Unit 2 - Complex Analysis

UNIT 2 – COMPLEX ANALYSIS
This set of Complex Analysis Multiple Choice Questions & Answers (MCQs) focuses on
“Functions of a Complex Variable”.
1. Find the domain of the function defined by f(z)=z/(z+z̅).
a) Im(z)≠0
b) Re(z)≠0
c) Im(z)=0
d) Re(z)=0
Answer: b
Explanation: Write z=x+iy ⇒ f(x+iy)=(x+iy)/(x+iy+x-iy)=(x+iy)/2x
=1/2+iy/2x ⇒ x≠0 ⇒ Re(z)≠0 .
2. Let f(z)=z+1/z. What will be the definition of this function in polar form?
a) (r+1/r)cosθ+i(r-1/r)sinθ
b) (r-1/r)cosθ+i(r+1/r)sinθ
c) (r+1/r)sinθ+i(r-1/r)cosθ
d) (r+1/r)sinθ+i(r-1/r)cosθ
Answer: a
Explanation: Write z=r(cosθ+isinθ), therefore, f(z)=z+1/z=r(cosθ+isinθ)+1/[r(cosθ+isinθ)]
=reiθ+(1/r)e-iθ=r(cosθ+isinθ)+1/r(cosθ-isinθ)=(r+1/r)cosθ+i(r-1/r)sinθ.
3. For the function f(z)=zi, what is the value of |f(ω)|+Arg f(ω), ω being the cube root of unity
with Im(ω)>0?
a) e-2π/3
b) e2π/3
c) e-2π/3+2π/3
d) e-2π/3-2π/3
Answer: a
Explanation: Let y=zi⇒ ln y=iln z=i(ln |z|+iarg z)=iln |z|-arg z
⇒ y=eiln |z|/earg z ⇒ |y|=earg z and Arg y=ln |z| ⇒ |f(ω)|+Arg f(ω)=e-2π/3+0=e-2π/3.
4. Let f(z)=(z2–z–1)7. If α2+α+1=0 and Im(α)>0, then find f(α).
a) 128α
b) -128α
c) 128α2
d) -128α2
Answer: c
Explanation: Note that α=ω. Therefore, f(α)=f(ω)=(ω2–ω–1)7
=(ω2+ω2)7=(2ω2)7=27ω14=128ω2=128α2.
5. For all complex numbers z satisfying Im(z)≠0, if f(z)=z2+z+1 is a real valued function,
then find its range.
a) (-∞, -1]
b) (-∞, 1/3)
c) (-∞, 1/2]
d) (-∞, 3/4)
Answer: d
Explanation: Let y=f(z). then z2+z+1=y has imaginary roots (∵Im(z)≠0)
⇒ D<0 ⇒ 1–4(1–y)<0 ⇒ 4y<3 ⇒ y<3/4 . Also, putting Re z=-1/2 and Im z=∞, we get, f(z)=-
∞.
6. Let x, y, z be integers, not all simultaneously equal. If ω is a cube root of unity with
Im(ω)≠1, and if f(z)=az2+bz+c, then find the range of |f(ω)|.
a) (0, ∞)
b) [1, ∞)
c) (√3/2, ∞)
d) [1/2, ∞)
Answer: b
Explanation: ω=-1/2+i√3/2. Therefore, |f(ω)|=|a+b(-1/2+i√3/2)+c(-1/2-i√3/2)|
=|(2a-b-c)/2+i(b√3-c√3)/2|=1/2[(2a-b-c)2+3(b-c)2]1/2={1/2[(a-b)2+(b-c)2+(c-a)2]}1/2.
Putting b=c=0 and a=1 gives us the minimum value=1, while, a=∞ gives us the maximum
value=∞.
7. Let f(z)=arg 1/(1 – z), then find the range of f(z) for |z|=1, z≠1.
a) (-∞, π/2)
b) (-π/2, π/2)
c) (-∞, ∞)
d) [0, π/2)
Answer: b
Explanation: Let y=1/(1-z) ⇒ z=1-1/y
|z|=1 ⇒ |1-1/y|=1 ⇒ |y-1|=|y| ⇒ locus of y is the perpendicular bisector of line segment

joining 0 and 1 ⇒ arg y ∈(-π/2, π/2).
8. Define f(z)=z2+bz−1=0 and g(z)=z2+z+b=0. If there exists α satisfying f(α)=g(α)=0,

which of the following cannot be a value of b?
a) √3i
b) -√3i
c) 0
d) √3i/2
Answer: d
Explanation: α2+bα−1=0 and α2+α+b=0 ⇒ (b−1)α−1−b=0 ⇒ α=(b+1)/(b-1)
⇒ (b+1)2/(b-1)2+(b+1)/(b-1)+b=0 ⇒ b=√3i, -√3i, 0.

9. Let f(z)=2(z+z̅)+3i(z-z̅) and g(z)=|z|. f(z)=2 divides the region g(z)≤6 into two parts. If
Q={(2+3i/4), (5/2+3i/4), (1/4-i/4), (1/8+i/4)}, then find the number of elements of Q lying
inside the smaller part.
a) 1
b) 2
c) 3
d) 4
Answer: b
Explanation: Writing z=x+iy, we get L as 2x–3y–1 and S as x2+y2–6, a point z1 lies in the
smaller region if L1>0 and S1<0. ∴ (2+3i/4) and (1/4-i/4) lie in the smaller region.
10. Find the range of the function defined by f(z)=Re[2iz/(1-z2)].
a) (−∞, 0) ⋃ (0, ∞)
b) [2, ∞)
c) (−∞, −1] ⋃ [1, ∞)
d) (−∞, 0] ⋃ [2, ∞)
Answer: c
Explanation: z=2i(x+iy)/(1-(x+iy)2)=2i(x+iy)/(1-(x2-y2+2ixy))
Using 1-x2=y2, z=(2ix-2y)/(2y2-2ixy)=-1/y
∵ –1≤y≤1 ⇒ –1/y≤-1 or -1/y≥1.
11. Let f(z)=|z|2+Re z(2(z+z̅)+3(z-z̅)/2i, the find the maximum value of |z|2/f(z).
a) 1
b) 2
c) 3
d) 4
Answer: b
Explanation: Write z=|z|(cosθ+isinθ) ⇒ |z|2/f(z)=1/(1+4cos2θ+3sinθcosθ)

=1/(1+4cos2θ+3/2sin2θ)=1/[2(1+cos2θ)+1+3/2sin2θ].
Now, 2(1+cos2θ)+1+3/2sin2θ=3+2cos2θ+3/2sin2θ≥3-(4+9/4)1/2=1/2.
Hence, maximum value is 2.
12. Consider a function f(z) of degree two, having real coefficients. If z1 and z2 satisfying
f(z1)=f(z2)=0 are such that Re z1=Re z2=0 and if z3 satisfies f(f(z3))=0, then select the
correct statement.
a) Re z3=0
b) Im z3=0
c) Re z3×Imz3≠0
d) Re z3=0 and Im z3=0
Answer: c
Explanation: f(z)=az2+b, with a, b of same sign ⇒ f(f(z))=a(az2+b)2+b
If z∈R or iz∈R ⇒ z2∈R ⇒ f(z)∈R ⇒ f(f(z))≠0 ⇒ Hence real or purely imaginary number
cannot satisfy f(f(z))=0.
13. Let f(z)=|1–z|, if zk=cos(2kπ/10)+isin(2kπ/10), then find the value of

f(z1)×f(z2)×…×f(z9).
a) 10
b) 15
c) 20
d) 30
Answer: a
Explanation: z10–1=(z-1)(z-z1)…(z-z9) ⇒ (z-z1)(z-z2) …(z-z9)=1+z+z2+…+z9.
Now, putting z=1, we get, (z-z1)(z-z2)…(z-z9)=f(z1)×f(z2)×…×f(z9)=10.

14. For a∈R, let f(z)=z5-5z+a. Select the correct statement for α satisfying f(α)=0.
a) α has exactly three possible real values for a>4
b) α has exactly one possible real value for a>4
c) α has exactly three possible real values for a<-4
d) α has exactly one possible real value for -4<a<4
Answer: b
Explanation: z5-5z+a=0 ⇒ z5-5z=-a ⇒ z(z-51/4)(z+51/4)(z2+51/2)=-a f'(z)=5z4–5=0 ⇒

(z2+1)(z2-1)=0 ⇒ (z-1)(z+1)(z2+1)=0 ⇒ α has exactly one possible real value for a>4 and
exactly three possible real values for -4<a<4.
15. Let f(z)=z4+a1z3+a2z2+a3z+a4=0; a1, a2, a3, a4 being real and non-zero. If f has a
purely imaginary root, then what is the value of the expression a3/(a1a2)+ a1a4/(a2a3) ?
a) 0
b) 1
c) -2
d) 2
Answer: b
Explanation: For real x(≠0), let ix be the root⇒x4-a1x3i- a2x2+a3xi+a4=0⇒x4-a2x2+a4=0

and a1x3-a3x=0
a1x3-a3x=0 ⇒ a1x2-a3=0 ⇒x2=a3/a1, putting this value in the equation,

a3/(a1a2)+a1a4/(a2a3)=1.
This set of Complex Analysis Objective Questions & Answers focuses on

“Differentiability”.
1. Which of the following is true?
a) Differentiability does not imply continuity
b) Differentiability implies continuity
c) Continuity implies differentiability

d) There is no relation between continuity and differentiable
Answer: b
Explanation: Any function that is differentiable is definitely continuous. If a function is not

continuous, it cannot be differentiable either. However, continuity cannot always imply
differentiability. Sometimes, functions that are continuous are not differentiable.
2. Which of the following is correct about the function f(x)=|x+5|?
a) Left and right limits are equal and hence it is differentiable
b) Left and right limits are not equal and hence it is differentiable
c) Left and right limits are equal and hence it is not differentiable
d) Left and right limits are not equal and hence it is not differentiable
Answer: d
Explanation: f(x)=(x+5), if x>-5 and f(x)=-(x+5), if x≤-5
Since the break is at x=-5, we calculate the limit at this point.
We know that, a function is not differentiable at point x=a, if either limh→0f(a+h)−f(a)hdoes

not exist or is infinity. We check limits for both the cases of the function.
Left limit: Here, a=-5.
limh→0f(−5+h)−f(−5)h
limh→0−(−5+h+5)−(−(−5+5))h
=-1
Right limit: Here, a=1.
limh→0f(1+h)−f(1)h
limh→0(1+h+5)−(1+5)h
=1
Since the two limits are not equal, the function is not differentiable.
3. Which of the following is correct about the function f(x)=|x2+18x+81|?
a) Left and right limits are equal and hence it is differentiable
b) Limits are not important to determine differentiability
c) Left and right limits are equal and hence it is not differentiable
d) Left and right limits are not equal and hence it is not differentiable
Answer: d
Explanation: f(x)=|(x+9)2|
Since the break is at x=-9, we frame for the function for both the sides of -9.
f(x)=(x+9)2, if x>-9 and f(x)=(x+9)2, if x≤-9
We know that, a function is not differentiable at point x=a, if either limh→0f(a+h)−f(a)hdoes

not exist or is infinity. We check limits for both the cases of the function.
Left limit: Here, a=-10.
limh→0f(−10+h)−f(−10)h
limh→0(−10+h+9)2−(−10+9)2h
limh→012+h2−2h−12h
=-2
Right limit: Here, a=10.
limh→0f(1+h)−f(1)h
limh→0(10+h+9)2−(10+9)2h
limh→0102+h2+20h−102h
=20
Since the two limits are not equal, the function is not differentiable.
4. Which of the following is true about f(z)=z2?
a) Continuous and differentiable
b) Continuous but not differentiable
c) Neither continuous nor differentiable

d) Differentiable but not continuous
Answer: a
Explanation: z=x+iy
In general the limits are discussed at origin, if nothing is specified.
f(x,y)=(x+iy)2
Left limit: limx→0y→0(x+iy)2
limy→0i2y2
=0
Right limit: limy→0x→0(x+iy)2
limx→0x2
=0
Both the limits are equal, therefore the function is continuous. To check differentiability,
f’(z)=limδz→0f(z+δz)−f(z)δzshould exist.
limδz→0(z+δz)2−(z)2δz
limδz→02z(δz)+(δz)2δz
=limδz→0(2z+(δz))
=2z
Since f’(z) exists, the function is differentiable as well.
5. Which of the following is true about f(z)=z+iz?
Answer: a
Explanation: z=x+iy

f(x, y)=x+iy+ix+i2y
f(x, y)=(x-y)+i(x+y)
Left limit: limx→0y→0(x-y)+i(x+y)
limy→0-y+iy
=0
Right limit: limy→0x→0x+iy+ix+i2y
limx→0x+ix
=0
f’(z)=limδz→0f(z+δz)−f(z)δz should exist.
limδz→0z+δz+iz+iδz−z+izδz
limδz→0δz+iδzδz
=z+i
6. Which of the following is true about f(z)=z2+2z?
Answer: a
Explanation: z=x+iy
f(x, y)=(x+iy)2+2(x+iy)
f(x, y)=x2-y2+2xiy+2x+2iy
Left limit: limx→0y→0x2-y2+2xiy+2x+2iy
limy→0-y2+2iy
=0
Right limit: limy→0x→0x2-y2+2xiy+2x+2iy
limx→0x2+2x
=0
f’(z)=limδz→0f(z+δz)−f(z)δz should exist.
limδz→0(z+δz)2+2(z+δz)−(z2+2z)δz
limδz→0z2+(δz)2+2z(δz)+2z+2(δz)−z2−2zδz
=2z+2
7. Which of the following is true about f(z)=z+izz2?
Answer: c
Explanation: z=x+iy
f(x, y)=x+iy+ix+i2y(x+iy)2
f(x, y)=i(x+y)+(x−y)(x+iy)2
Left limit: limx→0y→0i(x+y)+(x−y)(x+iy)2
limy→0i(y)+(−y)(iy)2
=does not exist
Since, the left limit itself does not exist, the function is not continuous. If a function is not
continuous, it cannot be differentiable as well.
8. Which of the following is true about f(z)=z2+(iz)2z2?
Answer: a
Explanation: z=x+iy
f(x, y)=(x+iy)2+(ix+i2y)2(x+iy)2
f(x, y)=(x+iy)2+(ix−y)2(x+iy)2
Left limit: limx→0y→0(x+iy)2+(ix−y)2(x+iy)2
limy→0(iy)2+(−y)2(iy)2
limy→0−(y)2+(y)2−(y)2
=−1+1−1
=0
Right limit: limy→0x→0(x+iy)2+(ix−y)2(x+iy)2
limx→0(x)2+(ix)2(x)2
limy→0(x)2−(x)2(x)2
=1−11
=0
f(z)=z2−z2z2=0
f’(z)=limδz→0f(z+δz)−f(z)δzshould exist.
limδz→00−0δz
=0

Example Question #1: Analytic And Harmonic Functions
Find a Harmonic Conjugate v(x,y) of u(x,y)=2x−x3+3xy2
Possible Answers:
v(x,y)=2y−3x2y+y3
v(x,y)=2−3x2+y3
v(x,y)=2y−x2y+y3
v(x,y)=2y−x3y+3xy2
v(x,y)=2x−x3+y3
Correct answer:
v(x,y)=2y−3x2y+y3
Explanation:
v(x,y)is said to be a harmonic conjugate of u(x,y) if there are both harmonic in their domain
and their first order partial derivatives satisfy the Cauchy-Riemann Equations. Computing the
partial derivatives
ux(x,y)=2−3x2+3y2=vy(x,y)
v(x,y)=2y−3x2y+y3+C
where C is any arbitrary constant.
Example Question #2: Analytic And Harmonic Functions
Given f(z)=z−z¯, where does f′(z) exist?
Possible Answers:
z=0
Nowhere
x=0
y=0
The Entire Complex Plane
Correct answer:
Nowhere
Explanation:
Rewriting f(z) in real and complex components, we have that
f(z)=z−z¯=(x+yi)−(x−yi)=2yi
So this implies that f(z)=f(x+yi)=u(x,y)+v(x,y)i
where u(x,y)=0 and v(x,y)=2y
Therefore, checking the Cauchy-Riemann Equations, we have that
ux(x,y)=0≠vy(x,y)=2
So the Cauchy-Riemann equations are never satisfied on the entire complex plane, so f(z) is
differentiable nowhere.
This set of Digital Signal Processing Multiple Choice Questions & Answers (MCQs)
focuses on “Bilinear Transformations”.
1. Bilinear Transformation is used for transforming an analog filter to a digital filter.
a) True
b) False
Answer: a
Explanation: The bilinear transformation can be regarded as a correction of the backward

difference method. The bilinear transformation is used for transforming an analog filter to a
digital filter.
2. Which of the following rule is used in the bilinear transformation?
a) Simpson’s rule
b) Backward difference
c) Forward difference
d) Trapezoidal rule
Answer: d
Explanation: Bilinear transformation uses trapezoidal rule for integrating a continuous time
function.
3. Which of the following substitution is done in Bilinear transformations?
a) s = 2T[1+z−11−z1]
b) s = 2T[1+z−11+]
c) s = 2T[1−z−11+z−1]
d) None of the mentioned
Answer: c
Explanation: In bilinear transformation of an analog filter to digital filter, using the

trapezoidal rule, the substitution for ‘s’ is given as
s = 2T[1−z−11+z−1].
4. What is the value of ∫nT(n−1)Tx(t)dt according to trapezoidal rule?
a) [x(nT)−x[(n−1)T]2]T
b) [x(nT)+x[(n−1)T]2]T
c) [x(nT)−x[(n+1)T]2]T
d) [x(nT)+x[(n+1)T]2]T
Answer: b
Explanation: The given integral is approximated by the trapezoidal rule. This rule states that
if T is small, the area (integral) can be approximated by the mean height of x(t) between the
two limits and then multiplying by the width. That is
∫nT(n−1)Tx(t)dt=[x(nT)+x[(n−1)T]2]T
5. What is the value of y(n)-y(n-1) in terms of input x(n)?
a) [x(n)+x(n−1)2]T
b) [x(n)−x(n−1)2]T
c) [x(n)−x(n+1)2]T
d) [x(n)+x(n+1)2]T
Answer: a
Explanation: We know that the derivative equation is

dy(t)/dt=x(t)
On applying integrals both sides, we get
∫nT(n−1)Tdy(t)=∫nT(n−1)Tx(t)dt
=> y(nT)-y[(n-1)T]=∫nT(n−1)Tx(t)dt
On applying trapezoidal rule on the right hand integral, we get
y(nT)-y[(n-1)T]=[x(nT)+x[(n−1)T]2]T
Since x(n) and y(n) are approximately equal to x(nT) and y(nT) respectively, the above
equation can be written as
y(n)-y(n-1)=[x(n)+x(n−1)2]T
6. What is the expression for system function in z-domain?
a) 2T[1+z−11−z1]
b) 2T[1+z−11−z1]
c) T2[1+z−11−z1]
d) T2[1−z−11+z−1]
Answer: c
Explanation: We know that
y(n)-y(n-1)= [x(n)+x(n−1)2]T
Taking z-transform of the above equation gives
=>Y(z)[1-z-1]=([1+z-1]/2).TX(z)
=>H(z)=Y(z)/X(z)=T2[1+z−11−z1].
7. In bilinear transformation, the left-half s-plane is mapped to which of the following in the
z-domain?
a) Entirely outside the unit circle |z|=1
b) Partially outside the unit circle |z|=1
c) Partially inside the unit circle |z|=1
d) Entirely inside the unit circle |z|=1

Answer: d
Explanation: In bilinear transformation, the z to s transformation is given by the expression
z=[1+(T/2)s]/[1-(T/2)s].
Thus unlike the backward difference method, the left-half s-plane is now mapped entirely
inside the unit circle, |z|=1, rather than to a part of it.
8. The equation s = 2T[1−z−11+z−1] is a true frequency-to-frequency transformation.
a) True
b) False
Answer: a
Explanation: Unlike the backward difference method, the left-half s-plane is now mapped
entirely inside the unit circle, |z|=1, rather than to a part of it. Also, the imaginary axis is
mapped to the unit circle. Therefore, equation s = 2T[1−z−11+z−1] is a true frequency-to-
frequency transformation.
9. If s=σ+jΩ and z=rejω, then what is the condition on σ if r<1?
a) σ > 0
b) σ < 0
c) σ > 1
d) σ < 1
Answer: b
Explanation: We know that if = σ+jΩ and z=rejω, then by substituting the values in the
below expression
s = 2T[1−z−11+z−1]
=>σ = 2T[r2−1r2+1+2rcosω]
When r<1 => σ < 0.

10. If s=σ+jΩ and z=rejω and r=1, then which of the following inference is correct?
a) LHS of the s-plane is mapped inside the circle, |z|=1
b) RHS of the s-plane is mapped outside the circle, |z|=1
c) Imaginary axis in the s-plane is mapped to the circle, |z|=1
d) None of the mentioned
Answer: c
Explanation: We know that if =σ+jΩ and z=rejω, then by substituting the values in the
below expression
s = 2T[1−z−11+z−1]
=>σ = 2T[r2−1r2+1+2rcosω]
When r=1 => σ = 0.
This shows that the imaginary axis in the s-domain is mapped to the circle of unit radius
centered at z=0 in the z-domain.
11. If s=σ+jΩ and z=rejω, then what is the condition on σ if r>1?
a) σ > 0
b) σ < 0
c) σ > 1
d) σ < 1
Answer: a
Explanation: We know that if = σ+jΩ and z=rejω, then by substituting the values in the
below expression
s = 2T[1−z−11+z−1]
=>σ = 2T[r2−1r2+1+2rcosω]
When r>1 => σ > 0.

12. What is the expression for the digital frequency when r=1?
a) 1Ttan(ΩT2)
b) 2Ttan(ΩT2)
c) 1Ttan−1(ΩT2)
d) 2Ttan−1(ΩT2)
Answer: d
Explanation: When r=1, we get σ=0 and
Ω = 2T[2sinω1+1+2cosω]
=>ω=2Ttan−1(ΩT2).
13. What is the kind of relationship between Ω and ω?
a) Many-to-one
b) One-to-many
c) One-to-one
d) Many-to-many
Answer: c
Explanation: The analog frequencies Ω=±∞ are mapped to digital frequencies ω=±π. The
frequency mapping is not aliased; that is, the relationship between Ω and ω is one-to-one. As
a consequence of this, there are no major restrictions on the use of bilinear transformation.
Q. The transformation technique in which there is one to one mapping from s-domain to z-
domain is
a. Approximation of derivatives
b. Impulse invariance method
c. Bilinear transformation method
d. Backward difference for the derivative
ANSWER: Bilinear transformation method

Question No. 38
The line integral of function F = yzi, in the counterclockwise direction, along the circle
x2+y2 = 1 at z = 1 is
(A) -2π
(B) -π
(C) π
(D) 2π
Answer : (B) -π
Question No. 236
Integration of the complex function f(z)=z2z2−1,in the counterclockwise direction, around

|z−1|=1,is
(A) -πi
(B) 0
(C) πi
(D) 2πi
Answer : (C) πi
Question No. 36
∫z2−4z2+4dz evaluated anticlockwise around the circle |z−i|=2 ,where i=−1−−−√, is
(A) −4π
(B) 0
(C) 2+π
(D) 2+2i
Answer : (A) −4π

Question No. 3
Given
f(z)= 1z+1−2z+3. If C is a counterclockwise path in the z-plane such that |z+1|=1, the value
of 12πj∮cf(z)dz is
(A) –2
(B) –1
(C) 1
(D) 2
Answer : (C) 1
Example Question #1 : Residue Theory
Cauchy's Residue Theorem is as follows:
Let C be a simple closed contour, described positively. If a function f is analytic inside C

except for a finite number of singular points zk inside C, then
∫Cf(z)dz=2πi∑k=1nResz=zkf(z)
Use Cauchy's Residue Theorem to evaluate the integral of f(z)=e−zz3 in the region |z|=4.
Possible Answers:
πi
3πi
−2πi
−πi2
−πi4
Correct answer:
πi
Explanation:
Note, for f(z)=e−zz3 a singularity exists where z=0. Thus, since where z=0 is the only
singularity for f(z) inside |z|=3, we seek to evaluate the residue for z=0.
Observe,
f(z) \newline = \frac{e^{-z}}{z^3} \newline = \frac{1}{z^3} \displaystyle

\sum_{k=0}^{\infty} \frac{(-1)^k z^k}{k!} \newline = \displaystyle \sum_{k=0}^{\infty}
\frac{(-1)^k z^{k-3}}{k!}
The coefficient of 1z is (−1)22!=12.
Thus,
Resz=0[f(z)]=Resz=0[e−zz3]=12.
Therefore, by Cauchy's Residue Theorem,
∫|z|=3f(z)dz=∫|z|=3e−zz2dz=2πiResz=0[e−zz2]=2πi(12)=πi
Hence,
∫|z|=3e−zz2dz=πi

Using Cauchy's Residue Theorem, evaluate the integral of f(z)=e−z(z−1)3 in the region |z|=4
Possible Answers:
−2πi(1+e)
πie
−e2πi
2πi
−2πi
Correct answer:
πie
Explanation:
Note, for f(z)=e−z(z−1)3 a singularity exists where z=1. Thus, since where z=1 is the only
singularity for f(z) inside |z|=4, we seek to evaluate the residue for z=1.
Observe,
f(z) \newline = \frac{e^{-z}}{(z-1)^3} \newline = \frac{e^{-u-1}}{u^3} \newline = \frac{

1}{e} * \frac{1}{u^3} \displaystyle \sum_{k=0}^{\infty} \frac{(-1)^k u^k}{k!} \newline =
\frac{1}{e}\displaystyle \sum_{k=0}^{\infty} \frac{(-1)^k u^{k-3}}{k!} \newline =
\frac{1}{e}\displaystyle \sum_{k=0}^{\infty} \frac{-1^{k}(z-1)^{k-3}}{k!}
The coefficient of 1z−1 is 1e∗(−1)22!=12e.
Thus,
Resz=0[f(z)]=Resz=0[e−z(z−1)2]=12e.
Therefore, by Cauchy's Residue Theorem,
∫|z|=3f(z)dz=∫|z|=3e−z(z−1)2dz=2πiResz=1[e−z(z−1)2]=2πi(12e)=πie
Hence,
∫|z|=3e−z(z−1)2dz=πie

Use Cauchy's Residue Theorem to evaluate the integral of f(z)=z3e1z in the region |z|=4.
Possible Answers:
−πi6
πi12
2πi
πi4
πi24
Correct answer:
πi12
Explanation:
Note, there is one singularity for f(z) where z=0.
Let u=1z
Then z=1u so f(z)=z3e1z=1u3eu.
Therefore, there is one singularity for f(z) where u=0. Hence, we seek to compute the residue
for f(z) where u=0=z
Observe,
f(z) \newline = z^{3} e^{\frac{1}{z}} \newline = \frac{1}{u^3} e^u \newline =

\frac{1}{u^3} \displaystyle \sum_{k=0}^{\infty} \frac{u^k}{k!} \newline = \displaystyle
\sum_{k=0}^{\infty} \frac{u^{k-3}}{k!} \newline = \displaystyle \sum_{k=0}^{\infty}
\frac{z^{3-k}}{k!}
So, when k=4, z3−k=z−1=1z.
Thus, the coefficient of 1z is 14!.
Therefore,
Resz=0[f(z)]=Resz=0[z2e1z]=14!
Hence, by Cauchy's Residue Theorem,
∫|z|=3z2e1zdz=2πiResz=0[z2e1z]=2πi∗14!=πi12
Therefore,
∫|z|=3z2e1zdz=πi12

For the following problem, use a modified version of the theorem which goes as follows:
Residue Theorem
If a function f is analytic everywhere in the finite plane except for a finite number of singular
points interior to a positively oriented simple closed contour C, then
∫Cf(z)dz=2πiResz=0[1z2f(1z)]
Use the Residue Theorem to evaluate the integral of f(z)=z61−z2 in the region |z|=3.
Possible Answers:
πi2
−πi3
πi
2πi
Correct answer:
Explanation:
Note,
1z2f(1z)=1z21z61−1z2=1z6z2−1=1z8−z6=−1z6∗11−z2
Thus, seeking to apply the Residue Theorem above for 1z2f(1z) inside |z|=3, we evaluate the
residue for z=0.
Observe,
\frac{1}{z^2}f(\frac{1}{z}) \newline = -\frac{1}{z^6}* \frac{1}{1-z^2} \newline = -

\frac{1}{z^6} \displaystyle \sum_{k=0}^{\infty} z^{2k} \newline = - \displaystyle
\sum_{k=0}^{\infty} z^{2k -6}
The coefficient of 1z is 0.
Thus,
Resz=0[1z2f(1z)]=0.
Therefore, by the Residue Theorem above,
∫|z|=3f(z)dz=2πiResz=0[1z2f(1z)]=2πi(0)=0
Hence,
∫|z|=3f(z)dz=0
Example Question 5: Residue Theory

Residue Theorem
Use the Residue Theorem to evaluate the integral of f(z)=5z+1z(z−2) in the region |z|=4.
Possible Answers:
πi
−2πi
10πi
−πi2
πi2
Correct answer:
10πi
Explanation:
Note,
1z2f(1z)=51z21z+11z−21z=51z+11−2z=5z(11−2z)+51−2z
residue for z=0.
Observe,
\frac{1}{z^2}f(\frac{1}{z}) \newline = \frac{5}{z} (\frac{1}{1 - 2z}) + \frac{5}{1 - 2z}

\newline = \frac{5}{z} \displaystyle \sum_{k=0}^{\infty} 2^{k} z^k + 5\displaystyle
\sum_{k=0}^{\infty} 2^{k} z^k \newline = 5\displaystyle \sum_{k=0}^{\infty} 2^{k} z^{k-
1} + 5\displaystyle \sum_{k=0}^{\infty} 2^{k} z^k \newline = 5\displaystyle
\sum_{k=0}^{\infty} 2^{k}( z^{k-1} + z^k)
The coefficient of 1z is 5∗20=5.
Thus,
Resz=0[1z2f(1z)]=5.
∫|z|=3f(z)dz=2πiResz=0[1z2f(1z)]=2πi(5)=10πi
Hence,
∫|z|=3f(z)dz=10πi
Example Question #6: Residue Theory

Residue Theorem
Use the Residue Theorem to evaluate the integral of f(z)=11+z3 in the region |z|=3.
Possible Answers:
−πi
4πi
−πi2
2πi
Correct answer:
0
Explanation:
Note,
1z2f(1z)=1z211+1z3=1z2+z−1=z1+z3
residue for z=0.
Observe,
\frac{1}{z^2}f(\frac{1}{z}) \newline = \frac{z}{1+z^2} \newline = z \displaystyle

\sum_{k=0}^{\infty} (-1)^k z^{3k} \newline = \displaystyle \sum_{k=0}^{\infty} (-1)^k
z^{3k+1}
Thus,
Resz=0[1z2f(1z)]=0.
∫|z|=2f(z)dz=2πiResz=0[1z2f(1z)]=2πi(0)=0
Hence,
∫|z|=2f(z)dz=0

Residue Theorem
Use the Residue Theorem to evaluate the integral of f(z)=2z in the region |z|=4.
Possible Answers:
πi
4πi
−πi4
−2πi
πi2
Correct answer:
4πi
Explanation:
Note,
1z2f(1z)=1z221z=1z2∗2z=2z
residue for z=0.
Observe, the coefficient of 2z is 2.
Thus,
Resz=0[1z2f(1z)]=2.
Hence,
∫|z|=2f(z)dz=4πi

Residue Theorem
Use the Residue Theorem to evaluate the integral of f(z)=7z−3z(z−1) in the region |z|=3.
Possible Answers:
14πi
2πi
\8 \pi i
−πi4
4πi
Correct answer:
14πi
Explanation:
Note,
1z2f(1z)=1z271z−31z(1z−1)=71z−31−z=7z(1−z)−31−z
residue for z=0.
Observe,
\frac{1}{z^2}f(\frac{1}{z}) \newline = \frac{7}{z (1-z)} - \frac{3}{1 -z} \newline =

\frac{7}{z}\displaystyle \sum_{k=0}^{\infty} z^{k} - 3\displaystyle \sum_{k=0}^{\infty}
z^{k} \newline = 7 \displaystyle \sum_{k=0}^{\infty} z^{k-1} - 3 \displaystyle
\sum_{k=0}^{\infty} z^{k}
Thus,
Resz=0[1z2f(1z)]=7.
Hence,
∫|z|=2f(z)dz=14πi
Find the residue of the function f(z)=cot(z)z6.
Possible Answers:
−2945
79
−145
−23
12
Correct answer:
−2945
Explanation:
Observe
f(z) \newline = \frac{cot(z)}{z^6} \newline = \frac{1}{z^6} (\frac{1}{z} - \frac{1}{3} z -

\frac{1}{45} z^3 - \frac{2}{945} z^5 - \cdots) \newline = \frac{1}{z^7} - \frac{1}{3}
\frac{1}{z^5} - \frac{1}{45} \frac{1}{z^3} - \frac{2}{945} \frac{1}{z} - \cdots
The coefficient of 1z is −2945.
Thus,
Resz=0[f(z)]=Resz=0[cot(z)z6]=−2945.
Find the residue at z=0 of f(z)=cos(1z).
Possible Answers:
0
14
−1
Correct answer:
Explanation:
Let u=1z.
Observe,
f(z) \newline = \cos (\frac{1}{z}) \newline =\cos (u) \newline =

\displaystyle\sum_{k=0}^{\infty} \frac{(-1)^{n}}{(2n)!} u^{2n} \newline =
\displaystyle\sum_{k=0}^{\infty} \frac{(-1)^{n}}{(2n)!} z^{-2n}
The coefficient of 1z is 0 since there is no 1z term in the sum.
Thus,
Resz=0[f(z))]=Resz=0[zcos(1z)]=0

Unit 2 - Complex Analysis

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Unit 2 - Complex Analysis

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UNIT 2 – COMPLEX ANALYSIS

1. Find the domain of the function defined by f(z)=z/(z+z̅).

Explanation: Write z=x+iy ⇒ f(x+iy)=(x+iy)/(x+iy+x-iy)=(x+iy)/2x

=1/2+iy/2x ⇒ x≠0 ⇒ Re(z)≠0 .

Explanation: Write z=r(cosθ+isinθ), therefore, f(z)=z+1/z=r(cosθ+isinθ)+1/[r(cosθ+isinθ)]

Explanation: Let y=zi⇒ ln y=iln z=i(ln |z|+iarg z)=iln |z|-arg z

⇒ y=eiln |z|/earg z ⇒ |y|=earg z and Arg y=ln |z| ⇒ |f(ω)|+Arg f(ω)=e-2π/3+0=e-2π/3.

4. Let f(z)=(z2–z–1)7. If α2+α+1=0 and Im(α)>0, then find f(α).

Explanation: Note that α=ω. Therefore, f(α)=f(ω)=(ω2–ω–1)7

Explanation: Let y=f(z). then z2+z+1=y has imaginary roots (∵Im(z)≠0)

Explanation: ω=-1/2+i√3/2. Therefore, |f(ω)|=|a+b(-1/2+i√3/2)+c(-1/2-i√3/2)|

Explanation: Let y=1/(1-z) ⇒ z=1-1/y

|z|=1 ⇒ |1-1/y|=1 ⇒ |y-1|=|y| ⇒ locus of y is the perpendicular bisector of line segment

8. Define f(z)=z2+bz−1=0 and g(z)=z2+z+b=0. If there exists α satisfying f(α)=g(α)=0,

Explanation: α2+bα−1=0 and α2+α+b=0 ⇒ (b−1)α−1−b=0 ⇒ α=(b+1)/(b-1)

⇒ (b+1)2/(b-1)2+(b+1)/(b-1)+b=0 ⇒ b=√3i, -√3i, 0.

10. Find the range of the function defined by f(z)=Re[2iz/(1-z2)].

c) (−∞, −1] ⋃ [1, ∞)

Using 1-x2=y2, z=(2ix-2y)/(2y2-2ixy)=-1/y

∵ –1≤y≤1 ⇒ –1/y≤-1 or -1/y≥1.

Explanation: Write z=|z|(cosθ+isinθ) ⇒ |z|2/f(z)=1/(1+4cos2θ+3sinθcosθ)

Hence, maximum value is 2.

d) Re z3=0 and Im z3=0

Explanation: f(z)=az2+b, with a, b of same sign ⇒ f(f(z))=a(az2+b)2+b

13. Let f(z)=|1–z|, if zk=cos(2kπ/10)+isin(2kπ/10), then find the value of

Explanation: z10–1=(z-1)(z-z1)…(z-z9) ⇒ (z-z1)(z-z2) …(z-z9)=1+z+z2+…+z9.

Now, putting z=1, we get, (z-z1)(z-z2)…(z-z9)=f(z1)×f(z2)×…×f(z9)=10.

a) α has exactly three possible real values for a>4

b) α has exactly one possible real value for a>4

c) α has exactly three possible real values for a<-4

d) α has exactly one possible real value for -4<a<4

Explanation: z5-5z+a=0 ⇒ z5-5z=-a ⇒ z(z-51/4)(z+51/4)(z2+51/2)=-a f'(z)=5z4–5=0 ⇒

Explanation: For real x(≠0), let ix be the root⇒x4-a1x3i- a2x2+a3xi+a4=0⇒x4-a2x2+a4=0

a1x3-a3x=0 ⇒ a1x2-a3=0 ⇒x2=a3/a1, putting this value in the equation,

This set of Complex Analysis Objective Questions & Answers focuses on

1. Which of the following is true?

a) Differentiability does not imply continuity

b) Differentiability implies continuity

c) Continuity implies differentiability

Explanation: Any function that is differentiable is definitely continuous. If a function is not

2. Which of the following is correct about the function f(x)=|x+5|?

a) Left and right limits are equal and hence it is differentiable

Explanation: f(x)=(x+5), if x>-5 and f(x)=-(x+5), if x≤-5

Since the break is at x=-5, we calculate the limit at this point.

We know that, a function is not differentiable at point x=a, if either limh→0f(a+h)−f(a)hdoes

Left limit: Here, a=-5.

Right limit: Here, a=1.

a) Left and right limits are equal and hence it is differentiable

b) Limits are not important to determine differentiability

f(x)=(x+9)2, if x>-9 and f(x)=(x+9)2, if x≤-9

We know that, a function is not differentiable at point x=a, if either limh→0f(a+h)−f(a)hdoes

Left limit: Here, a=-10.

Right limit: Here, a=10.

4. Which of the following is true about f(z)=z2?

a) Continuous and differentiable

b) Continuous but not differentiable

c) Neither continuous nor differentiable

In general the limits are discussed at origin, if nothing is specified.