Let f(z) be a function continuous at a point z0. To show that f(z) is also continuous at z0, we need to show: limz→z0 f(z) = f(z0) Since f(z) is given to be continuous at z0, by the definition of continuity: limz→z0 f(z) = f(z0) Therefore, if f(z) is continuous at a point z0, it automatically satisfies the condition for continuity at z0. Hence, f(z) is also continuous at z0. So in summary, if a function f(z) is continuous at a

1. Mathematics-3 (MAC331)
Complex Analysis
October 20, 2020
Md Firoz Ali
Department of Mathematics
National Institute of Technology Durgapur, India
Email- firoz.ali@maths.nitdgp.ac.in

2. 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
C
f(z) dz = 2πi
n
X
k=1
Res [f(z); zk ]
2 / 22 Md Firoz Ali Complex Analysis

3. 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
C
f(z) dz = 2πi
n
X
k=1
Res [f(z); zk ]
2 / 22 Md Firoz Ali Complex Analysis

4. 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
C
f(z) dz = 2πi
n
X
k=1
Res [f(z); zk ]
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,
Res [f(z); ∞] +
n
X
k=1
Res [f(z); zk ] = 0
2 / 22 Md Firoz Ali Complex Analysis

5. Residue
Example: Evaluate
ˆ
|z|=2
dz
z2(z2 − 1)
.
3 / 22 Md Firoz Ali Complex Analysis

6. Residue
Example: Evaluate
ˆ
|z|=2
dz
z2(z2 − 1)
.
Solution: The function f(z) =
1
z2(z2 − 1)
has double pole at z = 0 and simple pole at
z = ±1. Note that all the singular points of f(z) lies inside |z| = 2. Now
Res [f(z); 0] = lim
z→0
d
dz
[(z − 0)2
f(z)] = lim
z→0
d
dz
1
z2 − 1
= lim
z→0
−2z
(z2 − 1)2
= 0
Res [f(z); 1] = lim
z→1
(z − 1)f(z) = lim
z→1
1
z2(z + 1)
=
1
2
Res [f(z); −1] = lim
z→−1
(z + 1)f(z) = lim
z→−1
1
z2(z − 1)
= −
1
2
.
Then by Cauchy’s Residue theorem
ˆ
|z|=2
dz
z2(z2 − 1)
= 2πi(Res [f(z); 0] + Res [f(z); 1] + Res [f(z); −1]) = 0.
3 / 22 Md Firoz Ali Complex Analysis

7. Residue
Example: Evaluate
ˆ
|z−1|=1/2
dz
z2(z2 − 1)
.
4 / 22 Md Firoz Ali Complex Analysis

8. Residue
Example: Evaluate
ˆ
|z−1|=1/2
dz
z2(z2 − 1)
.
Solution: The function f(z) =
1
z2(z2 − 1)
has double pole at z = 0 and simple pole at
z = ±1. Note that only the pole z = 1 of f(z) lies inside |z − 1| = 1/2. Now
Res [f(z); 1] = lim
z→1
(z − 1)f(z) = lim
z→1
1
z2(z + 1)
=
1
2
Then by Cauchy’s Residue theorem
ˆ
|z|=2
dz
z2(z2 − 1)
dz = 2πi × Res [f(z); 1] = πi.
4 / 22 Md Firoz Ali Complex Analysis

9. Residue
Example: Evaluate
ˆ
|z−3|=1
1 − cos 2(z − 3)
(z − 3)3
dz.
5 / 22 Md Firoz Ali Complex Analysis

10. Residue
Example: Evaluate
ˆ
|z−3|=1
1 − cos 2(z − 3)
(z − 3)3
dz.
Solution: First we note that
f(z) =
1 − cos 2(z − 3)
(z − 3)3
=
1
(z − 3)3
"
1 − 1 +
4(z − 3)2
2!
−
16(z − 3)4
4!
+ · · ·
#
=
2
(z − 3)
−
16(z − 3)
4!
+ · · · .
Thus f(z) has a simple pole at z = 3.
5 / 22 Md Firoz Ali Complex Analysis

11. Residue
Example: Evaluate
ˆ
|z−3|=1
1 − cos 2(z − 3)
(z − 3)3
dz.
Solution: First we note that
f(z) =
1 − cos 2(z − 3)
(z − 3)3
=
1
(z − 3)3
"
1 − 1 +
4(z − 3)2
2!
−
16(z − 3)4
4!
+ · · ·
#
=
2
(z − 3)
−
16(z − 3)
4!
+ · · · .
Thus f(z) has a simple pole at z = 3.
The Laurent’s series is in the power of z − 3. The coefficient of 1
(z−3)
is 2. Hence,
Res [f(z); 3] = 2.
Then by Cauchy’s Residue theorem
ˆ
|z−3|=1
1 − cos 2(z − 3)
(z − 3)3
dz = 2πi × Res [f(z); 3] = 4πi.
5 / 22 Md Firoz Ali Complex Analysis

12. Residue
Example: Evaluate
ˆ
|z|=2
tan z dz.
6 / 22 Md Firoz Ali Complex Analysis

13. Residue
Example: Evaluate
ˆ
|z|=2
tan z dz.
Solution: The singularity of the function f(z) = tan z =
sin z
cos z
are given by
cos z = 0 =⇒ z = (2n + 1)
π
2
, n ∈ Z.
The singular points zn = (2n + 1) π
2
are simple poles of f(z).
6 / 22 Md Firoz Ali Complex Analysis

14. Residue
Example: Evaluate
ˆ
|z|=2
tan z dz.
Solution: The singularity of the function f(z) = tan z =
sin z
cos z
are given by
cos z = 0 =⇒ z = (2n + 1)
π
2
, n ∈ Z.
The singular points zn = (2n + 1) π
2
are simple poles of f(z).
Note that only the pole z0 = π
2
and z−1 = −π
2
of f(z) lies inside |z| = 2. Now
Res [f(z);
π
2
] = lim
z→ π
2
z −
π
2
tan z = lim
z→ π
2
z − π
2
cot z
= lim
z→ π
2
1
csc2 z
= 1
Res [f(z); −
π
2
] = lim
z→− π
2
z +
π
2
tan z = lim
z→− π
2
z + π
2
cot z
= lim
z→− π
2
1
csc2 z
= 1.
Then by Cauchy’s Residue theorem
ˆ
|z|=2
tan z dz = 2πi
Res [f(z);
π
2
] + Res [f(z); −
π
2
]
= 4πi.
6 / 22 Md Firoz Ali Complex Analysis

15. Residue
Example: Evaluate
ˆ
|z|=2
dz
(zn − 1)(z − 3)
.
7 / 22 Md Firoz Ali Complex Analysis

16. Residue
Example: Evaluate
ˆ
|z|=2
dz
(zn − 1)(z − 3)
.
Solution: The function f(z) =
1
(zn − 1)(z − 3)
has simple pole at z = 3 and at
z = zk = e2kπi/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
|z|=2
f(z) dz = 2πi
n
X
k=1
Res [f(z); zk ] = −2πi (Res [f(z); 3] + Res [f(z); ∞]) .
Now
Res [f(z); 3] = lim
z→3
(z − 3)f(z) = lim
z→1
1
zn − 1
=
1
3n − 1
,
Res [f(z); ∞] = −Res
1
z2
f
1
z
; 0
= Res
zn−1
(1 − 3z)(1 − zn)
; 0
#
= 0.
Therefore, ˆ
|z|=2
dz
(zn − 1)(z − 3)
dz =
2πi
1 − 3n
.
7 / 22 Md Firoz Ali Complex Analysis

17. Problem set
Example: Discuss continuity of the function
f(z) =



Re z
z
, z 6= 0
0, z = 0.
8 / 22 Md Firoz Ali Complex Analysis

18. Problem set
Example: Discuss continuity of the function
f(z) =



Re z
z
, z 6= 0
0, z = 0.
Solution: Let z0 6= 0. Then
lim
z→z0
f(z) = lim
z→z0
Re z
z
=
Re z0
z0
= f(z0).
8 / 22 Md Firoz Ali Complex Analysis

19. Problem set
Example: Discuss continuity of the function
f(z) =



Re z
z
, z 6= 0
0, z = 0.
Solution: Let z0 6= 0. Then
lim
z→z0
f(z) = lim
z→z0
Re z
z
=
Re z0
z0
= f(z0).
But
lim
z→0
f(z) = lim
z→0
Re z
z
=



1, if z = x → 0
0, if z = iy → 0.
Thus f(z) is continuous at all complex number except at the origin.
8 / 22 Md Firoz Ali Complex Analysis

20. Problem set
Example: Find the limit lim
z→0
f(z) if exist where f(z) =
xy
x2 + y2
+ 2xi.
9 / 22 Md Firoz Ali Complex Analysis

21. Problem set
Example: Find the limit lim
z→0
f(z) if exist where f(z) =
xy
x2 + y2
+ 2xi.
Solution: If f(z) = u(x, y) + iv(x, y) then u(x, y) =
xy
x2 + y2
and v(x, y) = 2x. Now
lim
(x,y)→(0,0)
u(x, y) = lim
(x,y)→(0,0)
xy
x2 + y2
= lim
(x,y)→(0,0)
mx2
x2 + m2x2
(along y = mx)
=
m
1 + m2
.
Since lim
(x,y)→(0,0)
u(x, y) does not exists, lim
z→0
f(z) does not exists.
9 / 22 Md Firoz Ali Complex Analysis

22. Problem set
Example: Determine where the following functions satisfy the CauchyŰRiemann
equations and where the functions are differentiable.
(i) f(z) = z2
(ii) f(z) = zRe z (iii) f(z) = 2z + 4z + 5 .
10 / 22 Md Firoz Ali Complex Analysis

23. Problem set
Example: Determine where the following functions satisfy the CauchyŰRiemann
equations and where the functions are differentiable.
(i) f(z) = z2
(ii) f(z) = zRe z (iii) f(z) = 2z + 4z + 5 .
Solution: (i) We have
f(z) = z2
=⇒ 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!)
10 / 22 Md Firoz Ali Complex Analysis

24. Problem set
Example: Determine where the following functions satisfy the CauchyŰRiemann
equations and where the functions are differentiable.
(i) f(z) = z2
(ii) f(z) = zRe z (iii) f(z) = 2z + 4z + 5 .
Solution: (i) We have
f(z) = z2
=⇒ 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
f(z) = zRe z =
z(z + z)
2
=⇒ fz (z) = z/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!)
10 / 22 Md Firoz Ali Complex Analysis

25. Problem set
Example: Determine where the following functions satisfy the CauchyŰRiemann
equations and where the functions are differentiable.
(i) f(z) = z2
(ii) f(z) = zRe z (iii) f(z) = 2z + 4z + 5 .
Solution: (i) We have
f(z) = z2
=⇒ 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
f(z) = zRe z =
z(z + z)
2
=⇒ fz (z) = z/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!)
(iii) We have
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.
10 / 22 Md Firoz Ali Complex Analysis

26. 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?
11 / 22 Md Firoz Ali Complex Analysis

27. 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.
11 / 22 Md Firoz Ali Complex Analysis

28. 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.
11 / 22 Md Firoz Ali Complex Analysis

29. 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(x2
+ y2
) + ibxy + c.
12 / 22 Md Firoz Ali Complex Analysis

30. 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(x2
+ y2
) + 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.
12 / 22 Md Firoz Ali Complex Analysis

31. 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(x2
+ y2
) + 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(x2
+ y2
) + ibxy + c is entire then it must satisfy the C-R equation at all
points. Here u(x, y) = a(x2 + y2) + c and v(x, y) = bxy. Then
ux = vy =⇒ 2ax = bx =⇒ 2a = b
uy = −vx =⇒ 2ay = −by =⇒ 2a = −b.
Thus a = b = 0.
12 / 22 Md Firoz Ali Complex Analysis

32. Problem set
Example: Let f(z) = u + iv be analytic. If
∂u
∂x
= u1(x, y) and
∂u
∂y
= u2(x, y) then
show that
f(z) =
ˆ
[u1(z, 0) − iu2(z, 0)] dz.
13 / 22 Md Firoz Ali Complex Analysis

33. Problem set
Example: Let f(z) = u + iv be analytic. If
∂u
∂x
= u1(x, y) and
∂u
∂y
= u2(x, y) then
show that
f(z) =
ˆ
[u1(z, 0) − iu2(z, 0)] dz.
Solution: We know that
f0
(z) = ux + ivx = ux − iuy = u1(x, y) − iu2(x, y).
Substituting y = 0, we get
f0
(x) = u1(x, 0) − iu2(x, 0).
Replacing x by z, we get
f0
(z) = u1(z, 0) − iu2(z, 0) =⇒
ˆ
[u1(z, 0) − iu2(z, 0)] dz.
13 / 22 Md Firoz Ali Complex Analysis

34. Problem set
Example: Let f(z) = u + iv be analytic. If
∂u
∂x
= u1(x, y) and
∂u
∂y
= u2(x, y) then
show that
f(z) =
ˆ
[u1(z, 0) − iu2(z, 0)] dz.
Solution: We know that
f0
(z) = ux + ivx = ux − iuy = u1(x, y) − iu2(x, y).
Substituting y = 0, we get
f0
(x) = u1(x, 0) − iu2(x, 0).
Replacing x by z, we get
f0
(z) = u1(z, 0) − iu2(z, 0) =⇒
ˆ
[u1(z, 0) − iu2(z, 0)] dz.
Example: Let f(z) = u + iv be analytic. If
∂v
∂y
= v1(x, y) and
∂v
∂x
= v2(x, y) then
show that
f(z) =
ˆ
[v1(z, 0) + iv2(z, 0)] dz.
13 / 22 Md Firoz Ali Complex Analysis

35. Problem set
Example: Let f(z) = u + iv be analytic. If
∂u
∂x
= u1(x, y) and
∂u
∂y
= u2(x, y) then
show that
f(z) =
ˆ
[u1(z, 0) − iu2(z, 0)] dz.
Solution: We know that
f0
(z) = ux + ivx = ux − iuy = u1(x, y) − iu2(x, y).
Substituting y = 0, we get
f0
(x) = u1(x, 0) − iu2(x, 0).
Replacing x by z, we get
f0
(z) = u1(z, 0) − iu2(z, 0) =⇒
ˆ
[u1(z, 0) − iu2(z, 0)] dz.
Example: Let f(z) = u + iv be analytic. If
∂v
∂y
= v1(x, y) and
∂v
∂x
= v2(x, y) then
show that
f(z) =
ˆ
[v1(z, 0) + iv2(z, 0)] dz.
Remark: This method of constructing an analytic function is called Milne-Thomson’s
method.
13 / 22 Md Firoz Ali Complex Analysis

36. 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.
14 / 22 Md Firoz Ali Complex Analysis

37. 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
f(z) =
ˆ
[u1(z, 0) − iu2(z, 0)] dz =
ˆ
[0 − i(ze−z
− e−z
)] dz = ize−z
+ c.
14 / 22 Md Firoz Ali Complex Analysis

38. Problem set
Example: If u − v = (x − y)(x2 + 4xy + y2) then find analytic function f(z) = u + iv
in terms of z.
15 / 22 Md Firoz Ali Complex Analysis

39. Problem set
Example: If u − v = (x − y)(x2 + 4xy + y2) 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)(x2
+ 4xy + y2
).
15 / 22 Md Firoz Ali Complex Analysis

40. Problem set
Example: If u − v = (x − y)(x2 + 4xy + y2) 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)(x2
+ 4xy + y2
).
Hence,
Ux = 3x2
+ 6xy − 3y2
= φ1(x, y), Uy = 3x2
− 6xy − 3y2
= φ2(x, y).
Therefore, by Milne’s method
F(z) =
ˆ
[φ1(z, 0) − iφ2(z, 0)] dz =
ˆ
[3z2
− i3z2
] dz = (1 − i)z3
+ c
Thus
F(z) = (1 + i)f(z) = (1 − i)z3
+ c =⇒ f(z) =
1 − i
1 + i
z3
+ c0
= −iz3
+ c0
15 / 22 Md Firoz Ali Complex Analysis

41. Problem set
Example: Find the radius of convergence of the power series
∞
X
n=1
nk
an
zn
.
16 / 22 Md Firoz Ali Complex Analysis

42. Problem set
Example: Find the radius of convergence of the power series
∞
X
n=1
nk
an
zn
.
Solution: Here an =
nk
an
.
1
R
= lim
n→∞

46. an+1
an

50. = lim
n→∞
(n + 1)k an
nk an+1
=
1
a
lim
n→∞
1 +
1
n
k
=
1
a
.
Thus the radius of convergence of the power series is R = a.
16 / 22 Md Firoz Ali Complex Analysis

51. Problem set
Example: Find the radius of convergence of the power series
∞
X
n=1
nk
an
zn
.
Solution: Here an =
nk
an
.
1
R
= lim
n→∞

55. an+1
an

59. = lim
n→∞
(n + 1)k an
nk an+1
=
1
a
lim
n→∞
1 +
1
n
k
=
1
a
.
Thus the radius of convergence of the power series is R = a.
Example: Find the radius of convergence of the power series
∞
X
n=1
n1/n
(z + i)n
.
16 / 22 Md Firoz Ali Complex Analysis

60. Problem set
Example: Find the radius of convergence of the power series
∞
X
n=1
nk
an
zn
.
Solution: Here an =
nk
an
.
1
R
= lim
n→∞

64. an+1
an

68. = lim
n→∞
(n + 1)k an
nk an+1
=
1
a
lim
n→∞
1 +
1
n
k
=
1
a
.
Thus the radius of convergence of the power series is R = a.
Example: Find the radius of convergence of the power series
∞
X
n=1
n1/n
(z + i)n
.
Solution: Here an = n1/n
.
1
R
= lim
n→∞
|an|1/n
= lim
n→∞
n1/n
1/n
= 10
= 1.
Thus the radius of convergence of the power series is R = 1.
16 / 22 Md Firoz Ali Complex Analysis

69. Problem set
Example: Find the radius of convergence of the power series
∞
X
n=1
n!
nn
zn
.
17 / 22 Md Firoz Ali Complex Analysis

70. Problem set
Example: Find the radius of convergence of the power series
∞
X
n=1
n!
nn
zn
.
Solution: Here an =
n!
nn
.
1
R
= lim
n→∞

74. an+1
an

78. = lim
n→∞
(n + 1)!nn
n!(n + 1)n+1
= lim
n→∞
nn
(n + 1)n
= lim
n→∞
1 +
1
n
−n
=
1
e
.
Thus the radius of convergence of the power series is R = e.
17 / 22 Md Firoz Ali Complex Analysis

79. Problem set
Example: Find the radius of convergence of the power series
∞
X
n=1
n!
nn
zn
.
Solution: Here an =
n!
nn
.
1
R
= lim
n→∞

83. an+1
an

87. = lim
n→∞
(n + 1)!nn
n!(n + 1)n+1
= lim
n→∞
nn
(n + 1)n
= lim
n→∞
1 +
1
n
−n
=
1
e
.
Thus the radius of convergence of the power series is R = e.
Example: Find the radius of convergence of the power series
∞
X
n=1
3n
n2 + 4n
z2n
.
17 / 22 Md Firoz Ali Complex Analysis

88. Problem set
Example: Find the radius of convergence of the power series
∞
X
n=1
n!
nn
zn
.
Solution: Here an =
n!
nn
.
1
R
= lim
n→∞

92. an+1
an

96. = lim
n→∞
(n + 1)!nn
n!(n + 1)n+1
= lim
n→∞
nn
(n + 1)n
= lim
n→∞
1 +
1
n
−n
=
1
e
.
Thus the radius of convergence of the power series is R = e.
Example: Find the radius of convergence of the power series
∞
X
n=1
3n
n2 + 4n
z2n
.
Solution: Let w = z2. Then the given series becomes
∞
X
n=1
3n
n2 + 4n
wn
. Here
an = 3n
n2+4n
.
1
R
= lim
n→∞

100. an+1
an

104. = 3 lim
n→∞
n2 + 4n
(n + 1)2 + 4(n + 1)
= 3.
Thus the series
∞
X
n=1
3n
n2 + 4n
wn
converges for |w| 1/3 and so the given series
converges for |z| 1/
√
3. Thus the radius of convergence of the given power series is
R = 1/
√
3.
17 / 22 Md Firoz Ali Complex Analysis

105. Problem set
Example: Find
ˆ
C
z dz along the following curves:
(i) z(t) = e2it
, t ∈ [−π, π] (ii) z(t) = t + it, t ∈ [0, 2].
18 / 22 Md Firoz Ali Complex Analysis

106. Problem set
Example: Find
ˆ
C
z dz along the following curves:
(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
C
z dz =
ˆ π
−π
z(t)z0
(t) dt =
ˆ π
−π
e−2it
(2i)e2it
dt = 2i
ˆ π
−π
dt = 4πi.
18 / 22 Md Firoz Ali Complex Analysis

107. Problem set
Example: Find
ˆ
C
z dz along the following curves:
(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
C
z dz =
ˆ π
−π
z(t)z0
(t) dt =
ˆ π
−π
e−2it
(2i)e2it
dt = 2i
ˆ π
−π
dt = 4πi.
(ii) If C : z(t) = t + it, t ∈ [0, 2] then
ˆ
C
z dz =
ˆ 2
0
z(t)z0
(t) dt =
ˆ 2
0
(t − it)(1 + i) dt = (1 + i)(1 − i)
ˆ 2
0
t dt = 4.
18 / 22 Md Firoz Ali Complex Analysis

108. Problem set
Example: Show that

113. ˆ
|z|=1
2z + 1
5 + z2
dz

118. ≤
3π
2
.
19 / 22 Md Firoz Ali Complex Analysis

119. Problem set
Example: Show that

124. ˆ
|z|=1
2z + 1
5 + z2
dz

129. ≤
3π
2
.
Solution: Let f(z) =
2z + 1
5 + z2
. On |z| = 1, we have
|f(z)| ≤
2|z| + 1
5 − |z|2
≤
3
4
.
Then by ML-formula

134. ˆ
|z|=1
2z + 1
5 + z2
dz

139. ≤
ˆ
|z|=1

143. 2z + 1
5 + z2

147. |dz| ≤
3
4
ˆ
|z|=1
|dz| =
3
4
× 2π =
3π
2
.
19 / 22 Md Firoz Ali Complex Analysis

148. Problem set
Example: Evaluate
ˆ
C
|z|2
dz along the square with
vertices 0, 1, 1 + i, i,
20 / 22 Md Firoz Ali Complex Analysis

149. Problem set
Example: Evaluate
ˆ
C
|z|2
dz along the square with
vertices 0, 1, 1 + i, i,
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 : z(t) = i(1 − t), t ∈ [0, 1]. x
y
C1
C2
C3
C4
20 / 22 Md Firoz Ali Complex Analysis

150. Problem set
Example: Evaluate
ˆ
C
|z|2
dz along the square with
vertices 0, 1, 1 + i, i,
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 : z(t) = i(1 − t), t ∈ [0, 1]. x
y
C1
C2
C3
C4
ˆ
C
|z|2
dz =
ˆ
C1
|z|2
dz +
ˆ
C2
|z|2
dz +
ˆ
C3
|z|2
dz +
ˆ
C4
|z|2
dz
=
ˆ 1
0
t2
dt +
ˆ 1
0
|1 + it|2
(i) dt +
ˆ 1
0
|(1 − t) + i|2
(−1) dt +
ˆ 1
0
|i(1 − t)|2
(−i) dt
=
ˆ 1
0
t2
dt + i
ˆ 1
0
(1 − t2
+ 2it) dt −
ˆ 1
0
(−2t + t2
+ 2i(1 − t)) dt − i
ˆ 1
0
(1 − t)2
dt
= ....
20 / 22 Md Firoz Ali Complex Analysis

151. Problem set
Example: Evaluate
ˆ
C
z|z| dz along the upper
semicircle |z| = R from R to −R and the line
segment [−R, R].
21 / 22 Md Firoz Ali Complex Analysis

152. Problem set
Example: Evaluate
ˆ
C
z|z| dz along the 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
y
21 / 22 Md Firoz Ali Complex Analysis

153. Problem set
Example: Evaluate
ˆ
C
z|z| dz along the 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
y
ˆ
C
z|z| dz =
ˆ
C1
z|z| dz +
ˆ
C2
z|z| dz
=
ˆ π
0
Reit
|Reit
|(iReit
) dt +
ˆ R
−R
t|t| dt
= iR3
ˆ π
0
e2it
dt +
ˆ 0
−R
(−t2
) dt +
ˆ R
0
(t2
) dt
= ....
21 / 22 Md Firoz Ali Complex Analysis

154. Problem set
Example: Evaluate
ˆ
C
4z3
dz along the following curves:
(i) z(t) = t2
+ it, t ∈ [0, 2] (ii) z(t) = t + it, t ∈ [0, 2] .
22 / 22 Md Firoz Ali Complex Analysis

155. Problem set
Example: Evaluate
ˆ
C
4z3
dz along the following curves:
(i) z(t) = t2
+ it, t ∈ [0, 2] (ii) z(t) = t + it, t ∈ [0, 2] .
Solution: (i) Since the function f(z) = 4z3 is analytic in C,
ˆ
C
4z3
dz 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) = z4 is the anti-derivative of f(z). Thus
ˆ
C
4z3
dz = F(z1) − F(z0) = (4 + 2i)4
22 / 22 Md Firoz Ali Complex Analysis

156. Problem set
Example: Evaluate
ˆ
C
4z3
dz along the following curves:
(i) z(t) = t2
+ it, t ∈ [0, 2] (ii) z(t) = t + it, t ∈ [0, 2] .
Solution: (i) Since the function f(z) = 4z3 is analytic in C,
ˆ
C
4z3
dz 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) = z4 is the anti-derivative of f(z). Thus
ˆ
C
4z3
dz = F(z1) − F(z0) = (4 + 2i)4
(ii) The initial and terminal points of the curve C are z0 = z(0) = 0 and
z1 = z(2) = 2 + 2i respectively. Thus
ˆ
C
4z3
dz =
h
z4
iz1
z0
= (2 + 2i)4
22 / 22 Md Firoz Ali Complex Analysis