Winter Semester 2021-22: Tutorial 2

MA 2002D: Mathematics IV

Department of Mathematics
National Institute of Technology
Calicut

1 / 101
Question 1

Evaluate the following complex line integrals:

2 / 101
Question 1(i)

R
c Re z dz where c is the straight line path from 1+i to 3+2i
R3
x(1+ 21 ) dx
R
c Re z dz = 1
2
=(1+ 2i )( x2 )31 = (1+ 2i )( 29 - 12 ) = 4+2i

3 / 101
Question 1(ii)

x
(a) Given y = 2 ⇒dx=2dy.
dz = dx + idy = 2dy + idy = (2 + i)dy .
R 2+i R 2+i
0 (x − iy )2 (dx + idy ) = 0 (x 2 − y 2 − i2xy )(dx + idy )
Substituting x = 2y and dz = (2 + i)dy , weget
R 2+i R1 10−i5
0 z̄dz = 0 (3y 2 − i4y 2 )(2 + i)dy = 2 .

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(b) We have to integrate along the real axis to 2 and then vertically to
(2 + i)(that is from (0, 0) to (2, 0) then vertically to (2, 1)).
Split the integral into two parts, from (0, 0) to (2, 0) and from (2, 0)
to (2, 1),then find lines equation in both parts and integrate by
similar methods in part (a).

(c) Given 2y 2 = x ⇒dx= 4ydy.

Then dz = (4y + i)dy .
R 2+i R1 8
0 (x 2 − y 2 − i2xy )(dx + idy ) = 0 (4y 4 − y 2 − i4y 3 )(4y + i)dy = 3 − i 41
15 .

5 / 101
question 1(iii)

R 2
z dz where c is the parabola y = x 2 from (0, 0) to (2, 4).

z = x + iy = x + ix 2 =⇒ dz=(dx+i2xdx)=dx(1+i2x), then
" #2
x3 ix 6 (88+26i)
z dz = 02 (x + ix 2 )2 (1 + i2x )dx ) =
R 2
+ ix 4 − x 5 −
R
3 3 = 3
3

6 / 101
Question 1(iv)

4 − z 3 + 2)dz around the square having vertives (0,0), (3,0), (3,3),

R
c (5z

(0,3).
Solution:
Let a = (0, 0), b = (3, 0), c = (3, 3), d = (0, 3).
R Rb Rc Rd Ra
c f (z)dz= a f (z)dz + b f (z)dz + c f (z)dz + d f (z)dz
Since C is a closed curve, and by cauchy theorem the integral will be equal
to zero.

7 / 101
question1(v)

z 2 + 2 dz where c is the boundary of the triangle with vertices

R

c

0,2 and (2+i) .

we have 3 boundaries,

▶ (0,0)to (2,0): x=0 → 2, y=0

▶ (2,0) to (2,1): y=0 → 1, x=2

▶ (2,1)to (0,0): x=2 → 0, y= x2

z 2 + 2 dz
R

c
R2 R1 R0
= 0 (x 2 + 2)dx + 0 [(2 + iy )2 + 2]idy + 2 [(x + i x2 )2 + 2](1 + 2i )dx = 0

8 / 101
Question 1(vi)

z̄ 2 dz where C is the circle |z| = 3.

R
C

Solution:
R Rb ′
We know that γ f (z)dz = a f (γ(t))γ (t)dt.
′
Parametrize C as γ(t) = 3e it , 0 ≤ t ≤ 2π. So γ (t) = 3ie it .
R 2π R 2π −it
z̄ 2 dz = 9e −2it 3ie it dt = 27i
R
c 0 0 e dt
= −27[e −it ]2π
0 = 27 − 27e
−2πi .

9 / 101
Question 2

Verify Cauchy’s theorem for the following integrals.

10 / 101
Question 2(i)
(3z 2 − 2z + 4)dz where c is the boundary of the square with vertices
R
c

2 ± 2i, −2 ± 2i.

Solution:
Let ABCD be the square with vertices −2 − 2i, 2 − 2i, 2 + 2i, and −2 + 2i
respectively.

Z Z Z
(3z 2 − 2z + 4)dz = (3z 2 − 2z + 4)dz + (3z 2 − 2z + 4)dz+
c AB BC
Z Z
2
(3z − 2z + 4)dz + (3z 2 − 2z + 4)dz
CD DA

Along AB, −2 ≤ x ≤ 2 and y = −2.

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Therefore,
Z Z 2
2
(3z − 2z + 4)dz = (3(x − 2i)2 − 2(x − 2i) + 4)dx =
AB −2
Z 2
(3x 2 − 12ix − 2x + 4i − 8)dx = −16 + 16i.
−2

Along BC, x = 2 and −2 ≤ y ≤ 2.

Therefore,
Z Z 2
2
(3z − 2z + 4)dz = (3(2 + yi)2 − 2(2 + yi) + 4)idy =
BC −2
Z 2
i (−3y 2 + 10yi + 12)dy = 32i.
−2

Similarly, along CD, x varies from 2 to -2 and y = 2.

(3z 2 − 2z + 4)dz = 16 + 16i.
R
Therefore, CD
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Also, along DA, x = −2 and y varies from 2 to -2y = 2.
2
R
Therefore, DA (3z − 2z + 4)dz = −64i.

(3z 2 − 2z + 4)dz = −16 + 16i + 32i + 16 + 16i − 64i = 0

R
Hence, c

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Question 2(ii)

2 − 2z
R
Varify Cauchy theorem for C (z + 3)dz where c is the circle | z |= 2 .

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Solution

Substituting Z = 2e iθ , dz = 2ie iθ dθ,−π < θ ≤ π

Rπ 2iθ
the integral becmes, −π (4e − 4e iθ + 3)(2ie iθ dθ)
Rπ Rπ Rπ
= −π 8ie 3iθ dθ − −π 8ie 2iθ dθ + −π 6ie iθ dθ
=0+0+0=0
Thus Cauchy’s theorem is varified

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Question 2(iii)

z 2 dz where C is the boundary of the

R
Varify Cauchy theorem for C

triangle with vertices 0, 2, 2i

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Solution
Let ABC denote the triangle with vertices A(0,0), B(2,0), C(0,2).
On AB, y = 0, z = x , 0 ≤ x ≤ 2,
On BC, x + y = 2, 2 ≤ x ≤ 0
On CA, x = 0, 2 ≤ y ≤ 0

Z Z Z Z
2 2 2
z dz = z dz + z dz + z 2 dz
C AB BC CA
Z 2 Z 0 Z 0
2 2
= x dx + (x + i(2 − x )) (1 − i)dx + (iy )2 idy
0 2 2

=0

Thus Cauchy’s theorem is varified

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Question 2(iv)

R dz
Using the integral C z+2 where C : |z| = 1. Prove that
Z 2π
1 + 2cosθ
dθ = 0.
0 5 + 4cosθ

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Solution

By Cauchy’s Theorem
dz
Z
=0
C z +2
Therefore it is enough to prove that
Z 2π
dz 1 + 2cosθ
Z
= dθ
C z +2 0 5 + 4cosθ

Take z = e iθ , then dz = ie iθ .
We have e iθ = cos(θ) + isin(θ)

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 Z 2π
dz cosθ + isinθ
Z
=i
C z +2 0 cosθ + 2 + isinθ
Z 2π
(cosθ + isinθ)(cosθ + 2 − isinθ)
=i
0 (cosθ + 2)2 + sin2 θ

On Simplification,
Z 2π
dz i(1 + 2cosθ) − 2sinθ
Z
= dθ
C z +2 0 5 + 4cosθ

Equating the imaginary part to zero we will get.

Z 2π
1 + 2cosθ
dθ = 0
0 5 + 4cosθ

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Question 3(i)

R (z 2 +1
Evaluate C z(2z+1) dz ; C: |z| = 1

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Solution

(z 2 +1 R (z 2 +1) 2 +1)
dz − C 2(z
R R
C z(2z+1) dz= C z 2z+1 dz
2 2 +1)
= C (z z+1) dz − C 12 2(z
R R
(z+ 1 )
dz
2

=2πi(0 + 1) − 2πi(( −1 2
2 ) + 1) ( By Cauchy’s integral formula)

=- πi2

22 / 101
Question 3(ii)

R 2z+1
Evaluate C z(z+2)(z−3) dz

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Solution

2z+1 A B C
Expressing as sum of partial fractions we get, z(z+2)(z−3) = z + z+2 + z+2
−1 −1 7
and on solving, we have A = 6 , B= 2 and C = 15 .

By Cauchy’s integral formula,

R −1/6 R −1/2 R 7/15
C z dz + C z+2 dz + C z−3 dz = 2πi[ −1
6 −
1
2 + 7
15 ] = −2πi
5

24 / 101
Question 3(iii)

Evaluate
dz
Z
, c : |z − 1| = 1
c (z − 3)3 (z− 3)(z − 4)

25 / 101
Solution

The singularities z = 3 and z = 4 lies outside c

therefore by Caushy Goursat theorem the integral becomes zero.

26 / 101
Question 3(iv)

Evaluate:

e −2z
Z
dz, C : |z| = 4
C (z + 3)4

27 / 101
Solution

By Generalized Cauchy’s Integral formula,

Let f (z) be analytic in a simply connected domain containing the simple
closed contour C . Then f (z) has derivatives of all orders at each point z0
inside C , with
n! f (z)
I
f (n) (z0 ) = dz
2πi γ (z − a)n+1
Hence,
e −2z 2πi −8e 6 πi
Z
′′′
dz = f (−3) =
|z|=4 (z + 3)4 3! 3

28 / 101
Question 3(v)

Evaluate:

5z 2 + 2z
Z
dz, C : |z| = 3
C (z − 2)3 (z − 1)(z + 4)

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Solution

5z 2 + 2z
The integrand dz has a pole of order 3 at z = 2
(z − 2)3 (z − 1)(z + 4)
and a simple poles at z = 1 and z = −4. Here z = 2 and z = 1 lie inside
|z| = 3.
Residue at z = 2
1 d2 3 5z 2 + 2z 4
= limz→2 2
(z − 2) 3
=
(3 − 1)! dz (z − 2) (z − 1)(z + 4) 3
Residue at z = 1
5z 2 + 2z −7
= limz→1 (z − 1) 3
=
(z − 2) (z − 1)(z + 4) 5
5z 2 + 2z −7
dz = 2πi( 43 +
R
Hence by Residue theorem, C 5 )
(z − 2)3 (z − 1)(z + 4)

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3(vi)

dz
Z
, c : |z − 1| = 1
c (z + 1)(z − 1)2

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Solution

dz
The integrand (z+1)(z−1)2
has a pole of order 2 at z = 1 and a simple pole
at z = −1. Here z = 1 lie inside |z − 1| = 1.
Residue at z = 1
d 1 −1
= limz→1 dz z+1 = limz→1 z2
= −1
R dz
Hence by Residue theorem, c (z+1)(z−1)2 = 2πi(−1) = −2πi

32 / 101
Question 3(vii)(a)

z+1
: |z − 21 | = 1
R
Evaluate C z(z−1)(z−2) dz, C 4

z +1 1 2 3
= − +
z(z − 1)(z − 2) 2z z − 1 2(z − 2)
z +1 1 1 1 3 1
Z Z Z Z
=⇒ dz = dz − 2 dz + dz
C z(z − 1)(z − 2) 2 Cz C z −1 2 C z −2
=0

33 / 101
Question 3(vii)(b)

R z+1 1
Evaluate C z(z−1)(z−2) dz, C : |z − 2| = 2
z+1
The integrand z(z−1)(z−2) has a simple pole at z = 2.
Hence by Residue theorem
z +1
Z
= 2πi Res(f , 2)
C z(z − 1)(z − 2)
z +1
= 2πi lim
z→2 z(z − 1)

= 3πi

34 / 101
Question 3(viii)

R (3z 2 +2z−4)dz
c (z 3 −4z)
, c : |z − i| = 3
Solution:

(3z 2 + 2z − 4) (3z 2 + 2z − 4) (3z 2 + 2z − 4)

Z Z Z
dz = dz = dz
c (z 3 − 4z) c z(z 2 − 4) c z(z + 2)(z − 2)

c : |z − i| = 3
Here z=0, z=2 and z=-2 lies inside c.

35 / 101
By Cauchy’s residue theorem,
n
(3z 2 + 2z − 4)
Z X
dz = 2πi Resz=zk f (z)
c z(z + 2)(z − 2) k=1

(3z 2 + 2z − 4) −4
Resz=0 f (z) = limz→0 z dz = =1
z(z + 2)(z − 2) −4
(3z 2 + 2z − 4) 12 + 4 − 4 3
Resz=2 f (z) = limz→2 (z − 2) dz = =
z(z + 2)(z − 2) 2×4 2
(3z 2 + 2z − 4) 12 − 4 − 4 1
Resz=−2 f (z) = limz→−2 (z + 2) dz = =
z(z + 2)(z − 2) −2 × −4 2
3 1
Z
f (z)dz = 2πi[1 + + ] = 6πi
2 2

36 / 101
Question 3(ix)

R z 3 +1
Evaluate the integral C (3z+1)3 dz, C : |z| = 1.
z 3 +1 1 z 3 +1
(3z+1)3
= 27 (z+1/3)3
−1
z= 3 is a singularity of f(z) which lies inside |z| = 1.
−1 z 3 +1
Therefore, z = 3 is pole of order 3 of f (z) = (3z+1)3
.
d 2 ((z+1/3)3 f (z))
Res(f (z), −1
3 )=
1
2 limz→ −1 dz 2
= −1
3
R z 3 +1 1 −2πi
Therefore, C (3z+1)3 dz = 2πi × 27 × −1 = 27 .

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Question 3(x)
R (z 2 +2z+5)dz
(x) c (2z−1)(z−1)2 (z−2) c : |z| = 2.5

Singularities are z = 21 , 1(order 2), 2

(z 2 +2z+5)
let f (z) = (2z−1)(z−1)2 (z−2) , we have the result
For pole of order m :
d m−1
Resz=z0 f (z) = limz→z0 1
(m−1)! dz m−1 ((z − z0 )m f (z))
so by using the above equation,
25
Resz= 1 f (z) = 3
2

Resz=1 f (z) = 4
13
Resz=2 f (z) = 3

So using the Cauchy’s Residue Theorem,

R (z 2 +2z+5)dz  25 13
  50 
c (2z−1)(z−1)2 (z−2) = 2πi 3 + 4 + 3 = 2πi 3 38 / 101
Question 3(xi)

2
R ez
Evaluate the integral C (z−i)4 dz, C : |z| = 2.
Solution:
z = i is the only singularity and is a pole of order 4 which lies inside
|z| = 2.
1 d3 z2 1 2 2
Res(f (z), i) = 6 limz→i dz 3
e = 6 limz→i 8z 3 e z + 12ze z

2i −4π
= =
3e 3
2
R ez 2i −4π
So by residue theorem C (z−i)4 dz = 2πi( 3e )= 3e

39 / 101
Question 3(xii)(a)

R (z 2 −3)dz
c (z 2 +2z+5) , where C is |z| = 1
Here the singularities and −1 − 2i are 1 + 2i which is lie out side of
|z| = 1. So the function we need to integrate is analytic inside he region.
So by Cauchy’s theorem

z 2 − 3 dz
Z

=0
c (z 2 + 2z + 5)

40 / 101
Question 3(xii)(b)

R (z 2 −3)
Evaluate the integral C (z 2 +2z+5) dz, where C is the circle |z + 1 − i| = 2.

Solution:
√
(z 2 −3) −2± 4−20
The singularities of f (z) = (z 2 +2z+5)
are z = 2 = −1 ± 2i. Here
−1 + 2i only lies inside the circle C , which is also a simple pole of f .
Therefore Res(f (z); −1 + 2i) = limz→−1+2i (z − (−1 + 2i))f (z) =
−3) 2
limz→1 (z − (−1 + 2i)) (z (z
2 +2z+5) = −1 +
3i
2.

By Cauchy’s Residue theorem,

R (z 2 −3) 3i
C (z 2 +2z+5) = 2πi(−1 + 2) = −2πi − 3π.

41 / 101
Question 3(xii)(c)

R (z 2 −3)
Evaluate the integral c (z 2 +2z+5) dz, where c is the circle |z − 1 + i| = 2.

Solution:
√
(z 2 −3) −2± 4−20
The singularities of (z 2 +2z+5)
are z = 2 = −1 ± 2i. Both these
points lie outside the circle c.
(z 2 −3)
Thus, (z 2 +2z+5)
is analytic inside and on the circle c.
By Cauchy-Goursat theorem, the integral becomes zero.

42 / 101
Question 4(i)

R 3z+2
Using Residue theorem, evaluate the integral c z(z−1)(z−2) dz where
C : |z| = 3

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Solution

3z+2
Let f (z) = z(z−1)(z−2) . The singularities of f (z) are 0, 1 and 2. All are
simple poles inside the cicle | z |= 3.
3z+2
Res(f (z), 0) = limz→0 zf (z) = limz→1 z z(z−1)(z−2) =1
3z+2
Res(f (z), 1) = limz→1 (z − 1)f (z) = limz→1 (z − 1) z(z−1)(z−2) = −5
3z+2
Res(f (z), 2) = limz→2 (z − 2)f (z) = limz→2 (z − 2) z(z−1)(z−2) =4
By Cauchy’s Residue theorem,
R 3z+2
C z(z−1)(z−2) = 2πi(1 − 5 + 4) = 0

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Question 4(ii)

R z 2 +3z+2
Using Residue theorem, evaluate the integral c z 2 (z−1) where

C : |z| = 1.5

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Solution
z 2 +3z+2
Let f (z) = z 2 (z−1)
. The singularities of f (z) are 0 and 1. Also, z = 1 is a
simple pole and z = 0 is a double pole.

z 2 + 3z + 2
Res(f (z), 1) = lim (z − 1)f (z) = lim (z − 1) =6
z→1 z→1 z 2 (z − 1)
d(z 2 f (z))
Res(f (z), 0) = lim
z→0 dz
d z 2 (z 2 + 3z + 2)
= lim ( )
z→0 dz z 2 (z − 1)
d z 2 + 3z + 2
= lim ( ) = −5
z→0 dz z −1
By Cauchy’s Residue theorem,
R z 2 +3z+2
C z 2 (z−1) = 2πi(6 − 5) = 2πi
46 / 101
Question 4(iii)

Using Residue theorem, evaluate the following integral

(4z 3 + 2z)
Z
dz, C : |z| = 2
C z4 − z3

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Solution

If the curve C encloses multiple poles, then the theorem

Z n
X
f (z)dz = 2πi Res(z=ak ) f (z)
C k=1

where z = ak is the poles of the f(z)

4z 3 + 2z 4z 2 + 2z
=
z4 − z3 z 2 (z − 1)

Therefore z=1 is a simple pole and z=0 is a pole of order 2.

The residue of f at z=a is a simple pole

− a (z − a)f (z)
Resz=a f (z) = limz →

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Therefore,
Resz=1 f (z) = 6

The residue of f at z=a is a pole of order n,

1 d n−1
Resz=a f (z) = ((z − a)n f (z))
n! dz n−1

Therefore,
Resz=0 f (z) = 2

Hence,
(4z 3 + 2z)
Z
dz = 16πi
C z4 − z3

49 / 101
Question iv

Using Residue theorem, evaluate the integral

R ez
C z 3 +z dz, C :—z—=2

50 / 101
Solution

z
Let f(z)= z 3e+z . The singularities of f(z) are 0 ,-i and i (simple poles), lies
inside the C.
e z
Res(f(z),0)=limz →
− 0 z z(z−i)(z+i) = 1
e z
−e i
− i (z − i) z(z−i)(z+i) = 2
Res(f(z),i)=limz →
e z −e −i
Res(f(z),-i)=limz →
− −i (z + i) z(z−i)(z+i) = 2

By Residue theorem
R ez
C z 3 +z dz=2πi[ Res(f;0)+Res(f;-i)+Res(f;i)]
−e −i −e i cos1
=2πi[1 + 2 + 2 ] =2πi(1 − 2 )

51 / 101
Question (v)

R sin z
Using Residue theorem, evaluate the following integral C (z−1)2 (z 2 +9)

52 / 101
Solution

sin z sin z
f (z) = (z−1)2 (z 2 +9)
= (z−1)2 (z+3i)(z−3i)

Therefore z = 3i is a simple pole which lies entirely in C .

Then by Residue theorem

sin z
Z  
dz = 2πi Res (f , 3i)
C (z − 1)2 (z 2 + 9) z=3i
π sinh 3
= 2πi lim (z − 3i)f (z) = 6i(4+3i) .
z→3i

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5(i)

z
Obtain Taylor series expansion of z−3 about z = 1. What is the region of
convergence?

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Solution
Note that f has a singularity at z=3

z z
f (z) = =
z −3 z −1−2
z
= z−1
2( 2 − 1)
−z z − 1 −1
= [1 − ( )]
2 2
It has a convergent geometric series representation when |z − 1| < 2, ie,
inside the disk centered at z = 1 with radius 2.
Hence the Taylor series expansion of f(z) in this region of convergence is
−z z −1 z −1 2
 
f (z) = 1+ +( ) + ...
2 2 2
55 / 101
Question 5(ii)

π
Find the Taylor series expansion of sin z about z = . Also find the
4
Maclaurin series expansion of sin z

56 / 101
Solution

If f (z) is an entire function, then its Taylor series representation

∞
X f n (b)
f (z) = (z − b)n
n=0
n!

is valid for all complex b and z.

sin z = !
1 π 1 π 2 1 π 3 1 π 4
       
√ 1+ z − − z− − z− + z− + ...
2 4 2! 4 3! 4 4! 4

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Solution 5(ii)

Maclaurin series expansion of f (z) is the representation

∞
X f n (0) n
f (z) = z
n=0
n!

z3 z5
sin z = z − + − ...
3! 5!

58 / 101
Question 5(iii)

z +3
Find the Taylor or Laurent series expansion of when
z(z 2− z − 2)
(a) |z| < 1

(b) 1 < |z| < 2 and

(c) |z| > 2

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Solution
z +3
Resolving f (z) = into partial fraction we get
z(z 2
− z − 2)
3 5 2
f (z) = − + +
2z 6(Z − 2) 3(z + 1)
(a) When |z| < 1 we have
3 5 2
f (z) = − + (1 − z2 )−1 + (z + 1)−1
2z 12 3
3 5 P∞ z
n 2 P∞
=− + + (−1)n z n which is the required
2z 12 n=0 2 3 n=0
expansion.

(b) 1 < |z| < 2 we have

3 5 2
f (z) = − + (1 − z2 )−1 + (1 + z1 )−1
2z 12 3z
3 5 P∞ z
n 2 P∞
=− + + (−1)n ( z1 )n which is the required
2z 12 n=0 2 3z n=0
expansion.
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(c) |z| > 2 then we have
3 5 2
f (z) = − + (1 − z2 )−1 + (1 + z1 )−1
2z 6z   3z
3 5 P∞ 2 n 2 P∞
=− + n=0 z + (−1)n ( z1 )n
2z 6z 3z n=0
3 5 P∞ 2n−1 2 P∞ 1
=− + n+1 + (−1)n z n+1 which is the required
2z 3 n=0 z 3 n=0
expansion.

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5(iv)

z 2 −z+5
Find Taylor or Laurent series expansions of (z+2)(z+3) that are valid when
(a) |z| < 2 (b) 2 < |z| < 3 and (c) |z| > 2.

62 / 101
Solution I
z 2 −z+5 7/5 11/5
Resolving f (z) = (z+2)(z+3) into partial fraction we get f (z) = z+2 + z+3

Hence f(z) is not analytic at z=-2 and z=-3

therefore, f(z) is analytic in |z| < 2, 2 < |z| < 3 and |z| > 2.
a)
for|z| < 2
since |z| < 2, |z| < 3
7/5 11/5
therefore f (z) = z+2 + z+3
7
= 10 (1 + z2 )−1 + 11
15 (1 + z3 )−1
b)
for 2 < |z| < 3

63 / 101
Solution II

we have | z2 | < 1 and | z3 | < 1

7/5z 11/15
therefore f (z) = 1 + 1+ z3
1+ z2
7
= 5z (1 + z2 )−1 + 11
15 (1 + z3 )−1
c)
for |z| > 3 we have —z—¿2
so that | z3 | > 1 and | z2 | > 1
the f (z) = 7
5z (1 + z2 )−1 + 11
5z (1 + z3 )−1

64 / 101
Question 5(v)

7z 2 −2 1 5 13
f (z) = z(z+1)(z−2) = z + 3(z+1) + 3(z−2)

a)|z + 1|< 1
1 1
z = − 1−(z+1) = −(1 + (z + 1) + (z + 1)2 + ...), |z + 1| < 1
1 1 (z+1)2
z−2 =− (z+1) = − 31 (1 + z+1
3 + 9 + ...), | z+1
3 |<1
3(1− 3 )
Substituting in f (z), we get the required series expansion with respect to
z +1
1 3
b)|z + 1| > 3 =⇒ | z+1 | < | z+1 |<1
Hence the series expansion wrt z + 1 can be found similarly as (a).

65 / 101
5 (vi)

Expand (z 2 − 1)(z 2 + 5z + 6)−1 as a Laurent series in 2 < |z| < 3.

Solution:

(z 2 − 1) 5z + 7
2
=1−
(z + 5z + 6) (z + 2)(z + 3)
 −3 8 
=1− +
z +2 z +3
 −3 h 2 i−1 8 h z i−1 
=1− 1+ + 1+
z z 3 3

66 / 101
Question 5(vii)

1+z
Find the Laurent series expansion of f (z) = (z 2 −2z 4 )
about z=0 and hence
find the residue at z=0.

67 / 101
Question 6(i)

Prove the integral

R 2π dθ
0 3cosθ+5 = π2 .

68 / 101
Solution

Let z = re iθ ; θ : 0 → 2π, that is r = 1.

z+ z1 dz
Then cosθ = 2 , dθ = iz .
R 2π dθ R dz
0 3cosθ+5 = c iz(3 z+1/z +5)
R2
= −2i
3
dz
c (z+1/3)(z+3) .
1
The singularity of f (z) = (z+1/3)(z+3) which lies inside c(|z| = 1) is
−1
z= 3 .

Resz= −1 f (z) = 38 .
3
R 2π
Then 0
dθ
3cosθ+5 = ( −2i
3 )2πi ×
3
8 = π2 .

69 / 101
6(ii)
prove the integral
R 2π cos 2θdθ π
0 5−4 cos θ = 6

Substitute z = re i θ: θ : 0 → 2π , so r=1, it is a unit circle.

e iθ +e −iθ z+ z1
cos(θ) = 2 == 2
z+ z1 2
cos(2θ) = 2cos 2 (θ) − 1 = 2( 2 ) −1
Substitute the above results in the integrals, find residues of the function
inside the integral and use Cauchy’s residue theorem you can simply prove
it.
1
sigularities are z = 2 and z = 2
5
residue at z=0 is 2
1 −17
residue at z = 2 is 6
70 / 101
6(iii)

prove the integral

R 2π sin2 θdθ π
0 5+4 cos θ = 4
z+ z1 z− z1
Substitute sin θ = 2i , cos θ = 2

Gives the singularities are o, - 12 , and 2

z+ z1 2
cos(2θ) = 2cos 2 (θ) − 1 = 2( 2 ) −1
residue at z=0 is - 25
residue at z=− 12 is 3
2

71 / 101
6(iv)
prove the integral
R 2π dθ π
0 15 sin2 θ+1
= 2

substitute z = re i θ: θ : 0 → 2π , so r=1, it is a unit circle.

e iθ −e −iθ z− z1
sin(θ) = 2i == 2i
15Z 4 −34z 2 +15
15sin2 (θ) + 1 = −4z 2

substitute these results in the integral,

R 2π dθ 1R −4z 2 dz
0 15 sin2 θ+1
= i |z|=1 z(15Z 4 −34z 2 +15)

find the residue of the function inside the integral and use Cauchy’s
residue theorem you can find the result.
q q
3 −3
sigularities are z = 5 and z = 5 .
−15
residue at both singularities is 32 .
72 / 101
Question 6(v)
R∞ dx π
Prove that −∞ (x 2 +1)(x 2 +4) = 6

Solution:

Let γ be the part of the circle |z| = R > 0 in the upper half plane and the
CR = [−R, R] ∪ γ.
R dz RR dx R dz
Then, CR (z 2 +1)(z 2 +4) = −R (x 2 +1)(x 2 +4) + γ (z 2 +1)(z 2 +4)

R dz
So that, when R → ∞, γ (z 2 +1)(z 2 +4) → 0 and we obtain
R∞ dx
−∞ (x 2 +1)(x 2 +4) .

73 / 101
 R∞ dx R dz
Therefore −∞ (x 2 +1)(x 2 +4) = limR→∞ CR (z 2 +1)(z 2 +4)

R dz
CR (z 2 +1)(z 2 +4) = 2πi[Res(f ; i) + Res(f ; 2i)].

1 −i
Res(f ; i) = limz→i (z+i)(z 2 +4)
= 6 .

1 i
Res(f ; 2i) = limz→2i (z 2 +1)(z+2i)
= 12 .

dz
= 2πi[ −i i
= π6 .
R
Therefore CR (z 2 +1)(z 2 +4) 6 + 12 ]

74 / 101
Question 6(vi)
R∞ x2 π
Prove that 0 (x 2 +16)2
= 16

Solution:

Let γ be the part of the circle |z| = R > 0 in the upper half plane and the
CR = [−R, R] ∪ γ.
R z2 RR x2 R z2
Then, CR (z 2 +16)2 dz = −R (x 2 +16)2 dx + γ (z 2 +16)2 dz

R z2 R∞ x2
So that, when R → ∞, γ (z 2 +16)2 dz → 0 and we obtain −∞ (x 2 +16)2 .

Now, (z 2 + 16)2 = (z + 4i)2 (z − 4i)2 . z = 4i lies inside CR and therefore

it is a pole.
75 / 101
Question 6(vii)

R∞ dx π
Prove that −∞ (x 2 +1)(x 2 +4) = 9

Similar to 6(v).
The integral will be equal to π6 , not π
9.

76 / 101
 z2 2
= 2πiResz=4i (z+4i)z2 (z−4i)2
R
(z 2 +16)2
dz
d z2 π
=2πi limz→4i dz (z+4i)2 = 8
R∞ x2 1 R∞ x2 π
Therefore, 0 (x 2 +16)2
= 2 −∞ (x 2 +16)2 = 16

77 / 101
Question 6(viii)

R∞ 2x 2 +3 7π
Prove that −∞ (x 2 +9)2 dx = 18

78 / 101
Solution
Let CR denote the region in the complex plane consisting of the real axis
axis from −R to R and the semicircle γ of radius R in the upper half
plane.

2z 2 + 3 2x 2 + 3 2z 2 + 3
Z Z R Z
dz = dx + dz (1)
CR (z 2 + 9)2 −R (x 2 + 9)2 γ (z 2 + 9)2
R 2z 2 +3
As R → ∞, by ML theorem, γ (z 2 +9)2 dz =0
R 2z 2 +3 R 2z 2 +3
Now, CR (z 2 +9)2 dz = CR (z+3i)2 (z−3i)2 dz
Since the singularities z = 3i lies inside CR and z = −3i lies outside CR ,
therefore by residue theorem,
R 2z 2 +3
CR (z 2 +9)2 dz = 2πiRes(f (z), 3i)
d 2z 2 +3 −7i
Res(f (z), 3i) = limz→3i 2 = 79 / 101
Question 6(ix)

Prove the following real integral

Z ∞
x sin xdx π
=
∞ x2 + 4 e2

80 / 101
Solution

ze iz
Let f(z) = z 2 +4

z 2 + 4 = (z + 2i)(z − 2i)
f is analytic inside and on the contour except at z=2i and z=-2i
f has simple pole at z=2i and z=-2i

Z Z R Z
f (z)dz = f (x )dx + f (z)dz
CR −R r

ze iz
Z Z
f (z)dz =
CR CR z2 + 4
X
= 2πi Resf (z)

81 / 101
 ze z
Res f (z) = lim (z − 2i)
z→2i z→2i (z − 2i)(z + 2i)
(2i)e i(2i)
=
4i
e −2
=
2
z=-2i is outside the curve therefore no need to consider it.

e −2
Z
f (z)dz = 2πi
CR 2
πi
=
e2

82 / 101
By ML Theorem,
ze iz dz |ze iz dz|
Z Z
| |≤ → 0 as R → ∞
r z2 + 4 r |z 2 + 4|
Z Z R
f (z)dz = f (x )dx
CR −R
πi
Z
f (z)dz =
CR e2
Z R
πi
lim f (x )dx = 2
R→∞ −R e
xe ix
Z ∞
πi
2
dx = 2
−∞ x + 4 e
Z ∞
x (cosx + isinx ) πi
2
= 2
−∞ x +4 e
Equating imaginary part,
Z ∞
xsinx π
=
−∞ x2 +4 e2
83 / 101
Question 6(x)

Prove
Z ∞
cos2x πcos2
dx =
−∞ x2 + 2x + 2 e2

84 / 101
Solution

i2z
e e i2z
Let f(z)= z 2 +2Z +2
= (z−(−1+i))(z−(−1−i))

Z Z R Z
f (z)dz = f (x )dx + f (z)dz
CR −R γ

e i2z
Z Z
f (z)dz = dz
CR CR (z − (−1 + i))(z − (−1 − i))
X
= 2πi Resf (z)

85 / 101
 e i2z
Res f (z) = lim (z − (−1 + i))
z→−1+i z→−1+i (z − (−1 + i))(z − (−1 − i))
e −2(i+1)
=
2i
e −2i
=
2ie 2
z=-1-i is outside the curve therefore no need to consider it.
−2(i+1)
f (z)dz = 2πi e = πe −2(i+1)
R
Thus, CR 2i

86 / 101
By ML Theorem,
e 2iz |e 2iz |
Z Z
| |≤ → 0 as R → ∞
γ z 2 + 2z + 2 γ |z 2 + 2z + 2|
Z Z R
f (z)dz = f (x )dx
CR −R
Z
f (z)dz = πe −2(i+1)
CR
Z R
lim f (x )dx = πe −2(i+1)
R→∞ −R
∞ e 2ix e −2i
Z
dx = π
−∞ x2
+ 2x + 2 e2
Z ∞
(cos2x + isin2x ) π(cos(−2) + i sin(−2))
2
dx =
−∞ x + 2x + 2 e2
Equating real part,
Z ∞
cos2x π cos 2
dx =
−∞ x2 + 2x + 2 e2
87 / 101
Question 6(xi)

Prove
Z ∞
cos x π cos 1
dx =
−∞ x2 − 2x + 5 2e 2

88 / 101
Solution

e iz e iz
Here, f (z) = z 2 −2z+5
= (z−(1+2i))(z−(1−2i))

Z Z R Z
f (z)dz = f (x )dx + f (z)dz
CR −R γ

e iz
Z Z
f (z)dz = dz
CR CR (z − (1 + 2i))(z − (1 − 2i))
X
= 2πi Resf (z)

89 / 101
 e iz
Res f (z) = lim (z − (1 + 2i))
z→1+2i z→1+2i (z − (1 + 2i))(z − (1 − 2i))
e i−2
=
4i
ei
=
4ie 2
z=1-2i is outside the curve therefore no need to consider it.
R e i πe i
Thus, CR f (z)dz = 2πi 4ie 2 = 2e 2

90 / 101
By ML Theorem,
e iz |e iz |
Z Z
| |≤ → 0 as R → ∞
γ z 2 − 2z + 5 γ |z 2 − 2z + 5|
Z Z R
f (z)dz = f (x )dx
CR −R
πe i
Z
f (z)dz =
CR 2e 2
πe i
Z R
lim f (x )dx = 2
R→∞ −R 2e
Z ∞
e ix πe i
2
dx =
−∞ x − 2x + 5 2e 2
Z ∞
(cosx + isinx ) π(cos 1 + i sin 1)
2
dx =
−∞ x − 2x + 5 2e 2
Equating real part,
Z ∞
cosx π cos 1
dx =
−∞ x 2 − 2x + 5 2e 2
91 / 101
6(xii)

Z ∞
xdx
=0
−∞ (x 2 + 9)2

92 / 101
Solution I
z z
The complex function f (z) = = has singularities at
(z 2 +9)2 (z+3i)2 (z−3i)2

z = ±3i. Let c be the contour consisting of the real axis from -R to R

(R > 3) followed by the semicircle in the upper half plane.
Then the only singularity of f(z) inside c is at z=3i and it’s a pole of order
2. Therefore its residue is
1 d
Res(f (z), 3i) = 1! limz−>3i dz (z − 3i)2 (z+3i)2z(z−3i)2
2 −2z(z+3i)
= limz−>3i (z+3i)(z+3i) 2 =0
R z
Hence c (z 2 +9)2 = 2πi × 0 = 0
The value of the integral is independent of R (R > 3)
z RR xdx Rπ Re iθ
iRe iθ dθ
R
Also c (z 2 +9)2 = −R (x 2 +9)2 = 0 2
( )
R e i2θ +9
2

93 / 101
Question 7

Let C be a simple closed contour with positive orientation and enclosing

the origin. Then, show that
∞
1
Z
1 X
e z+ z dz = 2πi
C n=0
n!(n + 1)!

94 / 101
Solution

1
The Laurent series representation of e z will give us

∞
1 1 1 ez
 
z+ z1
X
z
e =e 1+ + + + ... =
z 2!z 2 3!z 3 n=0
n!z n
By Generalized Cauchy’s Integral formula, Let f (z) be analytic in a simply
connected domain containing the simple closed contour C . Then f (z) has
derivatives of all orders at each point z0 inside C , with

n! f (z)
I
(n)
f (z0 ) = dz
2πi γ (z − a)n+1

95 / 101
Solution 7

Hence,

∞ ∞ ∞
ez 1 ez 1
Z Z X I
z+ z1
X X
e dz = = = 2πi
C C n=0 n!z n n=0
n! C zn n=0
n!(n + 1)!

96 / 101
Question 8

Discuss the nature of singularities of

(z − 2) 1
 
(i) f (z) = sin
z2 z −1

97 / 101
Solution (i)

Here limz→0 z 2 f (z) = 2sin(1) ̸= 0, hence z = 0 is a pole of 2 of f (z).

1
Zeros of f (z) are z = 2 and z = 1 + nπ , n = 1, 2, .....
The limit of the zeros is 1, hence z = 1 is an isolated essential singularity
of f (z).

98 / 101
8(ii)

1
f (z) = e z −1

99 / 101
Solution

ez − 1 = 0
=⇒ z = 2nπi n ∈ Z
d z
dz e − 1 = e z ) ̸= 0, at z = 2nπi
Therefore f(z) has simple pole at z = 2nπi

100 / 101
Question 8(iii)

Discuss the nature of singularities of

1
iii)f (z) = cos z−sin z
π
cosz − sinz = 0 =⇒ z = 4 + kπ, k ∈ Z
d 1 π
dz (cosz−sinz) = −(sinz + cosz) ̸= 0, atz = 4 + kπ
π
Therefore f(z) has simple pole at z = 4 + kπ.

101 / 101