Complex Integration

Pinaki Pal

Department of Mathematics
National Institute of Technology Durgapur
West Bengal, India
pinaki.pal@maths.nitdgp.ac.in

Pinaki Pal (NIT Durgapur) MA331; Odd Semester, 2022-23 1 / 84

Complex Integration
Complex Integration:

Let C : z(t), a ≤ t ≤ b be a contour and f

be any complex function defined on C.
Let P : a = t0 < t1 < ..... < tn = b be a
partition of [a, b].
Corresponding to the partition P, the
curve C is divided in n smaller arcs
σk = zk −1 → zk , k = 1, 2, ...., n where
zk = z(tk .)
Let ζk = z(sk ), tk −1 ≤ sk ≤ tk be an
arbitrary point in σk .
n
X Figure: Complex Integration
Let SP = f (ζk )(zk − zk −1 ).
k =1

Let ||P|| := max |tk − tk −1 |.

1≤k ≤n

Choose n → ∞ in such a way that ||P|| → 0.

Pinaki Pal (NIT Durgapur) MA331; Odd Semester, 2022-23 2 / 84
Complex Integration
If lim SP exists, f is said to be integrable and written as
||P||→0

ˆ n
X
f (z)dz = lim f (ζk )(zk − zk −1 ).
C ||P||→0
k =1
˛
If C is a closed path the integral is denoted by f (z)dz.
C

Theorem
If f is continuous on a contour C, then f is integrable along C.
If f = u + iv is continuous on a contour C : z(t) = x(t) + iy (t), t ∈ [a, b] then

ˆ ˆ b ˆ b ˆ b
0
f (z)dz = f (z(t))z (t)dt = (udx − vdy ) + i (vdx + udy ).
C a a a

The length of the contour C : z(t) = x(t) + iy (t), t ∈ [a, b] is given by

ˆ ˆ b ˆ bq
0
L(C) := |dz| = |z (t)|dt = x 2 (t) + y 2 (t) dt
C a a

Pinaki Pal (NIT Durgapur) MA331; Odd Semester, 2022-23 3 / 84

Complex Integration
ˆ ˆ ˆ
1 Linearity: [αf (z) + βg(z)]dz = α f (z)dz + β g(z)dz.
C C C
ˆ ˆ
2 f (z)dz = − f (z)dz.
−C C

Here if C is the curve joining the points from z0 to z1 then −C is the

curve joining the points from z1 to z0 .
ˆ ˆ ˆ
3 f (z)dz = f (z)dz + f (z)dz.
C1 +C2 C1 C2
4 M-L inequality: If L is the length of the curve C and M = max |f (z(t))|
t∈[a,b]
then ˆ ˆ
f (z)dz ≤ |f (z)||dz| ≤ ML.
C C
Hints:
ˆ ˆ b
0
f (z)dz = f (z(t))z (t)dt
C a
ˆ b ˆ
0
≤ |f (z(t))||z (t)|dt = |f (z)||dz|
a C
ˆ b
0
≤M |z (t)|dt = ML.
a
Pinaki Pal (NIT Durgapur) MA331; Odd Semester, 2022-23 4 / 84
Examples ˆ
dz 2π
Example: Show that ≤ where C : z(t) = 2eit , −π ≤ t ≤ π.
C z 2 + 10 3
Solution: For z ∈ C, |z + 10| ≥ 10 − |z|2 = 10 − |z(t)|2 = 10 − 4 = 6 and so
2

ˆ ˆ ˆ
dz |dz| 1 1 2π
2
= 2
≤ |dz| = × 4π = .
C z + 10 C |z + 10| 6 C 6 3

Theorem (Fundamental theorem of integration)

If a continuous function f has a primitive F in domain D i.e., F 0 (z) = f (z) for
all z ∈ D then for all paths C in D joining two points z0 and z1 in D, we have
ˆ
f (z) dz = F (z1 ) − F (z0 ).
C
Proof: Let z(t), t ∈ [a, b] be a parameterization of C with z0 = z(a), z1 = z(b). Then
ˆ ˆ b ˆ b ˆ b
0 0 0 dF (z(t))
f (z) dz = f (z(t))z (t) dt == F (z(t))z (t) dt = dt
C a a a dt

= F (z(b)) − F (z(a)) = F (z1 ) − F (z0 ).

Pinaki Pal (NIT Durgapur) MA331; Odd Semester, 2022-23 5 / 84

Examples
ˆ y
Example: Find z dz from z = 0 to z = 4 + 2i
C 2i 4 + 2i
along the curve C given by
(a) z(t) = t 2 + it (b) the line from z = 0 to 2i
and then line from z = 2i to 4 + 2i. x
Solution: (a) For the curve z(t) = t 2 + it the point z = 0 and
z = 4 + 2i corresponds t = 0 and 2 respectively. Therefore
ˆ ˆ 2 ˆ 2
8
z dz = 2
(t + it)(2t + i) dt = (2t 3 − it 2 + t) dt = 10 − i.
C 0 0 3
(b) Let C = C1 + C2 where C1 : x = 0, 0 ≤ y ≤ 2 and
C2 : y = 2, 0 ≤ x ≤ 4. Then
ˆ ˆ ˆ ˆ 2 ˆ 4
z dz = z dz + z dz = (iy )i dy + (x + 2i) dx
C C1 C2 0 0
ˆ 2 ˆ 4
= y dy + (x − 2i) dx = 2 + (8 − 8i) = 10 − 8i
0 0
Pinaki Pal (NIT Durgapur) MA331; Odd Semester, 2022-23 6 / 84
Examples

ˆ
Example: Find 3z 2 dz from z = 0 to z = 4 + 2i along the curve C
C
given by
(a) z(t) = t 2 + it (b) the line from z = 0 to 2i and then line from
z = 2i to 4 + 2i.
Solution: Let f (z) = 3z 2 and F (z) = z 3 . Then F 0 (z) = f (z). Thus
ˆ
3z 2 dz = F (4 + 2i) − F (0) = (4 + 2i)3
C

where C is any curve joining z = 0 and z = 4 + 2i.

Pinaki Pal (NIT Durgapur) MA331; Odd Semester, 2022-23 7 / 84

Cauchy’s Theorem

Theorem (Green’s Theorem)

Let M(x, y ) and N(x, y ) be continuous with continuous partial
derivative in a simply connected domain R whose boundary is a
simple closed contour. Then
ˆ ¨  
∂N ∂M
Mdx + Ndy = − dxdy
C R ∂x ∂y

where C is traversed in the positive sense.

Pinaki Pal (NIT Durgapur) MA331; Odd Semester, 2022-23 8 / 84

Cauchy’s Theorem
Theorem (Cauchy’s weak Theorem)
If f (z) is analytic with a continuous derivative in a simply
ˆ connected domain
D, and C is closed contour lying in D, then we have f (z) dz = 0.
C

Proof:
Let f (z) = u(x, y ) + iv (x, y ). By C-R equation, we have
ux = vy , & uy = −vx for x, y ∈ D.
Since f 0 (z) = ux + ivx is continuous, all these partial derivatives are
continuous.
Let C be a simple closed contour.
Then by Green’s Theorem
ˆ ˆ ˆ ˆ
f (z) dz = (u + iv )(dx + idy ) = (udx − vdy ) + i (vdx + udy )
C
¨C ¨C C

= (−vx − uy )dxdy + i (ux − vy )dxdy

R R
=0.
Pinaki Pal (NIT Durgapur) MA331; Odd Semester, 2022-23 9 / 84
Cauchy’s Theorem
Theorem (Cauchy’s Theorem or Cauchy-Goursat Theorem)
If f is analytic in˛a simply connected domain D and C is any closed contour
lying in D, then f (z)dz = 0.
C

Remark: The domain bounded by a simple closed contour is always simply

connected domain.
Corollary
˛
If f is analytic within and on a simple closed contour C then f (z)dz = 0..
C

Corollary
Let f be analytic ˆin a simply connected domain D and a ∈ D. Then the
z
function F (z) = f (ξ) dξ, z ∈ D is analytic in D such that F 0 (z) = f (z) for
a
all z ∈ D.

Pinaki Pal (NIT Durgapur) MA331; Odd Semester, 2022-23 10 / 84

Cauchy’s Theorem
Remark-1: Let ˆf be analytic in a SCD D and z0 and z1 be any two points
z1
inside D. Then f (z) dz is independent of the path in D joining the point z0
z0
and z1 .

Let C1 and C2 be two distinct curves

joining z0 and z1 .
Let C = C1 + (−C2 ). Then C is a closed
curve lying inside D.
˛
By Cauchy’s theorem f (z) dz = 0
C

ˆ
But ˆ ˆ
f (z) dz = f (z) dz + f (z) dz =
ˆC ˆC1 −C2

f (z) dz − f (z) dz
C1 C2
ˆ ˆ
Thus f (z) dz = f (z) dz.
C1 C2
Pinaki Pal (NIT Durgapur) MA331; Odd Semester, 2022-23 11 / 84
Cauchy’s Theorem
Remark-2: Suppose that f (z) is analytic
ˆ in a multiply connected domain D
and on its boundary C. Then we have f (z) dz = 0, where the integration is
C
performed along C in the positive sense.

Suppose we construct the line segment

AB, called a cross-cut, which connects
the outer boundary C1 with the inner
boundary C2 .
Then the domain bounded by the contour
C1 , the line segment AB, the contour C2 ,
and the line segment BA (traversed as
illustrated in Figure) is simply connected.
By Cauchy’s theorem ‰ ˆ fi ˆ
f (z) dz + f (z) dz + f (z) dz + f (z) dz = 0.
C1 AB C2 BA
ˆ ˆ
But f (z) dz = − f (z) dz.
AB BA
‰ fi ˆ
Thus f (z) dz + f (z) dz = 0 =⇒ f (z) dz = 0.
C1 C2 C

Pinaki Pal (NIT Durgapur) MA331; Odd Semester, 2022-23 12 / 84

Cauchy’s Theorem
Remark-3: Let f (z) be analytic in a domain D bounded
‰ by two ‰simple closed
contour C1 and C2 and also on C1 and C2 . Then f (z) dz = f (z) dz
C1 C2
where C1 and C2 are both traversed counterclockwise.
Theorem (Cauchy’s Theorem for multiply connected domains)
Let D be a multiply connected domain bounded externally by a simple closed contour C and internally by n simple closed
nonintersecting contours C1 , C2 , . . . , Cn . Let f be analytic on D ∪ C1 ∪ C2 ∪ . . . ∪ Cn . Then

‰ n ‰
X
f (z) dz = f (z) dz
C k =1 Ck

where C and C1 , C2 , . . . , Cn are all traversed counterclockwise.

Pinaki Pal (NIT Durgapur) MA331; Odd Semester, 2022-23 13 / 84

Cauchy’s Theorem
‰
dz
Example: Evaluate , n ∈ Z where C is any closed contour.
C (z − a)n

Solution:
If a lies outside C then 1/(z − a)n is analytic inside and on C.
‰
dz
By Cauchy’s theorem = 0.
C (z − a)n

If a lies inside C then consider a circle C 0 lying inside C of radius r with

center at z = a.
By Cauchy’s theorem for multiply connected domain,
‰ ‰
dz dz
= .
C (z − a)n C 0 (z − a)n

We know that
‰ (
dz 2πi, n=1
=
C0 (z − a)n 0, n 6= 1

Thus
‰ (
dz 2πi, n = 1
=
C (z − a)n 0, n=6 1

Pinaki Pal (NIT Durgapur) MA331; Odd Semester, 2022-23 14 / 84

Cauchy’s Theorem
˛
dz
Example: Evaluate where C is the circle
C z2 + 1
(a) |z − i| = 1 (b) |z + i| = 1 (c) |z| = 2 (d) |z − 1| = 1.
Solution: Let
˛ ˛ ˛
dz 1 dz 1 dz
I= 2
= +
C z +1 2i C z −i 2i C z +i
1
(a) Let C : |z − i| = 1. In this case, is y
z +i
analytic within and on C. Then
˛ ˛
1 dz 1 dz 1
I= + = × 2πi + 0 = π
2i C z − i 2i C z + i 2i i
1
(b) Let C : |z + i| = 1. In this case, is
z −i x
analytic within and on C. Then
˛ ˛ −i
1 dz 1 dz 1
I= + =0+ × 2πi = π
2i C z − i 2i C z + i 2i
Pinaki Pal (NIT Durgapur) MA331; Odd Semester, 2022-23 15 / 84
 Cauchy’s Theorem
(c) Let C : |z| = 2. In this case, both the
point i and −i lies inside C. The curve
C1 : |z − i| = 12 and C2 : |z + i| = 12 lies inside y
C. Then
‰ ‰
1 dz 1 dz i
I= +
2i C z − i 2i C z + i
‰ ‰ ‰ ‰ x
−i

1 dz dz dz dz
= + + +
2i C1 z − i C2 z − i C1 z + i C2 z + i
1
= (2πi + 0 + 0 + 2πi) = 2π.
2i
(d) Let C : |z − 1| = 1. In this case, both y
the points i and −i lies outside C and so
1
2
is analytic within and on C. Thus i
z +1
˛ 1 x
dz −i
I= 2
=0
C z +1
Pinaki Pal (NIT Durgapur) MA331; Odd Semester, 2022-23 16 / 84
Cauchy’s Integral Formula

Theorem (Cauchy’s Integral Formula)

Let f (z) be analytic in a simply connected
domain D. Then for any point a ∈ D and any
simple closed contour C enclosing a, we have
ˆ
1 f (z)
f (a) = dz
2πi C z − a

Theorem (Cauchy’s Integral Formula for derivatives)

Let f (z) be analytic in a simply connected domain D. Then for any
point a ∈ D and any simple closed contour C enclosing a, we have
ˆ
(n) n! f (z)
f (a) = dz n = 0, 1, 2, · · · .
2πi C (z − a)n+1

Pinaki Pal (NIT Durgapur) MA331; Odd Semester, 2022-23 17 / 84

Cauchy’s Integral Formula
˛
ez
Example: Evaluate dz where C is the circle (a) |z| = 3
C z −2
(b) |z| = 1.
ez
Solution: The integrand is not analytic at z = 2.
z −2
(a) Let C : |z| = 3 and so z = 2 lies inside C. If f (z) = ez then f (z)
is analytic within and on C. Then by Cauchy’s integral formula
˛ ˛
ez f (z)
dz = dz = 2πi × f (2) = 2πie2
C z − 2 C z − 2

ez
(b) Let C : |z| = 1 and so z = 2 lies outside C. If f (z) = then
z −2
f (z) is analytic within and on C. Then by Cauchy’s integral theorem
˛ ˛
ez
dz = f (z) dz = 0.
C z −2 C

Pinaki Pal (NIT Durgapur) MA331; Odd Semester, 2022-23 18 / 84

Cauchy’s Integral
˛ Formula
tan z
Example: Evaluate 2−1
dz where C is the circle |z| = 32 .
C z
tan z
Solution: The integrand 2 is not analytic at
z −1
z = i, −i, ± π2 , ± 3π
2 , . . ..

The points z = i, −i lies inside C : |z| = 23 . Then by Cauchy’s integral

formula
˛ ˛ y
tan z tan z
2
dz = dz
C z −1 C (z − 1)(z + 1)
˛ ˛
1 tan z 1 tan z i
= dz − dz
2 C (z − 1) 2 C (z + 1)
x
1 1 −i
= × 2πi tan 1 − × 2πi tan(−1)
2 2
= 2πi tan 1
Pinaki Pal (NIT Durgapur) MA331; Odd Semester, 2022-23 19 / 84
Cauchy’s Integral Formula
˛
e2z
Example: Evaluate dz where C is the circle |z| = 3.
C (z + 1)4
e2z
Solution: The integrand is not analytic at z = −1 and the
(z + 1)4
point z = −1 lies inside C.
Let f (z) = e2z . Then f (z) is analytic within and on C.
Then by Cauchy’s integral formula for derivatives we have
˛ ˛
3! f (z) 3 e2z
f (3) (−1) = 4
dz = dz.
2πi C (z + 1) πi C (z + 1)4

Now f 0 (z) = 2e2z , f 00 (z) = 4e2z , f 000 (z) = 8e2z .

˛
e2z πi 8πi
4
dz = f (3) (−1) = .
C (z + 1) 3 3e2

Pinaki Pal (NIT Durgapur) MA331; Odd Semester, 2022-23 20 / 84

Cauchy’s Integral
˛ Formula
z
e
Example: Evaluate dz where C is the circle |z − 1| = 3.
C (z + 1)2 (z − 2)
Solution: The integrand is not analytic at z = −1, 2 and these points lies
inside C.
By partial fraction
1 1/9 1/9 1/3
= − − .
(z + 1)2 (z − 2) (z − 2) (z + 1) (z + 1)2
Let f (z) = ez . Then f (z) is analytic within and on C.
Then by Cauchy’s integral formula for derivatives we have
˛ ˛ ˛ ˛
ez 1 ez 1 ez 1 ez
2
dz = dz − dz − dz
C (z + 1) (z − 2) 9 C (z − 2) 9 C (z + 1) 3 C (z + 1)2

1 1 1
= × 2πif (2) − × 2πif (−1) − × 2πif 0 (−1)
9 9 3
2πi 2 2πi 2
= (e − e−1 − 3e−1 ) = (e − 4e−1 )
9 9
Pinaki Pal (NIT Durgapur) MA331; Odd Semester, 2022-23 21 / 84
Cauchy’s Integral
˛ Formula
z +4
Example: Evaluate dz where C is the circle
C z2 + 2z + 5
|z + 1 − i| = 2.
Solution: The integrand is not analytic at z = −1 ± 2i. Note that the
point −1 + 2i lies inside C and the point −1 − 2i lies outside C.
z +4
Let f (z) = . Then f (z) is analytic within and on C.
z + 1 + 2i
Then by Cauchy’s integral formula we have
˛ ˛
z +4 z +4
2
dz = dz
C z + 2z + 5 C (z + 1 + 2i)(z + 1 − 2i)
˛
f (z)
= dz = 2πi f (−1 + 2i)
C (z + 1 − 2i)

−1 + 2i + 4 π
= 2πi = (3 + 2i).
−1 + 2i + 1 + 2i 2

Pinaki Pal (NIT Durgapur) MA331; Odd Semester, 2022-23 22 / 84

Cauchy’s Integral Formula
Theorem (Cauchy integral formula for multiply connected
domain)
Let D be a multiply connected domain bounded by two simple closed contour
C1 and C2 (C2 lying wholly within C1 ) and f (z) is analytic in D ∪ C1 ∪ C2 . If a
is any interior point of D, then
‰ ‰
n! f (z) n! f (z)
f (n) (a) = dz − dz, n = 0, 1, 2, . . . .
2πi C1 (z − a)n+1 2πi C2 (z − a)n+1

Pinaki Pal (NIT Durgapur) MA331; Odd Semester, 2022-23 23 / 84

Theorem
Let f be analytic in a domain D. Then the derivatives of all orders of f
exist and are analytic in D.

Converse of Cauchy’s Theorem:

Theorem (Morera’s Theorem)

¸Let f (z) be continuous in a domain D with the property that
C f (z) dz = 0 for every simple closed contour C. Then f (z) is analytic.

Pinaki Pal (NIT Durgapur) MA331; Odd Semester, 2022-23 24 / 84

Theorem (Cauchy Inequality)
Let f (z) be analytic in the open disk ∆(a, r ) = {z ∈ C : |z − a| < r } and
|f (z)| ≤ M for all z ∈ ∂∆(a, r ). Then

Mn!
|f (n) (a)| ≤ , n ∈ N.
rn
Proof: By Cauchy integral formula for derivatives, we have
˛
(n) n! f (z)
f (a) = dz, C : |z − a| = r
2πi (z − a)n+1
C

Then
˛ ˛
(n) n! |f (z)| Mn! Mn! Mn!
|f (a)| ≤ n+1
|dz| ≤ |dz| = 2πr = n .
2π |z − a| 2πr n+1 2πr n+1 r
C C

Pinaki Pal (NIT Durgapur) MA331; Odd Semester, 2022-23 25 / 84

Theorem (Liouville’s Theorem)
Any bounded entire function is constant.

Proof:
Let f (z) be a bounded entire function.
Then there exists a positive constant M such that |f (z)| ≤ M for all
z ∈ C.
Let a be any point of the complex plane and C be the
circumference of the circle |z − a| = R.
M
Then, by Cauchy’s inequality, we have |f 0 (a)| ≤ .
R
Since f (z) is an entire function, R may be taken arbitrarily large
and, therefore, |f 0 (a)| ≤ M
R → 0 as R → ∞.
Thus f 0 (a) = 0. Since a is arbitrary, f 0 (z) = 0 for all z ∈ C.
Thus f is constant.
Example:
The function sin z is an entire function and it is not bounded.
The function cos z is an entire function and it is not bounded.
Pinaki Pal (NIT Durgapur) MA331; Odd Semester, 2022-23 26 / 84
Power Series
A series of the form
∞
X
an (z − a)n = a0 + a1 (z − a) + a2 (z − a)2 + · · ·
n=0

is known as the power series about the point z = a where an and

a are fixed complex numbers and z is a complex variable.
P∞ n
The
P∞ power series n=0 an (z − a) is called absolutely convergent
if n=0 |an ||z − a|n is convergent.
For every power series ∞ n
P
n=0 an (z − a) there exist a real number
R such that for every z in |z − a| < R, the series is absolutely
convergent and for every z in |z − a| > R, the series is divergent.
The number R is called the radius of convergence of the power series and the circle |z − a| = R is called the circle of
convergence of the power series.

No general statement can be made about the convergence of a power series on the circle of convergence.

We write R = ∞ if the power series convergence for all z ∈ C and R = 0 if the power series convergence only at z = a.

Pinaki Pal (NIT Durgapur) MA331; Odd Semester, 2022-23 27 / 84

Power Series

Radius of convergence of a power series:

∞
X
If R is the radius of convergence of the power series an (z − a)n
n=0
then
1 an+1
= lim (Ratio Test)
R n→∞ an
or,
1
= lim |an |1/n (Root Test)
R n→∞

Pinaki Pal (NIT Durgapur) MA331; Odd Semester, 2022-23 28 / 84

Power Series ∞
X zn
Example: Find the radius of convergence of the power series .
n!
n=0
1
Solution: Here an = n! and a = 0. Then

1 an+1 n! 1
= lim = lim = lim = 0.
R n→∞ an n→∞ (n + 1)! n→∞ n + 1

Thus the radius of convergence is R = ∞ and the series converges for all
z ∈ C.
∞
X n
Example: Find the radius of convergence of the power series 2
zn.
4n + 1
n=0
n
Solution: Here an = 2
and a = 0. Then
4n + 1
1 an+1 n+1 4n2 + 1
= lim = lim 2
× = 1.
R n→∞ an n→∞ 4(n + 1) + 1 n
Thus the radius of convergence is R = 1 and the series converges for |z| < 1
and diverges for |z| > 1.
Pinaki Pal (NIT Durgapur) MA331; Odd Semester, 2022-23 29 / 84
Power Series

Example: Find the radius of convergence of the power series

X∞
(3 + i)−n z n .
n=0

Solution: Here an = (3 + i)−n and a = 0. Then

1 1 1
= lim |an |1/n = lim =√ .
R n→∞ n→∞ 3 + i 10
√
Thus the√ radius of convergence is R√ = 10 and the series converges
for |z| < 10 and diverges for |z| > 10.

Pinaki Pal (NIT Durgapur) MA331; Odd Semester, 2022-23 30 / 84

Power Series
∞
X
Example: Find the radius of convergence of the power series (3 + i)−n z 3n .
n=0
∞
X
Solution: Let w = z 3 . Then the series becomes an w n where
n=0
an = (3 + i)−n . Then

1 1/n 1 1
= lim |an | = lim =√ .
R n→∞ n→∞ 3 + i 10
∞
X √
Thus the series an w n convergence is converges for |w| < 10 and
n=0
√
diverges for |w| > 10.
∞
X
Hence the power series (3 + i)−n z 3n convergence is converges for
n=0
|z| < 101/6 and diverges for |z| > 101/6 . Thus the radius of convergence of
the given series is R = 101/6 .
Pinaki Pal (NIT Durgapur) MA331; Odd Semester, 2022-23 31 / 84
Power Series

Theorem
∞
X
Let R be the radius of convergence of the power series an (z − a)n
n=0
and it converges to the function f (z) in |z − a| < R. The f (z) is analytic
in |z − a| < R, i.e., a power series represents an analytic function
inside its circle of convergence.

Pinaki Pal (NIT Durgapur) MA331; Odd Semester, 2022-23 32 / 84

Taylor Series
Theorem (Taylor’s theorem)
Let f (z) be analytic in a domain D whose boundary is C. Then for all
z ∈ D we have
∞
X
f (z) = an (z − a)n , |z − a| < δ
n=0

f (n) (a)
where an = are called Taylor’s coefficients of f (z) and δ is the
n!
distance from a to the nearest point of C.

Remark:
The infinite series is called the Taylor’s series and
‰
f (n) (a) 1 f (z)
an = = dz.
n! 2πi |z−a|=δ (z − a)n+1
If a = 0 then the Taylor’s series is called Maclaurin’s series.
Pinaki Pal (NIT Durgapur) MA331; Odd Semester, 2022-23 33 / 84
Taylor Series
When D is a disk then Taylor’s theorem can be written as:
Taylor’s theorem: Let f (z) be analytic in the disk |z − a| < R. Then
for all z in the disk
∞ (n)
X f (a)
f (z) = (z − a)n , |z − a| < R.
n!
n=0
∞ (n)
X f (a)
The Taylor series (z − a)n converges to f (z) in the disk
n!
n=0
|z − a| < R.
Remark:
Any function f (z) which is analytic at a point z0 must have a
Taylor’s series about z0 valid in some nbd of z0 .
If f is an entire function then the radius of convergence can be
chosen arbitrary large, i.e., the region of validity of the Taylor’s
series becomes |z − z0 | < ∞.
Pinaki Pal (NIT Durgapur) MA331; Odd Semester, 2022-23 34 / 84
Taylor Series
1
Example: Find the Taylor series of f (z) = about z = 0.
1−z
Solution: Clearly,
0 1 00 1.2 000 1.2.3 (n) n!
f (z) = , f (z) = , f (z) = ,...,f (z) =
(1 − z)2 (1 − z)3 (1 − z)4 (1 − z)n+1

Thus f (n) (0) = n! and so the Taylor series of f (z) about z = 0 is

∞ (n) ∞
X f (0) X
f (z) = (z − 0)n = zn = 1 + z + z2 + z3 + · · · .
n!
n=0 n=0

If R is the radius of convergence of the Taylor series then

1 an+1 1
= lim = lim = 1.
R n→∞ an n→∞ 1

1
Example: The Taylor series of f (z) = about z = 0 is
1+z
∞ (n) ∞
X f (0) X
f (z) = (z − 0)n = (−1)n z n , |z| < 1.
n!
n=0 n=0

Pinaki Pal (NIT Durgapur) MA331; Odd Semester, 2022-23 35 / 84

Taylor Series

z +2
Example: Find the Taylor series of f (z) = about z = 0.
1 − z2
Solution: Here
z +2 3 1 1 1
f (z) = 2
= +
1−z 21−z 21+z
∞ ∞
! !
3 X n 1 X
= z + (−1)n z n , |z| < 1
2 2
n=0 n=0

= 2 + z + 2z 2 + z 3 + · · · , |z| < 1.

Pinaki Pal (NIT Durgapur) MA331; Odd Semester, 2022-23 36 / 84

Taylor Series
1
Example: Find the Taylor series of f (z) = about z = 0.
(z − 2)(z − 3)
Solution:
1 1 1 1 1 1 1
f (z) = = − =− +
(z − 2)(z − 3) z −3 z −2 3 1 − z/3 2 1 − z/2
∞ ∞
! !
1 X n 1 X
=− (z/3) + (z/2)n
3 2
n=0 n=0

where the first series is valid in |z| < 3 and the second series is valid in
|z| < 2. Thus both series are valid in |z| < 2. Hence
∞  
X 1 1
f (z) = − zn, |z| < 2.
2n+1 3n+1
n=0

Pinaki Pal (NIT Durgapur) MA331; Odd Semester, 2022-23 37 / 84

Laurent’s theorem: Laurent Series
Let f (z) be analytic in the annular region (annulus) D : R1 < |z − a| < R2 .
Then for each z ∈ D,
∞ ∞
X X bn
f (z) = an (z − a)n +
(z − a)n
n=0 n=1

with ‰
1 f (z)
an = dz and bn = a−n
2πi C (z − a)n+1
where C is any simple closed contour lying in D that makes a complete
counterclockwise revolution about a.
Alternatively, y
∞
R2
X
f (z) = cn (z − a)n ,
n=−∞
a C
where R1
‰ z
1 f (z)
cn = dz. x
2πi C (z − a)n+1
Pinaki Pal (NIT Durgapur) MA331; Odd Semester, 2022-23 38 / 84
Laurent Series

Remark:
Suppose f (z) is analytic inside the disk |z − a| < R1 . Then by
Cauchy’s theorem
‰
1 f (z)
bn = dz = 0
2πi C (z − a)−n+1

In this case, the Laurent series reduce to the Taylor series

∞
X
f (z) = an (z − a)n .
n=0

Pinaki Pal (NIT Durgapur) MA331; Odd Semester, 2022-23 39 / 84

Laurent Series
1
Example: Find the Taylor’s/Laurent’s series of f (z) = in
1−z
(i) |z| < 1 (ii) |z| > 1.
Solution: (i) Let |z| < 1. Then
∞
1 X
f (z) = = zn.
1−z
n=0

(ii) Let 1 < |z| < ∞. Then

∞  n
1 1 1X 1 1
f (z) = = 1

=− , <1
1−z −z 1 − z z z z
n=0

∞
X 1
=− , 1 < |z| < ∞.
z n+1
n=0
1
Example: Find the Laurent’s series of f (z) = in |z| > 1.
1+z
Solution:
∞  n ∞
1 1 1 X n 1 X n 1
f (z) = =   = (−1) = (−1) , 1 < |z| < ∞.
1+z z 1 + z1 z n=0 z n=0
z n+1

Pinaki Pal (NIT Durgapur) MA331; Odd Semester, 2022-23 40 / 84

Laurent Series
1
Example: Find the Taylor’s/Laurent’s series of f (z) = in
(z + 1)(z + 3)
(i) |z| < 1 (ii) 1 < |z| < 3 (iii) |z| > 3 (iv ) 0 < |z + 1| < 2
Solution: We have
1 1 1
f (z) = = −
(z + 1)(z + 3) 2(z + 1) 2(z + 3)
(i) Let |z| < 1. Then
1 1 1 1 z −1
f (z) = − = (1 + z)−1 − 1+
2(z + 1) 2(z + 3) 2 6 3

z2 z3
 
1 1 z
= (1 − z + z 2 − z 3 + · · · ) − 1− + − + ···
2 6 3 9 27
1 4 13 2
= − z+ z − ··· .
3 9 27
This is a Taylor’s series valid for |z| < 1.
Pinaki Pal (NIT Durgapur) MA331; Odd Semester, 2022-23 41 / 84
Laurent Series
(ii) Let 1 < |z| < 3. Then

−1
1 1 1 1 1 z −1
  
f (z) = − = 1+ − 1+
2(z + 1) 2(z + 3) 2z z 6 3

z2 z3
!
1 1 1 1 1 z
 
= 1− + − + ··· − 1− + − + ···
2z z z2 z3 6 3 9 27

2
z3
!
1 1 1 1 1 z z
 
= − + − + ··· + − + − + − ··· .
2z 2z 2 2z 3 2z 4 6 18 54 162

This is a Laurent’s series valid for 1 < |z| < 3.

(iii) Let |z| > 3. Then

1 1 1 1 −1 1 3 −1
   
f (z) = − = 1+ − 1+
2(z + 1) 2(z + 3) 2z z 2z z

1 1 1 1 1 3 9 27
   
= 1− + − + ··· − 1− + − + ···
2z z z 2 z 3 2z z z 2 z 3

1 4 13
= − + − ··· .
z2 z3 z4

This is a Laurent’s series valid for |z| > 3.

Pinaki Pal (NIT Durgapur) MA331; Odd Semester, 2022-23 42 / 84

Laurent Series

(iv ) Let 0 < |z + 1| < 2. We substitute u = z + 1. Then 0 < |u| < 2

and so
1 1 1 1 1 1 u −1
f (z) = − = − = − 1+
2(z + 1) 2(z + 3) 2u 2(u + 2) 2u 4 2

u u2 u3
 
1 1
= − 1− + − + ···
2u 4 2 4 8

(z + 1) (z + 1)2 (z + 1)3
 
1 1
= − 1− + − + ··· .
2(z + 1) 4 2 4 8

This is a Laurent’s series valid for 0 < |z + 1| < 2.

Pinaki Pal (NIT Durgapur) MA331; Odd Semester, 2022-23 43 / 84

Laurent Series
z
Example: Find the Taylor’s/Laurent’s series of f (z) = in
z 2 − 3z + 2

(i) |z| < 1 (ii) 1 < |z| < 2 (iii) |z| > 2.

Solution: We have
1 z 1 2
f (z) = = = − .
z 2 − 3z + 2 (z − 1)(z − 2) 1−z 2−z

(i) Let |z| < 1. Then

−1 ∞ ∞  n
1 2 z z

−1 n
X X
f (z) = − = (1 − z) − 1− = z − .
1−z 2−z 2 n=0 n=0
2

(ii) Let 1 < |z| < 2. Then

−1 −1 ∞  n ∞  n
1 2 1 1 z 1 X 1 z
  X
f (z) = − =− 1− − 1− =− − .
1−z 2−z z z 2 z n=0 z n=0
2

(iii) Let |z| > 2. Then

−1 −1 ∞  n ∞  n
1 2 1 1 2 2 1 X 1 2 X 2
 
f (z) = − =− 1− + 1− =− + .
1−z 2−z z z z z z n=0 z z n=0 z

Pinaki Pal (NIT Durgapur) MA331; Odd Semester, 2022-23 44 / 84

Laurent Series
Example: Suppose that f (z) is an entire function and that
|f (z)| ≤ M|z|k as |z| → ∞ for some k > 0. Then f (z) is a polynomial of
degree at most k .
Solution: Since f (z) is entire function, we have
∞ ∞ (n)
X X f (0)
f (z) = an z n = zn, z ∈ C.
n!
n=0 n=0

By Cauchy’s inequality, on the circle |z| = R (R is a very large

number), we have

f (n) (0) MR k M
|an | = ≤ = n−k
n! Rn R

Letting R → ∞, we see that an = 0 whenever n > k . Hence, f (z) is a

polynomial of degree at most k .
Pinaki Pal (NIT Durgapur) MA331; Odd Semester, 2022-23 45 / 84
Zeros of analytic function
Definition
Let f (z) be analytic at z0 . Then z0 is called a zero of f (z) if f (z0 ) = 0.

Theorem (The Fundamental Theorem of Algebra)

Every nonconstant polynomial has at least one zero in C.

Theorem
Every polynomial of degree n has exactly n (not necessarily distinct)
zeros in C.

Theorem (Zeros are isolated)

Suppose f (z) is analytic at a point z = z0 . Then either f (z) ≡ 0 in
some neighborhood of z0 , or there exists a real number r such that
f (z) 6= 0 in the punctured disk 0 < |z − z0 | < r .

Pinaki Pal (NIT Durgapur) MA331; Odd Semester, 2022-23 46 / 84

Zeros of analytic function
Definition
Let f (z) be analytic at z0 . Then f (z) has a Taylor series expansion in a
nbd of z0 as follows
∞
X
f (z) = an (z − z0 )n , |z − z0 | < δ
n=0

f (n) (z0 )
where an = .
n!
If a0 = a1 = a2 = · · · = am−1 = 0 but am = 6 0 then f (z) can be written
as
∞
X
f (z) = (z − z0 )m an (z − z0 )n−m = (z − z0 )m φ(z),
n=m

where φ(z) is analytic at z0 and φ(z0 ) 6= 0 and we say that f (z) has a
zero of order m at z = z0 .

Pinaki Pal (NIT Durgapur) MA331; Odd Semester, 2022-23 47 / 84

Singularity
Singular points: A point z0 is called a singular point of a function f (z) if
f (z) is not analytic at z0 (may even be undefined at z0 ) but is analytic at
some point in every neighborhood of z0 . The function f (z) is said to have
a singularity at z0 .
Isolated singular point: The singular point z0 is called isolated singular
point of f (z) if there exist a nbd of z0 containing no other singular point of
f (z).
Non-isolated singular point: The singular point z0 is called a
non-isolated singular point of f (z) if every neighborhood of z0 contains at
least one singularity of f (z) other than z0 .
Example: Consider the function f (z) = |z|2 = zz. This function is nowhere
analytic. It has no singular point.
sin z
Example: Consider the function f (z) = tan z = . Singular points of f (z)
cos z
are given by
π 3π
cos z = 0 =⇒ z = ± , ± , · · · .
2 2
All these singular points are isolated.
Pinaki Pal (NIT Durgapur) MA331; Odd Semester, 2022-23 48 / 84
Singularity
1 sin z1
Example: Consider the function f (z) = tan = .
z cos z1
Singular points of f (z) are given by

1 1 π 2
cos = 0 =⇒ = (2n + 1) =⇒ z = , n ∈ Z.
z z 2 (2n + 1)π

All these singular points are isolated.

Note that f (z) is not defined at z = 0 but f (z) is analytic in some nbd of
z = 0. Thus z = 0 is also a singular points of f (z).
2
Since lim = 0, every nbd of z = 0 contains many other
n→∞ (2n + 1)π
singular point of f (z).
Thus z = 0 is a non-isolated singular point of f (z).

Pinaki Pal (NIT Durgapur) MA331; Odd Semester, 2022-23 49 / 84

Singularity

Isolated singularity of a function f (z) at z = z0 can be further classified.

Let z = z0 be an isolated singularity of an analytic function f (z). Then in a deleted nbd of z = z0 , f (z) has a Laurent series
expansion of the form
∞ ∞
n −n
X X
f (z) = an (z − z0 ) + bn (z − z0 ) , 0 < |z − z0 | < δ.
n=0 n=1

∞
−n
X
The term bn (z − z0 ) is called the principal part of f (z) at z = z0 . Now there are three possibilities.
n=1

1. Removable singularity:

If the principal part in Laurent expansion of f (z) does not contain any term, that is bn = 0 for all n ∈ N then z0 is called a
removable singularity of f (z).

If a function f (z) is not defined at z0 but lim f (z) exist then z0 is called a removable singularity of f (z).
z→z0

In this case, if we define f (z) at z0 as equal to lim f (z) then f (z) will be analytic at z0 .
z→z0

Pinaki Pal (NIT Durgapur) MA331; Odd Semester, 2022-23 50 / 84

Singularity

2. Pole:
If the principal part has only a finite number of terms, that is bn = 0 for all n > m for some m ∈ N and bm 6= 0 then z0 is
called a pole of order m of f (z).

If m = 1 then z0 is called a simple pole. If m = 2 then z0 is called a double pole.

In this case, f (z) has the form
∞ m
n −n
X X
f (z) = an (z − z0 ) + bn (z − z0 ) .
n=0 n=1

If z0 is an isolated singularity and we can find a positive integer m such that

m
lim (z − z0 ) f (z) = A 6= 0
z→z0

then z0 is called a pole of order m of f (z).

φ(z)
An isolated singularity z0 of f (z) is a pole of order m iff f (z) can be expressed as f (z) = where φ(z) is
(z − z0 )m
analytic at z0 and φ(z0 ) 6= 0.
An isolated singularity z0 of f (z) is a pole of f (z) iff lim f (z) = ∞.
z→z0

Pinaki Pal (NIT Durgapur) MA331; Odd Semester, 2022-23 51 / 84

Singularity
3. Essential Singularity:
If the principal part in Laurent expansion of f (z) contains an infinite
number of terms, then z0 is called an isolated essential singularity of
f (z).
In this case, f (z) has the form
∞
X ∞
X
f (z) = an (z − z0 )n + bn (z − z0 )−n .
n=0 n=1

If z0 is a isolated singularity and there exist no positive integer m such

that
lim (z − z0 )m f (z) = A
z→z0

is called an essential singularity of f (z).

If z0 is a isolated singularity of f (z) and lim f (z) does not exist in C∞
z→z0
then z0 is called an essential singularity of f (z).
Remark: A point z0 is a pole of order m of f (z) iff z0 is a zero of order m of
1/f (z).
Pinaki Pal (NIT Durgapur) MA331; Odd Semester, 2022-23 52 / 84
Singularity
sin z
Example: Consider the function f (z) = .
z
sin z
Note that f (0) is not defined but lim = 1. Then z = 0 is a
z→0 z
removable singularity of f (z).
1 3
Example: Consider the function f (z) = 5
+ .
z(z − 2) (z − 2)2
f (z) has isolated singularity at z = 0, 2. Now,
 
1 3z 1
lim zf (z) = lim + =− .
z→0 z→0 (z − 2)5 (z − 2)2 32
 
5 1 3 1
lim (z − 2) f (z) = lim + 3z(z − 2) = .
z→0 z→2 z 2
Thus z = 0 is a simple pole and z = 2 is a pole of order 5 of f (z).
1
Example: The function f (z) = e z has essential singularity at z = 0
because
1 1 1 1
f (z) = e z = 1 + + + ··· .
z 2! z 2
Pinaki Pal (NIT Durgapur) MA331; Odd Semester, 2022-23 53 / 84
Singularity
sin z
Example: Find the singularities of the function f (z) = tan z =
cos z
and classify them.
Solution: Singularities of f (z) are given by
π
cos z = 0 =⇒ z = zn := (2n + 1) , n ∈ N.
2
Since lim f (z) does not exist, there are no removable singularity.
z→zn

z − zn 0
lim (z − zn )f (z) = lim ( form)
z→zn z→zn cot z 0
1
= lim = −1
z→zn − sec2 z
Thus all the singularities of f (z) are simple pole.

Pinaki Pal (NIT Durgapur) MA331; Odd Semester, 2022-23 54 / 84

Singularity
1
Example: Find the nature of the singularities of f (z) = .
z(ez − 1)
Solution: Singularities of f (z) are given by

z = 0 & ez = 1 =⇒ z = 0, 2nπi, n ∈ Z.

Thus z = 0 is a double pole. All other poles are simple.

1
Example: Find the nature of the singularities of f (z) = z sin .
z
Solution: The only singularity of f (z) is at z = 0. Note that
 
1 1 1 1 1 1
f (z) = z sin = z − 3
+ 5
− ··· = 1 − 2
+ − ···
z z 3!z 5!z 3!z 5!z 4

Since the series does not terminate, z = 0 is an essential singularity of

f (z).

Pinaki Pal (NIT Durgapur) MA331; Odd Semester, 2022-23 55 / 84

Singularity
Isolated singularity at ∞: A function f (z) has an isolated singularity at
z = ∞ if and only if f (1/z) has an isolated singularity at z = 0.
Moreover, we make the definition that the singularity of f (z) at z = ∞ is
removable, a pole, or essential according as the singularity of f (1/z) at z = 0
is removable, a pole, or essential.
Example:
1 The function f (z) = z 2 + 1 has a pole of order 2 at z = ∞ because
f (1/z) = (1/z 2 ) + 1 has a pole of order 2 at z = 0.
2 The function f (z) = ez has an isolated essential singularity at z = ∞
because f (1/z) = e1/z has an isolated essential singularity at z = 0.
1 z3
3 Let f (z) = . Then f (1/z) = 1+4z 2
which is analytic at z = 0.
z(z 2 + 4)
Thus f (z) is analytic at z = ∞.

Meromorphic function: A function f (z) is said to be meromorphic if it is

analytic in the finite complex plane C except possibly at a finite number of
poles.
Pinaki Pal (NIT Durgapur) MA331; Odd Semester, 2022-23 56 / 84
Residue
Residue at a finite point: We recall that if f (z) has an isolated
singularity at z0 , then in a deleted nbd of z = z0 , f (z) has a Laurent
series expansion of the
∞
X ∞
X
f (z) = an (z − z0 )n + bn (z − z0 )−n , 0 < |z − z0 | < δ.
n=0 n=1

with ‰
1 f (z)
an = dz and bn = a−n
2πi C (z − a)n+1
where C is any simple closed contour lying in the nbd of z0 that makes
a complete counterclockwise revolution about z0 .
The coefficient b1 is called the residue of f (z) at z0 and is denoted by
Res [f (z); z0 ]. Thus
‰
1
Res [f (z); z0 ] = b1 = f (z) dz.
2πi C

Pinaki Pal (NIT Durgapur) MA331; Odd Semester, 2022-23 57 / 84

Residue
Residue at ∞: If f (z) has an isolated singularity at ∞, then in a deleted nbd
of ∞, f (z) has a Laurent series expansion of the
∞
X ∞
X
f (z) = an z n + bn z −n , δ < |z| < ∞.
n=0 n=1

Let C be any simple closed contour lying in the nbd of ∞ that makes a
complete clockwise revolution about ∞. Then
 ∞  ∞ 
1 1 X 1 X
f (z) dz = an n
z dz + bn z −n dz = −b1 .
2πi C 2πi C 2πi C
n=0 n=1

Therefore, we define the residue of f (z) at z = ∞ as


1
Res [f (z); ∞] = f (z) dz
2πi C

= −b1 = −(coefficient of 1/z in the Laurent series expansion of f (z)).

Remark:    
1 1
Res [f (z); ∞] = −Res f ;0 .
z2 z

Pinaki Pal (NIT Durgapur) MA331; Odd Semester, 2022-23 58 / 84

Residue
Theorem (Residue at a pole)
If f(z) has a pole of order m at z = z0 , then

1 d m−1
Res [f (z); z0 ] = lim [(z − z0 )m f (z)].
(m − 1)! z→z0 dz m−1

In particular, if f (z) has a simple pole at z0 , then

Res [f (z); z0 ] = lim (z − z0 )f (z).

z→z0

Theorem (Residue at a pole)

Let f (z) and g(z) be analytic at z0 . If g(z) has a simple pole at z0 and
f (z0 ) 6= 0, then  
f (z) f (z0 )
Res ; z0 = 0 .
g(z) g (z0 )

Remark: Let z0 is a essential singularity of f (z). To find Res [f (z); z0 ], we use

the Laurent series expansion of f (z) about z0 .
Pinaki Pal (NIT Durgapur) MA331; Odd Semester, 2022-23 59 / 84
Residue
Example: Find the singularities in the complex plane and the residue
z 2 − 2z
at those singular points of the function
(z + 1)2 (z 2 + 4)
Solution: The function f (z) has a pole of order 2 at z = −1 and
simple poles at z = ±2i. Therefore
1 d d z 2 − 2z
Res [f (z); −1] = lim [(z + 1)2 f (z)] = lim
(2 − 1)! z→−1 dz z→−1 dz (z 2 + 4)

(z 2 + 4)(2z − 2) − (z 2 − 2z)(2z) 14
= lim =− .
z→−1 (z 2 + 4)2 25

z 2 − 2z 7+i
Res [f (z); 2i] = lim (z − 2i)f (z) = lim 2
= .
z→2i z→2i (z + 1) (z + 2i) 25

z 2 − 2z 7−i
Res [f (z); −2i] = lim (z + 2i)f (z) = lim 2
= .
z→−2i z→−2i (z + 1) (z − 2i) 25
Pinaki Pal (NIT Durgapur) MA331; Odd Semester, 2022-23 60 / 84
Residue
Example: Find the singularities in the complex plane and the residue
2
ez
at those singular points of the function .
(z − i)3
Solution: The function f (z) has a pole of order 3 at z = i. Therefore
1 d2 1 d2 2
Res [f (z); i] = lim 2 [(z − i)3 f (z)] = lim 2 ez
(3 − 1)! z→i dz 2 z→i dz
1 2 2 1
lim (2z 2 ez + ez ) = − .
=
2 z→i e
Example: Find the singularities in the complex plane and the residue
sin z − z
at those singular points of the functions f (z) = .
z3
Solution: The function f (z) has pole of order 3 at z = 0. Therefore,
Therefore
1 d2 3 1
Res [f (z); 0] = lim [z f (z)] = lim (sin z − z) = 0
2 z→0 dz 2 2 z→0
Pinaki Pal (NIT Durgapur) MA331; Odd Semester, 2022-23 61 / 84
Residue
1/z
Example:˛ Find the singularities in the complex plane and the residue at those singular points of the function f (z) = e and
1/z
evaluate e dz.
|z|=1

Solution: The function

1/z 1 1 1
f (z) = e =1+ + + ···
z 2! z 2

has an isolated essential singularity at z = 0. Thus Res [f (z); 0] = b1 = 1 and so

˛
1/z
e dz = 2πi × Res [f (z); 0] = 2πi.
|z|=1

1
Example: Find the singularities in the complex plane and the residue at those singular points of the function f (z) = sin and
˛ z2
1
 
evaluate sin dz.
|z|=1 z2

Solution: The function

1 1 1 1
f (z) = sin = − + ···
z2 z2 3! z 6

has an isolated essential singularity at z = 0. Thus Res [f (z); 0] = b1 = 0 and so

˛
1
sin dz = 2πi × Res [f (z); 0] = 0.
|z|=1 z2

Pinaki Pal (NIT Durgapur) MA331; Odd Semester, 2022-23 62 / 84

Residue
Example: Find the singularities in the extended complex plane and the residue at those singular points of the function
zn
f (z) = .
1+z
Solution: The function f (z) has an simple pole at z = −1. Now

1
f (1/z) =
z n−1 (1 + z)

which shows that z = ∞ is a pole of order n − 1 of f (z). Now

n n
Res [f (z); −1] = lim (z + 1)f (z) = lim z = (−1)
z→−1 z→−1

We know that
1 1
   
Res [f (z); ∞] = −Res f ;0 .
z2 z

1 1 1
 
If g(z) = f = then z = 0 is a pole of order n + 1 of g(z). Thus
z2 z z n+1 (1 + z)

1 dn n+1 1 dn −1
Res [f (z); ∞] = −Res [g(z); 0] = − lim [z g(z)] = − lim (1 + z)
n! z→0 dz n n! z→0 dz n

n+1 −(n+1) n+1

= (−1) lim (1 + z) = (−1) .
z→0

Remark: In the previous example, Res [f (z); −1] + Res [f (z); ∞] = 0.

Pinaki Pal (NIT Durgapur) MA331; Odd Semester, 2022-23 63 / 84

 Residue
Theorem (Cauchy’s Residue
Theorem)
Let f (z) be analytic inside and on a simple
closed contour C except for isolated
singularities at z1 , z2 , z3 , . . . , zn inside C. Then
ˆ n
X
f (z) dz = 2πi Res [f (z); zk ]
C k =1

Theorem (Residue Theorem for C∞ )

Suppose f(z) is analytic in C∞ except for isolated singularities at
z1 , z2 , z3 , . . . , zn , ∞. Then the sum of its residues (including the point at
infinity) is zero. That is,
n
X
Res [f (z); ∞] + Res [f (z); zk ] = 0
k =1

Pinaki Pal (NIT Durgapur) MA331; Odd Semester, 2022-23 64 / 84

Residue ˆ
dz
Example: Evaluate .
|z|=2 z 2 (z 2− 1)
1
Solution: The function f (z) = has double pole at z = 0
− 1) z 2 (z 2
and simple pole at z = ±1. Note that all the singular points of f (z) lies
inside |z| = 2. Now
d d 1 −2z
Res [f (z); 0] = lim [(z − 0)2 f (z)] = lim 2
= lim 2 =
z→0 dz z→0 dz z − 1 z→0 (z − 1)2
1 1
Res [f (z); 1] = lim (z − 1)f (z) = lim 2 =
z→1 z→1 z (z + 1) 2
1 1
Res [f (z); −1] = lim (z + 1)f (z) = lim 2 =− .
z→−1 z→−1 z (z − 1) 2
Then by Cauchy’s Residue theorem
ˆ
dz
2 (z 2 − 1)
= 2πi(Res [f (z); 0] + Res [f (z); 1] + Res [f (z); −1]) = 0.
|z|=2 z

Pinaki Pal (NIT Durgapur) MA331; Odd Semester, 2022-23 65 / 84

Residue
ˆ
dz
Example: Evaluate .
|z−1|=1/2 z 2 (z 2
− 1)
1
Solution: The function f (z) = has double pole at z = 0
− 1) z 2 (z 2
and simple pole at z = ±1. Note that only the pole z = 1 of f (z) lies
inside |z − 1| = 1/2. Now

1 1
Res [f (z); 1] = lim (z − 1)f (z) = lim =
z→1 z→1 z 2 (z+ 1) 2

Then by Cauchy’s Residue theorem

ˆ
dz
2 2
dz = 2πi × Res [f (z); 1] = πi.
|z|=2 z (z − 1)

Pinaki Pal (NIT Durgapur) MA331; Odd Semester, 2022-23 66 / 84

Residue ˆ
1 − cos 2(z − 3)
Example: Evaluate dz.
|z−3|=1 (z − 3)3
Solution: First we note that
1 − cos 2(z − 3)
f (z) =
(z − 3)3
4(z − 3)2 16(z − 3)4
 
1
= 1−1+ − + ···
(z − 3)3 2! 4!
2 16(z − 3)
= − + ··· .
(z − 3) 4!
Thus f (z) has a simple pole at z = 3.
1
The Laurent’s series is in the power of z − 3. The coefficient of (z−3) is
2. Hence, Res [f (z); 3] = 2.
Then by Cauchy’s Residue theorem
ˆ
1 − cos 2(z − 3)
dz = 2πi × Res [f (z); 3] = 4πi.
|z−3|=1 (z − 3)3
Pinaki Pal (NIT Durgapur) MA331; Odd Semester, 2022-23 67 / 84
Residue ˆ
Example: Evaluate tan z dz.
|z|=2

sin z
Solution: The singularity of the function f (z) = tan z = are given by
cos z
π
cos z = 0 =⇒ z = (2n + 1) , n ∈ Z.
2
The singular points zn = (2n + 1) π2 are simple poles of f (z).
π
Note that only the pole z0 = and z−1 = − π2 of f (z) lies inside |z| = 2. Now
2

z − π2


π  π 1
Res [f (z); ] = limπ z − tan z = limπ = limπ =1
2 z→ 2 2 z→ 2 cot z z→ 2 csc2 z

z + π2


π  π 1
Res [f (z); − ] = lim π z + tan z = lim π = lim π = 1.
2 z→− 2 2 z→− 2 cot z z→− 2 csc2 z

Then by Cauchy’s Residue theorem

ˆ  π π 
tan z dz = 2πi Res [f (z); ] + Res [f (z); − ] = 4πi.
|z|=2 2 2

Pinaki Pal (NIT Durgapur) MA331; Odd Semester, 2022-23 68 / 84

Residue ˆ
dz
Example: Evaluate .
|z|=2 (z n − 1)(z − 3)
1
Solution: The function f (z) = has simple pole at z = 3 and
(z n − 1)(z − 3)
at z = zk = e2k πi/n , k = 0, 1, 2, . . . , n − 1.
Note that the poles z = zk , k = 0, 1, 2, . . . , n − 1 of f (z) lies inside |z| = 2.
Then by Cauchy’s Residue theorem
ˆ Xn
f (z) dz = 2πi Res [f (z); zk ] = −2πi (Res [f (z); 3] + Res [f (z); ∞]) .
|z|=2 k =1

Now
1 1
Res [f (z); 3] = lim (z − 3)f (z) = lim = n ,
z→3 zn − 1 3 −1 z→1

z n−1
     
1 1
Res [f (z); ∞] = −Res f ; 0 = Res ; 0 = 0.
z2 z (1 − 3z)(1 − z n )
Therefore, ˆ
dz 2πi
n − 1)(z − 3)
dz = .
Pinaki Pal (NIT Durgapur) |z|=2 (zMA331; Odd Semester, 2022-23
1 − 3n 69 / 84
Problem set
Example: Discuss continuity of the function

 Re z , z 6= 0
z
f (z) =
0, z = 0.

Solution: Let z0 6= 0. Then

Re z Re z0
lim f (z) = lim = = f (z0 ).
z→z0 z→z0 z z0
But 
Re z 1, if z = x → 0
lim f (z) = lim =
z→0 z→0 z 0, if z = iy → 0.

Thus f (z) is continuous at all complex number except at the origin.

Pinaki Pal (NIT Durgapur) MA331; Odd Semester, 2022-23 70 / 84

Problem set
xy
Example: Find the limit lim f (z) if exist where f (z) = + 2xi.
z→0 + y2 x2
xy
Solution: If f (z) = u(x, y ) + iv (x, y ) then u(x, y ) = 2 and
x + y2
v (x, y ) = 2x. Now
xy
lim u(x, y ) = lim
(x,y )→(0,0) (x,y )→(0,0) x 2 + y2
mx 2
= lim (along y = mx)
(x,y )→(0,0) x 2 + m2 x 2
m
= .
1 + m2
Since lim u(x, y ) does not exists, lim f (z) does not exists.
(x,y )→(0,0) z→0

Pinaki Pal (NIT Durgapur) MA331; Odd Semester, 2022-23 71 / 84

Problem set

Example: Determine where the following functions satisfy the Cauchy Riemann equations and where the functions are
differentiable.
2
(i) f (z) = z (ii) f (z) = zRe z (iii) f (z) = 2z + 4z + 5 .

Solution: (i) We have

2
f (z) = z =⇒ fz (z) = 2z.

Thus f (z) satisfy the C-R equation only at the origin. Hence f (z) is not differentiable at all non zero points.

At z = 0, we have to check it separately by definition. (Home work!)

(ii) We have
z(z + z)
f (z) = zRe z = =⇒ fz (z) = z/2.
2
Thus f (z) satisfy the C-R equation only at the origin. Hence f (z) is not differentiable at all non zero points.

At z = 0, we have to check it separately by definition. (Home work!)

f (z) = 2z + 4z + 5 =⇒ fz (z) = 4.

Thus f (z) does not satisfy the C-R equation at any points of the complex plane. Hence f (z) is not differentiable at at any points of

the complex plane.

Pinaki Pal (NIT Durgapur) MA331; Odd Semester, 2022-23 72 / 84

Problem set

Example: If f (z) is continuous at a point z0 , show that f (z) is also

continuous at z0 . Is the same true for differentiability at z0 ?
Solution: Let f (z) = u(x, y ) + iv (x, y ). If f (z) is continuous at a point
z0 then u(x, y ) and v (x, y ) is continuous at a point z0 . Thus
f (z) = u(x, y ) − iv (x, y ) is also continuous at z0 .
The same is not true for differentiability at z0 . For example, let
f (z) = z. Then f (z) is differentiable at z = 0 but f (z) = z is not
differentiable at z = 0.

Pinaki Pal (NIT Durgapur) MA331; Odd Semester, 2022-23 73 / 84

Problem set
Example: Find the values of the constants a, b, c so that the following
functions becomes entire function:
(i) f (z) = x + ay − i(bx + cy ) (ii) f (z) = a(x 2 + y 2 ) + ibxy + c.
Solution: (i) If f (z) = x + ay − i(bx + cy ) is entire then it must satisfy the
C-R equation at all points. Here u(x, y ) = x + ay and v (x, y ) = −(bx + cy ).
Then

ux = vy =⇒ 1 = −c =⇒ c = −1
uy = −vx =⇒ a = b.

(ii) If f (z) = a(x 2 + y 2 ) + ibxy + c is entire then it must satisfy the C-R
equation at all points. Here u(x, y ) = a(x 2 + y 2 ) + c and v (x, y ) = bxy . Then

ux = vy =⇒ 2ax = bx =⇒ 2a = b
uy = −vx =⇒ 2ay = −by =⇒ 2a = −b.

Thus a = b = 0.

Pinaki Pal (NIT Durgapur) MA331; Odd Semester, 2022-23 74 / 84

Problem set
∂u ∂u
Example: Let f (z) = u + iv be analytic. If = u1 (x, y ) and = u2 (x, y ) then show that
∂x ∂y

ˆ
f (z) = [u1 (z, 0) − iu2 (z, 0)] dz.

Solution: We know that

0
f (z) = ux + ivx = ux − iuy = u1 (x, y ) − iu2 (x, y ).

Substituting y = 0, we get
0
f (x) = u1 (x, 0) − iu2 (x, 0).

Replacing x by z, we get ˆ
0
f (z) = u1 (z, 0) − iu2 (z, 0) =⇒ [u1 (z, 0) − iu2 (z, 0)] dz.

∂v ∂v
Example: Let f (z) = u + iv be analytic. If = v1 (x, y ) and = v2 (x, y ) then show that
∂y ∂x

ˆ
f (z) = [v1 (z, 0) + iv2 (z, 0)] dz.

Remark: This method of constructing an analytic function is called Milne-Thomson’s method.

Pinaki Pal (NIT Durgapur) MA331; Odd Semester, 2022-23 75 / 84

Problem set

Example: If u = e−x (x sin y − y cos y ) then find analytic function

f (z) = u + iv in terms of z.
Solution: We have,

ux = e−x sin y − xe−x sin y + ye−x cos y = u1 (x, y ),

uy = xe−x cos y + ye−x sin y − e−x cos y = u2 (x, y ).

Therefore, by Milne’s method

ˆ ˆ
f (z) = [u1 (z, 0) − iu2 (z, 0)] dz = [0 − i(ze−z − e−z )] dz = ize−z + c.

Pinaki Pal (NIT Durgapur) MA331; Odd Semester, 2022-23 76 / 84

Problem set
Example: If u − v = (x − y )(x 2 + 4xy + y 2 ) then find analytic function
f (z) = u + iv in terms of z.
Solution: If f (z) = u + iv then if (z) = −v + iu. Thus
(1 + i)f (z) = (u − v ) + i(u + v ) = U + iV = F (z)
is analytic function. Here
U = u − v = (x − y )(x 2 + 4xy + y 2 ).
Hence,
Ux = 3x 2 + 6xy − 3y 2 = φ1 (x, y ), Uy = 3x 2 − 6xy − 3y 2 = φ2 (x, y ).
Therefore, by Milne’s method
ˆ ˆ
F (z) = [φ1 (z, 0) − iφ2 (z, 0)] dz = [3z 2 − i3z 2 ] dz = (1 − i)z 3 + c

Thus
1−i 3
F (z) = (1 + i)f (z) = (1 − i)z 3 + c =⇒ f (z) = z + c 0 = −iz 3 + c 0
1+i
Pinaki Pal (NIT Durgapur) MA331; Odd Semester, 2022-23 77 / 84
Problem set
∞
X nk
Example: Find the radius of convergence of the power series zn.
an
n=1
k
n
Solution: Here an = .
an
k
(n + 1)k an

1 an+1 1 1 1
= lim = lim = lim 1 + = .
R n→∞ an n→∞ nk an+1 a n→∞ n a

Thus the radius of convergence of the power series is R = a.

∞
X
Example: Find the radius of convergence of the power series n1/n (z + i)n .
n=1

Solution: Here an = n1/n .

1  1/n
1/n
= lim |an | = lim n1/n = 10 = 1.
R n→∞ n→∞

Thus the radius of convergence of the power series is R = 1.

Pinaki Pal (NIT Durgapur) MA331; Odd Semester, 2022-23 78 / 84

Problem set
Example: Find the radius of convergence of the power series
∞
X n! n
z .
nn
n=1
n!
Solution: Here an = .
nn
(n + 1)!nn nn 1 −n
 
1 an+1
= lim = lim = lim = lim 1 +
R n→∞ an n→∞ n!(n + 1)n+1 n→∞ (n + 1)n n→∞ n
Thus the radius of convergence of the power series is R = e.
Example: Find the radius of convergence of the power series
∞
X 3n
z 2n .
n2 + 4n
n=1
∞
X 3n
Solution: Let w = z 2 . Then the given series becomes w n.
n2 + 4n
n=1
3n
Here an = n2 +4n
.
1
Pinaki Pal (NIT Durgapur) an+1 n2 + 4n
MA331; Odd Semester, 2022-23 79 / 84
Problem set
ˆ
Example: Find z dz along the following curves:
C

(i) z(t) = e2it , t ∈ [−π, π] (ii) z(t) = t + it, t ∈ [0, 2].

Solution: (i) If C : z(t) = e2it , t ∈ [−π, π] then
ˆ ˆ π ˆ π ˆ π
0 −2it 2it
z dz = z(t)z (t) dt = e (2i)e dt = 2i dt = 4πi.
C −π −π −π

(ii) If C : z(t) = t + it, t ∈ [0, 2] then

ˆ ˆ 2
z dz = z(t)z 0 (t) dt
C 0
ˆ 2 ˆ 2
= (t − it)(1 + i) dt = (1 + i)(1 − i) t dt = 4.
0 0

Pinaki Pal (NIT Durgapur) MA331; Odd Semester, 2022-23 80 / 84

Problem set

ˆ
2z + 1 3π
Example: Show that dz ≤ .
|z|=1 5 + z2 2
2z + 1
Solution: Let f (z) = . On |z| = 1, we have
5 + z2
2|z| + 1 3
|f (z)| ≤ 2
≤ .
5 − |z| 4

Then by ML-formula
ˆ ˆ ˆ
2z + 1 2z + 1 3 3 3π
dz ≤ |dz| ≤ |dz| = ×2π = .
|z|=1 5 + z2 |z|=1 5+z 2 4 |z|=1 4 2

Pinaki Pal (NIT Durgapur) MA331; Odd Semester, 2022-23 81 / 84

 Problem set
y

ˆ
2
Example: Evaluate |z| dz along the square with vertices 0, 1, 1 + i, i,
C C3
Solution: Let C = C1 + C2 + C3 + C4 where C1 : z(t) = t, t ∈ [0, 1],
C2 : z(t) = 1 + it, t ∈ [0, 1], C3 : z(t) = (1 − t) + i, t ∈ [0, 1], C4 C2
C4 : z(t) = i(1 − t), t ∈ [0, 1].

C1 x

ˆ ˆ ˆ ˆ ˆ
2 2 2 2 2
|z| dz = |z| dz + |z| dz + |z| dz + |z| dz
C C1 C2 C3 C4

ˆ 1 ˆ 1 ˆ 1 ˆ 1
2 2 2 2
= t dt + |1 + it| (i) dt + |(1 − t) + i| (−1) dt + |i(1 − t)| (−i) dt
0 0 0 0

ˆ 1 ˆ 1 ˆ 1 ˆ 1
2 2 2 2
= t dt + i (1 − t + 2it) dt − (−2t + t + 2i(1 − t)) dt − i (1 − t) dt
0 0 0 0

= ....

Pinaki Pal (NIT Durgapur) MA331; Odd Semester, 2022-23 82 / 84

 Problem set ˆ
y
Example: Evaluate z|z| dz along the
C
upper semicircle |z| = R from R to −R
and the line segment [−R, R].
Solution: Let C = C1 + C2 where
C1 : z(t) = Reit , t ∈ [0, π],
C2 : z(t) = t, t ∈ [−R, R]. x
ˆ ˆ ˆ
z|z| dz = z|z| dz + z|z| dz
C C1 C2
ˆ π ˆ R
= Reit |Reit |(iReit ) dt + t|t| dt
0 −R
ˆ π ˆ 0 ˆ R
3 2it 2
= iR e dt + (−t ) dt + (t 2 ) dt
0 −R 0

= ....
Pinaki Pal (NIT Durgapur) MA331; Odd Semester, 2022-23 83 / 84
Problem set ˆ
Example: Evaluate 4z 3 dz along the following curves:
C

(i) z(t) = t 2 + it, t ∈ [0, 2] (ii) z(t) = t + it, t ∈ [0, 2] .

ˆ
Solution: (i) Since the function f (z) = 4z 3 is analytic in C, 4z 3 dz
C
is independent of the path.
The initial and terminal points of the curve C are z0 = z(0) = 0 and
z1 = z(2) = 4 + 2i respectively.
Note that F (z) = z 4 is the anti-derivative of f (z). Thus
ˆ
4z 3 dz = F (z1 ) − F (z0 ) = (4 + 2i)4
C
(ii) The initial and terminal points of the curve C are z0 = z(0) = 0
and z1 = z(2) = 2 + 2i respectively. Thus
ˆ h iz1
4z 3 dz = z 4 = (2 + 2i)4
C z0

Pinaki Pal (NIT Durgapur) MA331; Odd Semester, 2022-23 84 / 84