z0 ) = 2
dz
2πi |z−z0 |=R (z − z0 )

for all R > 0. Since

f (z) ln(|z| + 1) ln(R + |z0 | + 1)
2
≤ ≤
(z − z0 ) R2 R2
for |z − z0 | = R,
1 f (z) ln(R + |z0 |) + 1
Z
| f 0 (z0 )| = 2
dz ≤ .
2πi |z−z0 |=R (z − z0 ) R
And since
ln(R + |z0 | + 1) 1
lim = lim = 0,
R→∞ R R→∞ R + |z0 | + 1

by L’Hospital (see Problem 8, section 2.2), we conclude that | f 0 (z0 )| = 0 and hence
f 0 (z0 ) = 0 for every z0 ∈ C. Therefore, f (z) is a constant. 
3.2 Cauchy Integral Theorem and Cauchy Integral Formula 63

13. Let CN be the boundary of the square

{|x| ≤ Nπ, |y| ≤ Nπ},

where N is a positive integer. Show that

dz
Z
lim =0
N→∞ CN z3 cos z

Proof. When z = x + yi ∈ CN , either x = ±Nπ or y = ±Nπ. When x = ±Nπ,

| cos z|2 = (cos x)2 + (sinh y)2 ≥ (cos x)2 = (cos(Nπ))2 = 1

When y = ±Nπ,

| cos z|2 = (cos x)2 + (sinh y)2 ≥ (sinh y)2 = (sinh(Nπ))2 > 1

Therefore, | cos z| ≥ 1 when z ∈ CN . We also have |z| ≥ Nπ when z ∈ CN . Therefore,

1 1
≤
z3 cos z N 3π 3

and
dz 1 8Nπ 8
Z Z
3
≤ 3 3 |dz| = 3 3
= 2 2
CN z cos z N π CN N π N π

Since
8
lim =0
N→∞ N 2 π 2

we conclude
dz
Z
lim =0
N→∞ CN z3 cos z

14. Let CN be the boundary of the square

n π πo
|x| ≤ Nπ + , |y| ≤ Nπ +
2 2
oriented counterclockwise, where N is a positive integer. Show that

dz
Z
lim = 0.
N→∞ CN z2 sin z

[Refer to: problem 6 in section 4.3]

64 Chapter 3. Complex Integrals

Proof. When z = x + yi ∈ CN , either x = ±(Nπ + π/2) or y = ±(Nπ + π/2). When

x = ±(Nπ + π/2),

| sin z|2 = (sin x)2 + (sinh y)2 ≥ (sin x)2 = (sin(Nπ + π/2))2 = 1

When y = ±(Nπ + π/2),

| sin z|2 = (sin x)2 + (sinh y)2 ≥ (sinh y)2

= (sinh(Nπ + π/2))2 ≥ (sinh(3π/2))2 > 1.

Therefore, | sin z| ≥ 1 when z ∈ CN . We also have |z| ≥ Nπ + π/2 when z ∈ CN .

Consequently,
1 1
≤
z2 sin z (N + 1/2)2 π 2
and
dz 1 8(N + 1/2)π 8
Z Z
2
≤ |dz| = = .
CN z sin z (N + 1/2)2 π 2 CN
2
(N + 1/2) π 2 (N + 1/2)π
And since
8
lim =0
N→∞ (N + 1/2)π

we conclude
dz
Z
lim =0
N→∞ CN z2 sin z


15. Compute the contour integral

z2011
Z
2011 + z2010 + z2009 + 1
dz,
Cz

where C is the circle |z| = 2 oriented counter-clockwise.

Solution. First, we show that all roots of

z2011 + z2010 + z2009 + 1 = 0

lie inside |z| < 2. Otherwise, suppose that

z2011 + z2010 + z2009 + 1 = 0

for some |z| ≥ 2. Then

1 1 1
1 + + 2 + 2011 = 0
z z z
3.2 Cauchy Integral Theorem and Cauchy Integral Formula 65

and hence
1 1 1 1 1 1
1 = − − 2 − 2011 ≤ + 2 + 2011 .
z z z |z| |z| |z|

When |z| ≥ 2,
1 1 1 1 1 1
+ 2 + 2011 ≤ + + 2011 < 1.
|z| |z| |z| 2 4 2
This is a contradiction. Therefore, all roots of

z2011 + z2010 + z2009 + 1 = 0

lie inside |z| < 2. It follows that

z2011 z2011
Z Z
2011 + z2010 + z2009 + 1
dz = 2011 + z2010 + z2009 + 1
dz
Cz |z|=R z

for all R > 2 by CIT.

We have
z2011 1 z2009 − z + 1
= 1 − + .
z2011 + z2010 + z2009 + 1 z z(z2011 + z2010 + z2009 + 1)
Since
z2009 − z + 1 R2009 + R + 1
≤
z(z2011 + z2010 + z2009 + 1) R(R2011 − R2010 − R2009 − 1)

for |z| = R,

z2009 − z + 1 2π(R2009 + R + 1)
Z
2011 + z2010 + z2009 + 1)
dz ≤
|z|=R z(z R2011 − R2010 − R2009 − 1

and hence
z2009 − z + 1
Z
lim 2011 + z2010 + z2009 + 1)
dz = 0.
R→∞ |z|=R z(z

And we have
dz
Z Z
dz = 0 and = 2πi.
|z|=R |z|=R z

Therefore,

z2011
Z
2011 + z2010 + z2009 + 1
dz = −2πi.
Cz


66 Chapter 3. Complex Integrals

16. Calculate
z2008
Z
2009 + z + 1
dz
Cz

where C is the circle |z| = 2 oriented counter-clockwise.

Solution. First, we prove that all zeroes of z2009 + z + 1 lie inside the circle |z| = 2.
Otherwise, z2009 + z + 1 = 0 for some |z| ≥ 2. Then
1 1
z2009 + z + 1 = 0 ⇒ 1 + + =0
z2008 z2009
On the other hand,
1 1 1 1 1 1
1+ + ≥ 1− − ≥ 1− − >0
z2008 z2009 |z|2008 |z|2009 22008 22009

for |z| ≥ 2. Contradiction. So all zeroes of z2009 + z + 1 lie inside the circle |z| = 2
and hence z2008 /(z2009 + z + 1) is analytic in |z| ≥ 2. Therefore,

z2008 z2008
Z Z
2009 + z + 1
dz = 2009 + z + 1
dz
|z|=2 z |z|=R z

for all R > 2 by Cauchy Integral Theorem.

We observe that
z2008 1 z+1
2009
− = − 2009 .
z +z+1 z z(z + z + 1)
For |z| = R > 2,
z+1 R+1
≤
z(z2009 + z + 1) R(R2009 − R − 1)
and hence
z+1 2π(R + 1)
Z
dz ≤ .
|z|=R z2009 + z + 1 R2009 − R − 1
It follows that
z+1
Z
lim dz = 0.
R→∞ |z|=R z2009 + z + 1
Therefore,
z2008 z2008
Z Z
2009 + z + 1
dz = lim 2009 + z + 1
dz
|z|=2 z R→∞ |z|=R z
dz
Z
= lim = 2πi.
R→∞ |z|=R z


3.2 Cauchy Integral Theorem and Cauchy Integral Formula 67

17. Let C be the circle |z| = 1 oriented counter-clockwise.

(a) Compute
1
Z
dz
C z2 − 8z + 1
(b) Use or not use part (a) to compute
1
Z π
dθ
0 4 − cos θ

Solution. The function

1 1
= √ √
z2 − 8z + 1 (z − 4 − 15)(z − 4 + 15)
√
has a singularity in |z| < 1 at z = 4 − 15. Therefore,
1 1
Z
dz = 2πi lim√ √ √
C z2 − 8z + 1 z=4− 15 (z − 4 − 15)(z − 4 + 15)
 
1 πi
= 2πi √ √ =−
√
z − 4 − 15 z=4− 15 15
That is,

1 deiθ
Z Z π
2
dz = 2iθ − 8eiθ + 1
C z − 8z + 1 −π e
eiθ
Z π
=i 2iθ − 8eiθ + 1
dθ
−π e
1
Z π
=i iθ −iθ − 8
dθ
−π e + e
i π 1
Z
=− dθ
2 −π 4 − cos θ
1
Z π
= −i dθ
0 4 − cos θ
Therefore,
1
Z π
π
dθ = √
0 4 − cos θ 15


18. Compute the integral

dx
Z π
.
−π 2 − (cos x + sin x)
68 Chapter 3. Complex Integrals

Solution. Let z = eix . Then dz = ieix dx, dx = −idz/z and hence

dx dx
Z π Z π
=
−π 2 − (cos x + sin x) −π 2 − (eix + e−ix )/2 − (eix − e−ix )/(2i)
−idz
Z
= 2 2
|z|=1 2z − (z + 1)/2 − (z − 1)/(2i)
dz
Z
= (i − 1) 2
|z|=1 z − 2(1 + i)z + i
dz
Z
= (i − 1)
|z|=1 (z − z1 )(z − z2 )
2πi(i − 1) √
= = 2π
z2 − z1
√ √ √ √
where z1 = (1 + 2/2) + (1 + 2/2)i and z2 = (1 − 2/2) + (1 − 2/2)i.


19. Let f (z) be an entire function satisfying

| f (z1 + z2 )| ≤ | f (z1 )| + | f (z2 )|

for all complex numbers z1 and z2 . Show that f (z) is a polynomial of degree at most
1.

Proof. We have
n n
∑ f (zk ) = f (z1) + f (z2) + ∑ f (zk )
k=1 k=3
n
= f (z1 + z2 ) + ∑ f (zk )
k=3
n
= f (z1 + z2 ) + f (z3 ) + ∑ f (zk )
k=4
n
= f (z1 + z2 + z3 ) + ∑ f (zk )
k=4
!
n
= . . . = f (z1 + z2 + . . . + zn ) = f ∑ xk .
k=1

Therefore,
!
n n
∑ f (zk ) = f (z1) + f (z2) + . . . + f (zn) = f (z1 + z2 + . . . + zn) = f ∑ xk
k=1 k=1

for all complex numbers z1 , z2 , . . . , zn . Particularly, this holds for z1 = z2 = . . . =

zn = z/n:
z
nf = f (z)
n
3.2 Cauchy Integral Theorem and Cauchy Integral Formula 69

for all z ∈ C and all positive integer n. Let M be the maximum of | f (z)| for |z| = 1.
Then
z
| f (z)| = n f ≤ nM
n
for all z satisfying |z| = n.
By Cauchy Integral Formula,
1 f (z)
Z
00
f (z0 ) = 3
dz
πi |z|=n (z − z0 )

for |z0 | < n. Since

f (z) | f (z)| nM
3
= 3
≤
(z − z0 ) |z − z0 | (n − |z0 |)3

for |z| = n and |z0 | < n,

1 f (z) 2n2 M
Z
3
dz ≤ .
πi |z|=n (z − z0 ) (n − |z0 |)3
And since
2n2 M 2M/n
lim = lim = 0,
n→∞ (n − |z0 |)3 n→∞ (1 − |z0 |/n)3

we conclude that f 00 (z0 ) = 0 for all z0 . Therefore, f 0 (z) ≡ a is a constant and

f (z) = az + b is a polynomial of degree at most 1. 

20. Let f (z) be an entire function satisfying that | f (z)| ≤ |z|2 for all z. Show that
f (z) ≡ az2 for some constant a satisfying |a| ≤ 1.

Proof. For every z0 ∈ C, we have

3! f (z)
Z
f 000 (z0 ) = 4
dz
2πi |z−z0 |=R (z − z0 )

for all R > 0. Since

f (z) |z|2 (R + |z0 |)2
≤ ≤
(z − z0 )4 R4 R4

for |z − z0 | = R,

3! f (z) 6(R + |z0 |)2

Z
000
| f (z0 )| = 4
dz ≤ .
2πi |z−z0 |=R (z − z0 ) R3
And since

6(R + |z0 |)2 |z0 | 2

 
6
lim = lim 1+ = 0,
R→∞ R3 R→∞ R R
70 Chapter 3. Complex Integrals

we conclude that | f 000 (z0 )| = 0 and hence f 000 (z0 ) = 0 for every z0 ∈ C. Therefore,
f 000 (z) ≡ 0, f 00 (z) ≡ 2a, f 0 (z) ≡ 2az + b and f (z) ≡ az2 + bz + c for some constants
a, b and c.
Since | f (z)| ≤ |z|2 , |az2 + bz + c| ≤ |z|2 for all z. Take z = 0 and we obtain |c| ≤ 0.
Hence c = 0. Therefore, |az2 + bz| ≤ |z|2 and hence |az + b| ≤ |z| for all z. Take
z = 0 again and we obtain |b| ≤ 0. Hence b = 0. So |az2 | ≤ |z|2 and hence |a| ≤ 1.
In conclusion, f (z) = az2 with a satisfying |a| ≤ 1. 

21. Let f (z) be a complex polynomial of degree at least 2 and R be a positive number
such that f (z) 6= 0 for all |z| ≥ R. Show that
dz
Z
= 0.
|z|=R f (z)

[Refer to: problem 5 in section 4.3]

Proof. Let f (z) = a0 + a1 z + · · · + an zn , where an 6= 0 and n = deg f . Since f (z) 6= 0

for |z| ≥ R, 1/ f (z) is analytic in |z| ≥ R. Hence
dz dz
Z Z
=
|z|=R f (z) |z|=r f (z)

for all r ≥ R by Cauchy Integral Theorem.

Since

| f (z)| ≥ |an ||z|n − |an−1 ||z|n−1 − · · · − |a0 |

we have
1 1
≤
f (z) |an |r − |an−1 |rn−1 − · · · − |a0 |
n

for |z| = r sufficiently large. It follows that

dz 2πr
Z
≤
|z|=r f (z) |an |rn − |an−1 |rn−1 − · · · − |a0 |

And since n ≥ 2,
2πr
lim
r→∞ |an |rn − |a
n−1 |r
n−1 − · · · − |a |
0
2π
= lim =0
r→∞ n|an |r n−1 − (n − 1)|an−1 |r n−1 − · · · − |a1 |

by L’Hospital. Hence
dz dz
Z Z
= lim = 0.
|z|=R f (z) r→∞ |z|=r f (z)


 3.3 Improper integrals 71

3.3 Improper integrals

1. Compute the integral
xdx
Z ∞
.
0 x3 + 1

Solution. Consider the contour integral of z/(z3 + 1) along LR = [0, R], CR = {z =

Reit : 0 ≤ t ≤ 2π/3} and MR = {te2πi/3 : 0 ≤ t ≤ R}. By Cauchy Integral Formula,
zdz zdz zdz zdz
Z Z Z Z
3 +1
+ 3 +1
− 3 +1
=
LR z CR z MR z |z−eπi/3 |=1/2 z3 + 1
By Cauchy Integral Formula,
zdz 2πi exp(πi/3)
Z
=
|z−eπi/3 |=1/2 z3 + 1 (exp(πi/3) + 1)(exp(πi/3) − exp(−πi/3))
2π exp(πi/3)
= √ .
(exp(πi/3) + 1) 3
For z lying on CR ,
z R
≤
z3 + 1 R3 − 1
and hence
zdz 2πR
Z
≤
CR z3 + 1 3(R3 − 1)
It follows that
zdz
Z
lim =0
R→∞ CR z3 + 1
And
Z R
zdz xdx
Z
3
= exp(4πi/3)
MR z + 1 0 x3 + 1
Therefore, we have
xdx 2π exp(πi/3)
Z ∞
(1 − exp(4πi/3)) = √
0 x3 + 1 (exp(πi/3) + 1) 3
and hence
xdx 2π
Z ∞
= √ .
0 x3 + 1 3 3

72 Chapter 3. Complex Integrals

2. Compute the integral

cos x
Z ∞
dx.
0 x4 + 1

Solution. Since cos x/(x4 + 1) is even,

cos x 1 cos x
Z ∞ Z ∞
dx = dx.
0 x4 + 1 2 −∞ x4 + 1
Actually, we have
cos x 1 eix
Z ∞ Z ∞
dx = dx
0 x4 + 1 2 4
−∞ x + 1

since eix = cos x + i sin x.

Consider the contour integral of eiz /(z4 + 1) along the path LR = [−R, R] and CR =
{|z| = R, Im(z) ≥ 0}, oriented counterclockwise. By Cauchy Integral Theorem, we
have
eiz eiz
Z Z
4
dz + 4
dz
LR z + 1 CR z + 1
eiz eiz
Z Z
= 4
dz + 4
dz
|z−eπi/4 |=1/2 z + 1 |z−e3πi/4 |=1/2 z + 1

By Cauchy Integral Formula,

√ √
eiz 2πiei( 2+i 2)/2
Z
dz =
4
|z−eπi/4 |=1/2 z + 1 (eπi/4 − e3πi/4 )(eπi/4 − e−πi/4 )(eπi/4 − e−3πi/4 )
√ √
π(1 − i) exp((− 2 + i 2)/2)
= √
2 2
and similarly,
√ √
eiz π(1 + i) exp((− 2 − i 2)/2)
Z
4
dz = √
|z−e3πi/4 |=1/2 z + 1 2 2
Therefore,
√
Z
eiz
Z
eiz πe− 2/2 √ √
4
dz + 4
dz = √ (cos( 2/2) + sin( 2/2)).
LR z + 1 CR z + 1 2
For z lying on CR , y = Im(z) ≥ 0 and hence |eiz | = e−y ≤ 1. Hence

eiz 1
4
≤ 4
z +1 R −1
and it follows that
eiz
Z
πR
4
dz ≤ 4
CR z + 1 R −1
3.3 Improper integrals 73

Since
πR
lim = 0,
R→∞ R4 − 1

we conclude that
eiz
Z
lim 4
dz = 0.
R→∞ CR z + 1

Therefore,

cos x 1 ∞ eix
Z ∞ Z
dx = dx
0 x4 + 1 2 −∞ x4 + 1
1 eiz
Z
= lim dz
2 R→∞ LR z4 + 1
√ √ ! √ !!
πe− 2/2 2 2
= √ cos + sin .
2 2 2 2


 4. Series

4.1 Taylor and Laurent series

1. Find the Taylor series of the following functions and their radii of convergence:
(a) z sinh(z2 ) at z = 0;
(b) ez at z = 2;
z2 + z
(c) at z = −1.
(1 − z)2

Solution. (a) Since ez = ∑∞ n

n=0 z /n!,

2 2
!
2 ez − e−z
z sinh(z ) = z
2
!
z ∞
z2n ∞
nz
2n
= ∑ − ∑ (−1)
2 n=0 n! n=0 n!
∞
1 − (−1)n z2n+1
= ∑
n=0 2 n!
∞
z4m+3
= ∑
m=0 (2m + 1)!

where we observe that (1 − (−1)n )/2 = 0 if n = 2m is even and 1 if n = 2m + 1

is odd. Since f (z) is entire, the radius of convergence is ∞.
76 Chapter 4. Series

(b) Let w = z − 2. Then z = w + 2 and

∞ 2
∞
wn e (z − 2)n
ez = ew+2 = e2 ew = e2 ∑ = ∑ n! .
n=0 n! n=0
Since f (z) is entire, the radius of convergence is ∞.
(c) Let w = z + 1. Then
z2 + z w2 − w 3 2
= = 1 − + .
(1 − z)2 (2 − w)2 2 − w (2 − w)2
We have
3 3 1 3 ∞ wn ∞
3wn
− =− = − ∑ n = − ∑ n+1
2−w 2 1 − (w/2) 2 n=0 2 n=0 2
and
 0  0
2 2 1
= =
(2 − w)2 2−w 1 − (w/2)
!0
∞
wn nwn−1
∞
= ∑ 2n = ∑ n
n=0 n=0 2
∞
(n + 1)wn
= ∑ 2n+1 .
n=0
Therefore,
z2 + z ∞
3wn ∞
(n + 1)wn
= 1 − ∑ + ∑ .
(1 − z)2 n=0 2
n+1
n=0 2n+1
∞
(n − 2)(z + 1)n
= ∑ .
n=1 2n+1
Since f (z) is analytic in |z + 1| < 2 and has a singularity at z = 1, the radius of
convergence is 2.

2. Find the Taylor series of (cos z)2 at z = π.

Solution. Let w = z − π. Then

(cos z)2 = (cos(z + π))2 = (cos w)2
2
e + e−iw
 iw
1
= = (e2iw + e−2iw + 2)
2 4
1 ∞ n
(2i) wn 1 ∞ (−2i)n wn 1
= ∑ + ∑ +
4 n=0 n! 4 n=0 n! 2
1 1 ∞ (2i)2n w2n 1 ∞
(−1)n 22n−1 w2n
= + ∑ = +∑
2 2 n=0 (2n)! 2 n=0 (2n)!
∞
(−1)n 22n−1 w2n ∞
(−1)n 22n−1 (z − π)2n
= 1+ ∑ = 1+ ∑
n=1 (2n)! n=1 (2n)!

4.1 Taylor and Laurent series 77

3. Let f (z) be a function analytic at 0 and g(z) = f (z2 ). Show that g(2n−1) (0) = 0 for
all positive integers n.

(z) = ∑∞ n
Proof. Since f (z) is analytic at 0, f√ n=0 an z in some disk |z| < r. Therefore,
2 ∞ 2n
g(z) = f (z ) = ∑n=0 an z in |z| < r and hence
∞
g(m) (0) m ∞
∑ m! z = ∑ anz2n
m=0 n=0

And since the power series representation of an analytic function is unique, we must
have g(m) (0) = 0 for m is odd, i.e., m = 2n − 1 for all positive integers n. 
1
4. Find a power-series expansion of the function f (z) = about the point 4i, and
3−z
calculate the radius of convergence.

Solution. Notice that

1 1
=
3−z (3 − 4iz) − (z − 4i)
1 1
= ·
3 − 4i z − 4i
1−
3 − 4i 
1 ∞ 
z − 4i n z − 4i
= ∑ 3 − 4i for <1
3 − 4i n=0 3 − 4i

That is, for |z − 4i| < |3 − 4i| < 5. Thus

1 ∞
(z − 4i)n
= ∑ n+1
3−z n=0 (3 − 4i)

with radius of convergence 5. 

5. Find a Laurent-series expansion of the function f (z) = z−1 sinh(z−1 ) about the point
0, and classify the singularity at 0.
Solution. For g(z) = sinh z we know that
(
(n) sinh z, when n is even;
g (z) =
cosh z, when n is odd

Thus
(
(n) sinh(0) = 0, when n is even;
g (0) =
cosh(0) = 1, when n is odd

In this case, the Maclaurin series for g(z) = sinh z is:

z3 z5
z+ + +···
3! 5!
78 Chapter 4. Series

The Laurent series for g(z−1 ) = sinh z−1 is:

1 1 1
+ + +···
z 3!z3 5!z5
Thus Laurent series for f (z) = z−1 g(z−1 ) = z−1 sinh z−1 is
1 1 1
2
+ 4 + 6 +···
z 3!z 5!z
Notice that f (z) = z−1 sinh z−1 is analytic for z 6= 0, which means that z = 0 is an
isolated singularity and is, in fact, an essential singularity.


6. Consider the function

sin z
f (z) = .
cos(z3 ) − 1
Classify the singularity at z = 0 and calculate the residue.
Solution. Notice that f has an isolated singularity at z = 0. Thus, expanding numer-
ator and denominator in Taylor series we have
z3 z5 z7
z− + − +···
f (z) = 3! 5! 7!
z6 z12 z18
− + − ···
2! 4! 6!
z2 z4 z6

z 1− + − +···
3! 5! 7!
= 6 2z6 2z12
 
z
− 1− + −···
2! 4! 6!
z2 z4 z6
−2 1 − + − +···
= · 3! 5! 7!
z5 2z6 2z12
1− + −···
4! 6!
z2 z4 z6
 
−2 1
= 1− + − +··· ·
z5 3! 5! 7! 2z6 2z12
1− + −···
4! 6!
Let
2 2z6 2z12
g(z) = − + −···
4! 6! 8!
Thus we have
z2 z4 z6
 
−2 1
f (z) = 5
1− + − +··· · (4.1)
z 3! 5! 7! 1 − z6 g(z)
where g(z) is analytic and nonzero at z = 0, in fact, g(0) = 1/12. So for ε sufficiently
small, if |z| < ε, then |g(z)| < 1. Thus for |z| < min{ε, 1}, we can expand
1
1 − z6 g(z)
4.1 Taylor and Laurent series 79

in a geometric series. Hence

z2 z4 z6
 
−2 6 12 2

f (z) = 1 − + − + · · · 1 + z g(z) + z (g(z)) + · · ·
z5 3! 5! 7!
 
−2 2 2 2z 6 12 2

= + − + + · · · 1 + z g(z) + z (g(z)) + · · ·
z5 3!z3 5!z 7!
−2 −1
So the residue of f at z = 0 is = . 
5! 60

7. Prove that the coefficients cn in the expansion

∞
1
= ∑ cnzn
1 − z − z2 n=0

satisfy the recurrence relation c0 = c1 = 1, cn = cn−1 + cn−2 for n ≥ 2. What is the

radius of convergence of the series? What would be a good name for the cn ’s?
∞
Solution. Let S = ∑ cnzn for z such that the series converges.
n=0
Thus
∞ ∞
zS = ∑ cnzn+1 = ∑ cn−1zn
n=0 n=1

and
∞ ∞
z2 S = ∑ cnzn+2 = ∑ cn−2zn
n=0 n=2

Then we have
∞ ∞ ∞
S − zS − z2 S = ∑ cnzn − ∑ cn−1zn − ∑ cn−2zn
n=0 n=1 n=2
∞ ∞ ∞
= c0 + c1 z + ∑ cn zn − c0 z − ∑ cn−1 zn − ∑ cn−2 zn
n=2 n=2 n=2
∞
= c0 + (c1 − c0 )z + ∑ (cn − cn−1 − cn−2 )zn
n=2

Now, we know that

∞
1
= ∑ cn zn = S
1 − z − z2 n=0
then

S − zS − z2 S = 1

Thus
c0 = 1, c1 − c0 = 0, and cn − cn−1 − cn−2 = 0 for n ≥ 2.
Therefore c0 = c1 = 1, and cn = cn−1 + cn−2 for n ≥ 2. 
80 Chapter 4. Series

8. Find the Laurent series of the function

z+4
f (z) =
z2 (z2 + 3z + 2)
in
(a) 0 < |z| < 1;
(b) 1 < |z| < 2;
(c) |z| > 2;
(d) 0 < |z + 1| < 1.

Solution. We write f (z) as a sum of partial fractions:

z+4 5 2 3 1
=− + 2+ − .
z2 (z2 + 3z + 2) 2z z z + 1 2(z + 2)

For 0 < |z| < 1,

∞
3
= 3 ∑ (−1)n zn
z+1 n=0

and
1 1 1 1 ∞ (−1)n zn
− =− =− ∑ .
2(z + 2) 4 1 + (z/2) 4 n=0 2n

Therefore,
5 2 ∞
1 ∞ (−1)n zn
f (z) = − + 2 + 3 ∑ (−1)n zn − ∑
2z z n=0 4 n=0 2n
∞
2 5
= − + ∑ (−1)n(3 − 2−n−2)zn
z2 2z n=0

For 1 < |z| < 2,

3 3 1 3 ∞ (−1)n
= = ∑
z + 1 z 1 + (1/z) z n=0 zn

and
1 1 1 1 ∞ (−1)n zn
− =− =− ∑ .
2(z + 2) 4 1 + (z/2) 4 n=0 2n

Therefore,
5 2 3 ∞ (−1)n 1 ∞ (−1)n zn
f (z) = − + 2+ ∑ − ∑
2z z z n=0 zn 4 n=0 2n
∞
3(−1)n 1 1 ∞
= ∑ zn+1 z2 2z ∑ (−1)n2−n−2zn.
− + −
n=2 n=0
4.1 Taylor and Laurent series 81

For 2 < |z| < ∞,

3 3 1 3 ∞ (−1)n
= = ∑
z + 1 z 1 + (1/z) z n=0 zn

and
1 1 1 1 ∞ (−1)n 2n
− =− =− ∑ .
2(z + 2) 2z 1 + (2/z) 2z n=0 zn

Therefore,

5 2 3 ∞ (−1)n 1 ∞ (−1)n 2n
f (z) = − + 2+ ∑ − ∑
2z z z n=0 zn 2z n=0 zn
∞
(−1)n+1 (3 − 2n−2 )
= ∑ .
n=3 zn

For 0 < |z + 1| < 1, we let w = z + 1 and then

z+4 5 2 3 1
= + + − .
z2 (z2 + 3z + 2) 2(1 − w) (1 − w)2 w 2(w + 1)

For 0 < |w| < 1,

5 5 ∞
= ∑ wn ,
2(1 − w) 2 n=0
 0 !
∞ ∞
2 2
(1 − w)2
=
1−w
= ∑ 2wn = ∑ 2(n + 1)wn
n=0 n=0

and
1 1 ∞
− = − ∑ (−1)n wn .
2(w + 1) 2 n=0

Therefore,
∞ 
9 (−1)n

3
f (z) = + ∑ 2n + − wn
w n=0 2 2
∞ 
9 (−1)n

3
= + ∑ 2n + − (z + 1)n .
z + 1 n=0 2 2


82 Chapter 4. Series

9. Write the two Laurent series in powers of z that represent the function
1
f (z) =
z(1 + z2 )
in certain domains and specify these domains.

Solution. Since f (z) is analytic at z 6= 0, ±i, it is analytic in 0 < |z| < 1 and 1 <
|z| < ∞.
For 0 < |z| < 1,
 
1 1 1
f (z) = =
z(1 + z2 ) z 1 − (−z2 )
1 ∞ ∞
= ∑ (−1)n z2n = ∑ (−1)n z2n−1
z n=0 n=0

and for 1 < |z| < ∞,

 
1 1 1
f (z) = =
z(1 + z2 ) z3 1 − (−z−2 )
1 ∞ ∞
= 3 ∑ (−1)n z−2n = ∑ (−1)n z−2n−3
z n=0 n=0

10. Let
z2
f (z) =
z2 − 3z + 2
Find the Laurent series of f (z) in each of the following domains:
(a) 1 < |z| < 2
(b) 1 < |z − 3| < 2

Solution. First, we write f (z) as a sum of partial fractions:

z2 3z − 2 4 1
2
= 1+ = 1+ −
z − 3z + 2 (z − 2)(z − 1) z−2 z−1

In 1 < |z| < 2,

z2 2 1 1
2
= 1− −
z − 3z + 2 1 − z/2 z 1 − 1/z
∞
1 ∞
= 1 − 2 ∑ 2−n zn − ∑ z−n
n=0 z n=0
∞ ∞
= −1 − ∑ 21−n zn − ∑ z−n
n=1 n=1
4.1 Taylor and Laurent series 83

In 1 < |z − 3| < 2,
z2 4 1
2
= 1+ −
z − 3z + 2 (z − 3) + 1 2 + (z − 3)
   
4 1 1 1
= 1+ −
z − 3 1 + 1/(z − 3) 2 1 + (z − 3)/2
4 ∞ −n 1 ∞
= 1+ ∑
z − 3 n=0
(−1)n
(z − 3) − ∑ (−1)n2−n(z − 3)n
2 n=0
∞ ∞
1
= + 4 ∑ (−1)n+1 (z − 3)−n − ∑ (−1)n 2−n−1 (z − 3)n
2 n=1 n=1

11. Let
z2
f (z) =
z2 − z − 2
Find the Laurent series of f (z) in each of the following domains:
(a) 1 < |z| < 2
(b) 0 < |z − 2| < 1

Solution. First, we write f (z) as a sum of partial fractions:

z2 z+2 4 1
= 1+ = 1+ −
z2 − z − 2 (z − 2)(z + 1) 3(z − 2) 3(z + 1)
In 1 < |z| < 2,
z2 2 1 1 1
2
= 1− −
z −z−2 3 1 − z/2 3z 1 + 1/z
2 ∞ −n n 1 ∞
= 1 − ∑ 2 z − ∑ (−1)n z−n
3 n=0 3z n=0
1 1 ∞ 1−n n 1 ∞
= − ∑ 2 z + ∑ (−1)n z−n
3 3 n=1 3 n=1
In 0 < |z − 2| < 1,
z2 4 1 1
2
= 1+ −
z −z−2 3(z − 2) 3 3 + (z − 2)
4 1 1
= 1+ −
3(z − 2) 9 1 + (z − 2)/3
4 1 ∞
= 1+ − ∑ (−1)n 3−n (z − 2)n
3(z − 2) 9 n=0
∞
8 4
= + + ∑ (−1)n+1 3−n−2 (z − 2)n
9 3(z − 2) n=1

84 Chapter 4. Series

12. Find the Laurent series of

1
z 2
e −1
√
in z up to z6 and show the series converges in 0 < |z| < 2π.

Solution. Let f (z) = 1/(ez − 1). Since

∞
zn ∞ n
z ∞
zn
ez − 1 = ∑ n! − 1 = = z
∑ n! ∑ (n + 1)!
n=0 n=1 n=0

ez − 1 has a zero at 0 of multiplicity one and hence f (z) has pole at 0 of order 1. So
the Laurent series of f (z) is given by
∞
a−1
f (z) = ∑ an zn = + a0 + a1 z + a2 z2 + a3 z3 + ∑ an zn
n=−1 z n≥4

in 0 < |z| < r for some r > 0.

Since (ez − 1) f (z) = 1, we have
!
1= a−1 + a0 z + a1 z2 + a2 z3 + a3 z4 + ∑ an−1 zn
n≥5
!
z z2 z3zn z4
1+ + + + +∑ .
2 6 24 120 n≥5 (n + 1)!

Comparing the coefficients of 1, z, z2 , z3 and z4 on both sides, we obtain



 a−1 = 1

 a−1

 a0 + =0
2




 a0 a−1
a1 + + =0
2 6

 a1 a0 a−1
a2 + + + =0


2 6 24



 a3 + a2 + a1 + a0 + a−1 = 0



2 6 24 120
Solving it, we have a−1 = 1, a0 = −1/2, a1 = 1/12, a2 = 0 and a3 = −1/720.
Hence
1 1 z z3
f (z) = − + − + ∑ an zn
z 2 12 720 n≥4

and
1 2 1 1 z2 z6
z 2 = f (z ) = 2
− + − + ∑ an z2n .
e −1 z 2 12 720 n≥4

Note that f (z) is analytic in {z : ez − 1 6= 0} = {z 6= 2nπi}. So it is analytic

√ in
2
0 < |z| < 2π. Therefore, f (z ) is√ 2
analytic in 0 < |z | < 2π, i.e., 0 < |z| < 2π. So
the series converges in 0 < |z| < 2π. 
 4.2 Classification of singularities 85

4.2 Classification of singularities

1. For each of the following complex functions, do the following:
• find all its singularities in C;
• write the principal part of the function at each singularity;
• for each singularity, determine whether it is a pole, a removable singularity, or
an essential singularity;
• compute the residue of the function at each singularity.
1
(a) f (z) =
(cos z)2
Solution. f (z) is singular at cos z = 0, i.e., z = nπ +π/2. Let w = z−nπ −π/2.
Then
1 1 1
2
= 2
= .
(cos z) (cos(w + nπ + π/2)) (sin w)2

Since sin w has a zero of multiplicity one at w = 0, f (z) has a pole of order 2
at z = nπ + π/2. So

1 a−2 a−1
2
= 2 + + ∑ an wn .
(sin w) w w n≥0

Since
!2
2
∞
(−1)n w2n+1 ∞
(sin w) = ∑ (2n + 1)! = w2 + ∑ bn wn
n=0 n=4

we have
! !
∞
a−2 a−1
1= + + ∑ an wn w2 + ∑ bn wn .
w2 w n≥0 n=4

Comparing the coefficients of 1 and w on both sides, we obtain a−2 = 1 and

a−1 = 0. So the principal part of f (z) at z = nπ + π/2 is

1
(z − nπ − π/2)2

with residue 0. 
86 Chapter 4. Series
 
3 1
(b) f (z) = (1 − z ) exp
z
z
Solution. Since e = ∑∞ n
n=0 z /n!,
  ∞
3 1 1
(1 − z ) exp = (1 − z3 ) ∑ n
z n=0 n!z
∞ ∞
1 1
= ∑ n!zn − ∑ n!zn−3
n=0 n=0
∞ ∞ 3 3−n
1 1 z
= − −
∑ n!zn ∑ n!zn−3 ∑ n!
n=0 n=4 n=0
∞ ∞ 3 3−n
1 1 z
= 1+ ∑ n
− ∑ n
− ∑
n=1 n!z n=1 (n + 3)!z n=0 n!
∞   3 3−n
1 1 1 z
=∑ − n
+ 1 − ∑
n=1 n! (n + 3)! z n=0 n!

Therefore, f (z) has an essential singularity at z = 0 with principal part

∞  
1 1 1
∑ n! − (n + 3)! zn
n=1

and residue
1 1 23
Res f (z) = − = .
z=0 1! 4! 24

sin z
(c) f (z) = 2010
z
Solution. Since
sin z 1 ∞ (−1)n z2n+1
= ∑ (2n + 1)!
z2010 z2010 n=0
∞
(−1)n z2n−2009
= ∑
n=0 (2n + 1)!
1004
(−1)n z2n−2009 ∞
(−1)n z2n−2009
= ∑ + ∑
n=0 (2n + 1)! n=1005 (2n + 1)!

f (z) has a pole of order 2009 at z = 0 with principal part

1004
(−1)n z2n−2009
∑ (2n + 1)!
n=0

and with residue

sin z (−1)1004 1
Res 2010
= = .
z=0 z (2 · 1004 + 1)! 2009!

4.2 Classification of singularities 87
ez
(d) f (z) =
1 − z2
Solution. Since 1 − z2 = (1 − z)(1 + z), f (z) has poles of order 1 at 1 and −1.
Therefore,

ez ez e
Res = =−
z=1 1 − z2 (1 − z2 )0 z=1 2

and
ez ez 1
Res = = .
z=−1 1 − z2 (1 − z2 )0 z=−1 2e

And the principal parts of f (z) at z = 1 and z = −1 are

e 1
− and
2(z − 1) 2e(z + 1)

respectively.

 
2 1
(e) f (z) = (1 − z ) exp
z
Solution. The function has a singularity at 0 where
  ∞
2 1 1
(1 − z ) exp = (1 − z2 ) ∑ n
z n=0 (n!)z
∞ ∞
1 1
= ∑ (n!)zn − ∑ (n!)zn−2
n=0 n=0
∞ ∞
1 1 1
= 1+ ∑ n
− ∑ n−2
− (z2 + z + )
n=1 (n!)z n=3 (n!)z 2
∞ ∞
1 1 1
= −z2 − z + + ∑ n
− ∑ n
2 n=1 (n!)z n=1 (n + 2)!z
∞  
1 1 1
= −z2 − z + + ∑ − z−n
2 n=1 n! (n + 2)!

So the principal part is

∞  
1 1
∑ n! − (n + 2)! z−n
n=1

the function has an essential singularity at 0 and

1 1 5
Res f (z) = − =
z=0 1! 3! 6

88 Chapter 4. Series
1
(f) f (z) =
(sin z)2
Solution. The function has singularities at kπ for k ∈ Z. At z = kπ, we let
w = z − kπ and then
!−1
1 1 ∞
(−1)n w2n+1
= = ∑
(sin z)2 (sin w)2 n=0 (2n + 1)!
!−1
1 ∞
(−1)n w2n
= 2 ∑
w n=0 (2n + 1)!
!−1
1 ∞
(−1)n+1 w2n
= 2 1− ∑
w n=1 (2n + 1)!
!m
1 ∞ ∞
(−1)n+1 w2n
= 2 ∑ ∑
w m=0 n=1 (2n + 1)!
!
∞
1
= 2 1 + ∑ an wn
w n=2

So the principal part at kπ is

1
(z − kπ)2

the function has a pole of order 2 at kπ and

Res f (z) = 0
z=kπ


1 − cos z
(g) f (z) =
z2
Solution. The function has a singularity at 0 where
!
1 − cos z 1 ∞
(−1)n z2n 1 ∞ (−1)n+1 z2n
= 1 − ∑ = ∑ (2n)!
z2 z2 n=0 (2n)! z2 n=1
∞
(−1)n+1 z2n−2
= ∑ (2n)!
n=1

So the principal part is 0, the function has a removable singularity at 0 and

Res f (z) = 0
z=0


4.2 Classification of singularities 89
ez
(h) f (z) =
z(z − 1)2
Solution. The function has two singularities at 0 and 1. At z = 0,
! !
ez 1 ∞ zn ∞

z(z − 1)2
= ∑ n! ∑ (n + 1)zn
z n=0 n=0
!
∞
1 n
= 1 + ∑ an z
z n=1

So the principal part at 0 is 1/z, the function has a pole of order 1 at 0 and
Res f (z) = 1
z=0

At z = 1, we let w = z − 1 and then

! !
ez e1+w e ∞
wn ∞
n n
= = ∑ n! ∑ (−1) w
z(z − 1)2 (1 + w)w2 w2 n=0 n=0
! !
e wn∞ ∞
= 2 1+w+ ∑ 1 − w + ∑ (−1)n wn
w n=2 n! n=2
!
∞
e
= 1 + ∑ an wn
w2 n=2

So the principal part at 1 is

e
(z − 1)2
the function has a pole of order 2 at 1 and
Res f (z) = 0
z=1


(i) f (z) = tan z
Solution. The function has singularities at {cos z = 0} = {z = kπ + π/2 : k ∈ Z.
At z = kπ + π/2, we let w = z − kπ − π/2 and then
 π  π cos w
tan z = tan w + kπ + = tan w + =−
2 2 sin w
Since sin w has a zero of multiplicity 1 at w = 0, tan z has a pole of order 1 at
z = kπ + π/2. Therefore
 cos w  cos w
Res tan z = Res − =− = −1
z=kπ+π/2 w=0 sin w (sin w)0 w=0
and the principal part of tan z at z = kπ + π/2 is
1 1
− =− .
w z − kπ − π/2

90 Chapter 4. Series
 
2 1
(j) f (z) = (1 − z ) sin
z
Solution. The function has a singularity at 0 where
(−1)n
  ∞
2 1
(1 − z ) sin = (1 − z2 ) ∑ 2n+1
z n=0 ((2n + 1)!)z
∞
(−1)n ∞
(−1)n
= ∑ 2n+1
− ∑ 2n−1
n=0 ((2n + 1)!)z n=0 ((2n + 1)!)z
∞
(−1)n ∞
(−1)n
= −z + ∑ 2n+1
− ∑ 2n−1
n=0 ((2n + 1)!)z n=1 (2n + 1)!)z
(−1)n
∞ ∞
(−1)n+1
= −z + ∑ 2n+1
− ∑ 2n+1
n=0 ((2n + 1)!)z n=0 ((2n + 3)!)z
∞  
1 1
= −z + ∑ (−1)n + z−2n−1
n=0 (2n + 1)! (2n + 3)!

So the principal part is

∞  
1 1
∑ (−1) (2n + 1)! + (2n + 3)! z−2n−1.
n
n=0

Therefore, the function has an essential singularity at 0 and

1 1 7
Res f (z) = + = .
z=0 1! 3! 6

ez
(k) f (z) =
z2011
Solution. The function has a singularity at 0 where
ez 1 ∞
zn
=
z2011 z2011 ∑ n!
n=0
n−2011 2010 n−2011
∞
z z ∞
zn−2011
= ∑ = ∑ + ∑ n!
n=0 n! n=0 n! n=2011

Therefore, the principal part of f (z) at z = 0 is

2010 n−2011
z
∑ n!
n=0

and f (z) has a pole of order 2011 and residue

1
Res f (z) =
z=0 2010!
at z = 0. 
 4.3 Applications of residues 91
cos z
(l) f (z) =
z2 − z3

Solution. The function has singularities at {z2 − z3 = 0} = {z = 0, 1}. At

z = 0, z2 − z3 has a zero of multiplicity 2 and hence f (z) has a pole of order 2.
Suppose that the Laurent series of f (z) at z = 0 is given by
cos z a−2 a−1
= 2 + + ∑ an zn .
z2 − z3 z z n≥0

Hence
!
a−2 a−1 ∞
(−1)n z2n
(z2 − z3 ) + + ∑ an zn = cos z = 1 + ∑ .
z2 z n≥0 n=1 (2n)!

Comparing the coefficients of 1 and z on both sides, we obtain that a−2 = 1

and a−1 − a−2 = 0 and hence a−1 = a−2 = 1. So the principal part of f (z) at
z = 0 is
1 1
+
z2 z
with residue

Res f (z) = 1.
z=0

At z = 1, z2 − z3 has a zero of multiplicity 1 and hence f (z) has a pole of order

1. Hence
cos z cos z
Res = = − cos(1)
z=1 z2 − z3 (z2 − z3 )0 z=1

and the principal part of f (z) at z = 1 is

cos(1)
− .
z−1


4.3 Applications of residues

1. Calculate
8−z
Z
dz,
C z(4 − z)

where C is the circle of radius 7, centre 0, negatively oriented.

Solution. Observe that

8−z 8 − 2z + z 2 1 2 1
f (z) = = = + = −
z(4 − z) z(4 − z) z (4 − z) z (z − 4)
92 Chapter 4. Series

The function f has two singularities on C, z = 0 and z = 4. Both are inside C. At

z = 4, −1/(z − 4) is analytic and then
Res f (z) = 2
z=0

Similarly, since 2/z is analytic at z = 4,

Res f (z) = −1
z=4

From Cauchy’s residue theorem, and considering that C is negatively oriented, we

have that
 
8−z
Z
dz = −2πi Res f (z) + Res f (z)
C z(4 − z) z=0 z=4
= −2πi(2 − 1) = −2πi

2. Compute the integral
dθ
Z π
.
0 2 − cos θ

Solution. Since 1/(2 − cos θ ) is even,

dθ 1 dθ
Z π Z π
= .
0 2 − cos θ 2 −π 2 − cos θ
Let z = eiθ . Then cos θ = (z + z−1 )/2 and dθ = −iz−1 dz. Hence
dθ 1 dθ
Z π Z π
=
0 2 − cos θ 2 −π 2 − cos θ
−idz
Z
= −1
|z|=1 2z(2 − (z + z )/2)
dz
Z
=i 2
.
|z|=1 z − 4z + 1

The function
1 1
2
= √ √
z − 4z + 1 (z − 2 − 3)(z − 2 + 3)
√
has a singularity in |z| < 1 at z = 2 − 3. Therefore,
dz 1
Z
= 2πi Res√
|z|=1 z2 − 4z + 1 z=2− 3 z2 − 4z + 1
1 πi
= 2πi √ = −√ .
(z2 − 4z + 1)0 z=2− 3 3
Therefore,
dθ
Z π
π
=√
0 2 − cos θ 3

4.3 Applications of residues 93

3. Let a, b ∈ R such that a2 > b2 . Calculate the integral

dθ
Z π
.
0 a + b cos θ
π
Answer: √
a − b2
2

4. Evaluate the contour integral of the following functions around the circle |z| = 2011
oriented counterclockwise:
1
(a) ;
sin z
1
(b) 2z .
e − ez

Solution. (a) f (z) = 1/ sin z is analytic in {z 6= nπ : n ∈ Z}. It has a pole of order

one at nπ since (sin z)0 |z=nπ = cos(nπ) = (−1)n 6= 0. So

1 1
Res = = (−1)n .
z=nπ sin z cos(nπ)

Therefore,
dz 1
Z
= 2πi ∑ Res
|z|=2011 sin z |nπ|<2011
z=nπ sin z

= 2πi ∑ (−1)n = 2πi.

|n|≤640

(b) f (z) = 1/(e2z − ez ) is analytic in

{e2z − ez 6= 0} = {ez 6= 1} = {z 6= 2nπi : n ∈ Z}.

Since (e2z − ez )0 |z=2nπi = 1 6= 0, f (z) has a pole of order one at 2nπi. So

1 1
Res = = 1.
z=2nπi e2z − ez 2e2z − ez z=2nπi

Therefore,
dz 1
Z
= 2πi ∑ Res
|z|=2011 e2z − ez |2nπi|<2011
z=2nπi e2z − ez

= 2πi ∑ 1 = 2πi ∑ 1 = 1282πi.

|2nπi|<2011 |n|≤320


94 Chapter 4. Series

5. Let
f (z) = (z − a1 )(z − a2 )...(z − an )
be a complex polynomial with n ≥ 2 distinct roots a1 , a2 , ..., an .
(a) Prove that
n
dz 1
Z
= 2πi ∑
|z|=R f (z) k=1 ∏ j6=k (ak − a j )

for R > |ak | (k = 1, 2, ..., n).

(b) Use (a) and Cauchy Integral Theorem to prove that
n
1
∑ ∏ j6=k (ak − a j ) = 0
k=1

for all distinct complex numbers a1 , a2 , ..., an .

Proof. By Residue theorem,

n
dz 1
Z
= 2πi ∑ Res .
|z|=R f (z) k=1
z=ak f (z)

At each ak , 1/ f (z) has a pole of order one and

−1
1 ∏ j6=k (z − a j ) 1
Res = Res = .
z=ak f (z) z=ak z − ak ∏ j6=k k − a j )
(a
Therefore,
n
dz 1
Z
= 2πi ∑ .
|z|=R f (z) k=1 ∏ j6=k (ak − a j )

Since deg( f (z)) = n ≥ 2,

dz
Z
=0
|z|=R f (z)
by Problem 21 in Section 3.2. Therefore,
n
1
∑ ∏ j6=k (ak − a j ) = 0.
k=1


6. Use Cauchy Integral Theorem or Residue Theorem to show that
N
1 dz 1 (−1)n
Z
= + 2 ∑ 2 2
2πi CN z2 sin z 6 n=1 n π

and conclude that

π2 ∞
(−1)n+1 1 1 1
=∑ 2
= 1− 2 + 2 − 2 +···
12 n=1 n 2 3 4
4.3 Applications of residues 95

Solution. The function f (z) = 1/(z2 sin z) has singularities at z = nπ for n ∈ Z. So

n
1 dz 1
Z
2
= ∑ Res 2
2πi CN z sin z n=−N z=nπ z sin z

by Residue Theorem.
At z = nπ for a nonzero integer n,

z2 6= 0 and (sin z)0 6= 0.

nπ nπ

Therefore, 1/(z2 sin z) has a pole of order 1 at nπ for n 6= 0. It follows that

1 1 1 (−1)n
Res = 2 = =
2
z=nπ z sin z z (sin z)0 z=nπ n2 π 2 cos(nπ) n2 π 2

for n 6= 0.
At z = 0, z2 sin z has a zero multiplicity 3 and hence 1/(z2 sin z) has a pole of order
3. Suppose that the Laurent series of f (z) at z = 0 is given by
a−3 a−2 a−1
+ 2 + + ∑ an zn .
z3 z z n≥0

Then
!
2 a−3 a−2 a−1
z sin z + 2 + + ∑ an zn
z3 z z n≥0
! !
z2
= 1 − + ∑ bn zn a−3 + a−2 z + a−1 z2 + ∑ an zn = 1.
3! n≥3 n≥3

Comparing the coefficients of 1, z and z2 on both sides, we have



 a−3 = 1

a−2 = 0

 a
 a−1 − −3 = 0
6
Solving the equation, we obtain a−1 = 1/6, a−2 = 0 and a−3 = 1. So

1 1
Res = .
z=0 z2 sin z 6
Therefore,
n
1 dz 1
Z
2
= ∑ Res 2
2πi CN z sin z n=−N z=nπ z sin z
1 n=−1 (−1)n n=N (−1)n
= + ∑ 2 2 +∑ 2 2.
6 n=−N n π n=1 n π
 96 Chapter 4. Series

We observe that
(−1)n (−1)−n
=
n2 π 2 (−n)2 π 2
and hence we obtain
N
1 dz 1 (−1)n
Z
= + 2 ∑ 2 2.
2πi CN z2 sin z 6 n=1 n π

By Problem 14 in section 3.2, we have

1 dz
Z
lim =0
N→∞ 2πi CN z2 sin z
Consequently,

1 ∞
(−1)n
+2 ∑ 2 2 = 0
6 n=1 n π

That is,

π2 ∞
(−1)n ∞
(−1)n+1
=−∑ 2
= ∑ n2 .
12 n=1 n n=1

4.3.1 Improper integrals

cos x
Z ∞
1. Compute the integral 4 + x2 + 1
dx
−∞ x

Solution. since eix = cos x + i sin x,

eix
Z ∞ 
cos x
Z ∞
4 2
dx = Re 4 2
dx
−∞ x + x + 1 −∞ x + x + 1

Actually, we have

cos x eix
Z ∞ Z ∞
4 2
dx = 4 2
dx
−∞ x + x + 1 −∞ x + x + 1

in this case since sin x/(x4 + x2 + 1) is odd.

Consider the contour integral of eiz /(z4 + z2 + 1) along the path LR = [−R, R] and
CR = {|z| = R, Im(z) ≥ 0}, oriented counterclockwise. By CIT or residue theorem,
we have
eiz eiz
Z Z
4 2
dz + 4 2
dz
LR z + z + 1 CR z + z + 1
n
eiz
= 2πi ∑ Res 4 2
z=zk z + z + 1
k=1
4.3 Applications of residues 97

where z1 , z2 , . . . , zn are the singularities of eiz /(z4 + z2 + 1) inside the region {|z| <
R, Im(z) > 0}.
We find the singularities of eiz /(z4 + z2 + 1) by solving z4 + z2 + 1 = 0: we observe
that (z2 − 1)(z4 + z2 + 1) = z6 − 1. So the function has four singularities ±eπi/3 and
±e2πi/3 . Two of them eπi/3 and e2πi/3 lie above the real axis. Therefore,

eiz eiz
Z Z
4 2
dz + 4 2
dz
LR z + z + 1 CR z + z + 1
eiz eiz
 
= 2πi Res 4 2 + Res 4 2 .
z=eπi/3 z + z + 1 z=e2πi/3 z + z + 1

Since all zeros of z4 + z2 + 1 have multiplicity one, all poles of

eiz /(z4 + z2 + 1)

have order one. Therefore,

√
eiz eiz exp((− 3 + i)/2)
Res 4 2 = 4 2 = √
z=eπi/3 z + z + 1 (z + z + 1)0 z=eπi/3 3i − 3

and
√
eiz eiz exp((− 3 − i)/2)
Res 4 2 = 4 2 = √ .
z=e2πi/3 z + z + 1 (z + z + 1)0 z=e2πi/3 3i + 3

Hence

eiz eiz √
    
1 1
Z Z
π
4 2
dz + 4 2
dz = 3 cos + 3 sin .
LR z + z + 1 CR z + z + 1 3 2 2

For z lying on CR , y = Im(z) ≥ 0 and hence |eiz | = e−y ≤ 1. Hence

eiz 1
4 2
≤ 4
z +z +1 R − R2 − 1

and it follows that

eiz
Z
πR
4 2
dz ≤ 4
CR z + z + 1 R − R2 − 1

Since
πR
lim = 0,
R→∞ R4 − R2 − 1

we conclude that

eiz
Z
lim 4 2
dz = 0.
R→∞ CR z + z + 1
98 Chapter 4. Series

Therefore,
cos x eix
Z ∞ Z ∞
4 2
dx = 4 2
dx
−∞ x + x + 1 −∞ x + x + 1
eiz
Z
= lim dz
R→∞ LR z4 + z2 + 1
π √
    
1 1
= 3 cos + 3 sin .
3 2 2

2. Compute the integral
sin x
Z ∞
dx.
−∞ x2 + 2x + 2

Solution. since eix = cos x + i sin x,

eix
Z ∞ 
sin x
Z ∞
2 2
dx = Im 2 2
dx
−∞ x + 2x + 2 −∞ x + 2x + 2

Consider the contour integral of eiz /(z2 + 2z2 + 2) along the path LR = [−R, R] and
CR = {|z| = R, Im(z) ≥ 0}, oriented counterclockwise.
Since eiz /(z2 + 2z + 2) has two isolated singularities at −1 ± i with −1 + i lying
inside the curve LR ∪CR , we have
eiz eiz
Z Z
2 2
dz + 2 2
dz
LR z + 2z + 2 CR z + 2z + 2
eiz
= 2πi Res 2
z=−1+i z + 2z + 2
eiz
= 2πi 2
(z + 2z + 2)0 z=−1+i
2πi exp(−i − 1) π
= = (cos(1) − i sin(1))
2i e
by Cauchy Integral Theorem or residue theorem.
For z lying on CR , y = Im(z) ≥ 0 and hence |eiz | = e−y ≤ 1. Hence

eiz 1
≤ 2
z2 + 2z2 + 2 R − 2R2 − 2
and it follows that
eiz
Z
πR
2
dz ≤ 2
CR z + 2z + 2 R − 2R2 − 2
Since
πR
lim = 0,
R→∞ R2 − 2R2 − 2
4.3 Applications of residues 99

we conclude that
eiz
Z
lim 2 2
dz = 0.
R→∞ CR z + 2z + 2

Therefore,

eix
Z 
sin x
Z ∞ ∞
2 2
dx = Im 2 2
dx
−∞ x + 2x + 2 −∞ x + 2x + 2
eiz
 Z 
= Im lim dz
R→∞ LR z2 + 2z2 + 2
π
= − sin(1).
e

 Bibliography

[1] Brown, J. W. & Churchill, R. V. (2009). Complex Variables and Applications. 8th
Edition. New York: McGraw-Hill Higher Education.

[2] Needham, T. (1997). Visual Complex Analysis. New York: The Clarendon Press,
Oxford University Press.

[3] Spiegel, M. R. (1981). Theory and problems of Complex variables. Singapore:

McGraw-Hill.

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