# Complex Integration Questions and Answers – Zeros and Poles

This set of Complex Integration Multiple Choice Questions & Answers (MCQs) focuses on “Zeros and Poles”.

1. Which of the following zeroes is obtained by evaluating the function f(z)= $$\frac{z^2+1} {1-z^2}$$?
a) z = i; z = -i
b) z = 1; z = -1
c) z = 2; z = -2
d) z = 0

Explanation: The zeroes of f (z) are explained by f (z) = 0
i.e. f(z)=$$\frac{\left(z+i\right) (z-i)} {1-z^2} =0$$
(z+i)(z-i)=0
z = i is a simple zero.
z = -i is a simple zero.

2. Which of the following zeroes is obtained by evaluating f(z)=$$\frac{z^3-1} {z^3+1}$$?
a) 1, ω, ω2
b) f(z) has no zeroes
c) ±1, ±2, ±3
d) ±i

Explanation: The zeroes of f(z) are explained by f(z) = 0
$$\frac{z^3-1} {z^3+1} =0$$
z3-1 = 0
z = $${(1)} ^ {\frac {1}{3}}$$ = 1,ω,ω2 (cube rots of unity)
Thus, the zeroes of $$\frac{z^3-1} {z^3+1}$$ are 1,ω,ω2

3. Which of the following zeroes can be obtained by evaluating sin$$\frac {1} {z-a}$$?
a) z = $$a+\ \frac {1} {n\pi}$$, where n= ±1, ±2, ±3, …
b) z = $$a-\ \frac {1} {n\pi}$$, where n= ±1, ±2, ±3, …
c) z = $$a+\ \frac {1} {n\pi}$$, where n=1,2,3, …
d) z = a+nπ,where n=1,2,3, …

Explanation: The zeroes are given by f (z) = 0
sin$$\frac {1} {z-a}$$ =0
$$\frac {1} {z-a}$$ =nπ,n = ±1, ±2, ±3, …
(z-a) n π=1
Thus, the zeroes are $$z=a+\frac {1} {n\pi}$$, n= ±1, ±2, ±3, …

4. Which of the following zeroes is obtained by evaluating $$sin\frac{z-z} {{z}^3}$$?
a) f(z) has no zeroes
b) 1,ω,ω2
c) ±1, ±2, ±3, …
d) ±i

Explanation: Let f(z) = $$sin\frac{z-z} {{z}^3}$$
Zeroes of f (z) are given by f (z) = 0
f(z)=$$sin\frac{z-z} {{z}^3}$$ Zeroes of f(z) are given by f(z) = 0
f(z) = $$sin\frac{z-z} {{z}^3} = \frac {\left [z-\ \frac{z^3} {3!} +\frac{z^5} {5!}-\dots \right]-\ z} {z^3} =\ \frac{\frac{-z^3} {3!} +\frac{z^5} {5!}-\dots} {z^3}$$
f(z)=$$\frac {-1} {3!} +\frac{z^2} {5!}-\dots$$
Now, $$\lim_{{z \to 0}} \frac {sin z-z} {{z}^3} =\frac {-1} {3!} \neq 0$$
Therefore, f (z) has no zeroes.

5. Which of the following zeroes is obtained by evaluating$$\frac{1-e^{2x}} {z^4}$$?
a) 0, ±iπ,±i2π, …
b) 0, π,2π, …
c) 1, ±π, ±2π, …
d) 1, ±iπ,±i2π, …

Explanation: Let f (z) = $$\frac{1-e^{2z}} {z^4}$$
Zeroes of f (z) are given by f (z) = 0
$$\frac{1-e^{2z}} {z^4} =0$$
1-e2z=0
e2z = e2inπ
2z=2inπ
z=inπ
z=0, ±iπ,±i2π, …
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6. Which of the following zeroes is obtained by evaluating $$\frac{z^2-1} {1+\ z^2}$$?
a) f(z) has no zeroes
b) ±1
c) ±i, ±2i, …
d) 1

Explanation: The zeroes of f(z) are explained by f(z) = 0
i.e. f(z) = $$\frac{\left(z+1\right) (z-1)} {1+z^2} =0\$$
(z+1)(z-1) =0
z = 1 is a simple zero
z = -1 isa simple zero

7. Which of the following zeroes is obtained by evaluating cos$$\frac{1} {z-a}$$?
a) Z =0, ±iπ,±i2π, …
b) z= ±i, ±i2, …
c) z= a-$$\frac {1} {n\pi}$$, where n= ±1, ±2, ±3, …
d) z=a+$$\frac{2\pi} {n}$$, n = ±1, ±2, ±3 …

Explanation: The zeroes are given by f (z) = 0
cos$$\frac {1} {z-a}$$ =0
$$\frac {1} {z-a} =n\frac {\pi}{2}$$, n= ±1, ±3, …
(z-a) $$\frac {\pi}{2}$$ =1
Thus, the zeroes are z=a+$$\frac{2\pi} {n}$$, n= ±1, ±3, …

8. Which of the following zeroes is obtained by evaluating$$\frac{z^4-1} {z^3+1}$$?
a) z= ±i, ±2i, …
b) z= ±1, ±i, …
c) z= ±1, ±2, …
d) z = 1

Explanation: Given that f (z) = $$\frac{z^4-1} {z^3+1}$$ The zeroes are given by f (z) = 0
z4-1=0
z4=1
z=- 1, + 1, -i, +i
Therefore, the zeroes of this function are ±1, ±i

9. Which of the following zeroes is obtained by evaluating cos$$\frac{z-1} {{z}^2}$$?
a) z= ±i, ±2i, …
b) z= ±1, ±i, …
c) f(z) has no zeroes
d) z = 1

Explanation: Let f(z)=cos$$\frac{z-1} {{z}^2}$$
Zeroes of f(z) are given by f(z) = 0
f(z)=cos$$\frac{cosz-1} {{z}^2}$$ = $$\frac{\left [ 1 – \frac{z^{2}}{2!}+ \frac{z^{2}}{4!}+ \frac{z^{2}}{6!} \right ]-1}{z^{2}}$$
$$\frac{\frac{-z^2} {2!} +\frac{z^4} {4!}-\frac{z^6} {6!} +\dots} {z^2} =\ \frac {-1} {2!} +\ \frac{z^2} {4!}-\dots$$
Now, $$\lim_{z\rightarrow 0} \frac{cosz-1} {{z}^2} =\frac {-1} {2!} \neq 0$$

f(z) has no zeroes.

10. Which of the following zeroes is obtained by evaluating$$\frac{e^{\frac{1}{z}}} {{(z-a)}^2}$$?
a) f(z) has no zeroes
b) z= ±i, ±2i, …
c) z= ±1, ±i, …
d) z = 1

Explanation: Given: $$f\left (z \right) =\ \frac{e^{\frac{1}{z}}} {{(z-a)} ^2}$$
The zeroes of f (z) are given by f (z) = 0
Now, $$\lim_{z\rightarrow 0} \frac{{e}^\frac{1}{z}} {{(z-a)} ^2} =\frac {\infty}{{a}^2} = \infty \neq 0$$
f(z) has no zeroes.

11. Which of the following poles is obtained by evaluating$$\frac{1-e^{2z}} {z^4}$$?
a) z = 1 is a pole of order 2
b) z = 0 is a pole of order 3
c) z = 2 is a pole of order 4
d) z = 3 is a pole of order 4

Explanation: Given: f(z)=$$\frac {1-\ e^{2z}} {z^4}$$
$$=\ \frac {1-\left [1+\ \frac{2z} {1!} +\frac{{\left(2z\right)} ^2} {2!} +\frac{{\left(2z\right)} ^3} {3!} +\dots \right]} {z^4}$$
$$=\ \frac {-\left [\frac {2} {1!} +\frac{4z} {2!} +\frac{8z^2} {3!} +\dots \right]} {z^3}$$
Here, z = 0 is a pole of order 3

12. Which of the following poles is obtained by evaluating f(z)=$$\frac{z}{{(z-1)} ^2}$$?
a) z = 1 is a pole of order 2
b) z = 0 is a pole of order 1
c) z = 1 is a pole of order 3
d) z = 0 is a pole of order 2

Explanation: Given: f(z)=$$\frac{z}{{(z-1)} ^2}$$
Poles of f(z) are obtained by equating the denominator to zero.
(z-1) 2=0
We observe that
from the above given function,
z = 1 is a pole of order 2

13. Which of the following poles is obtained by evaluating f(z)$$\frac{{e}^z} {z^2+4}$$?
a) z = ±2i
b) z = ±i
c) z = 0
d) z = 1

Explanation: The poles of f(z) are obtained by equating the denominator to zero.
z2+4=0
z2=-4
z = ±2i
z = +2i is a simple pole
z = -2i is an an another simple pole

14. If a function f(z) is analytic in a region R, is zero at a point z=z0 in R, then z0 is called a zero of f(z)
a) True
b) False

Explanation: The zero of a function f(z) is given by f(z) = 0. If a function f(z) is analytic in a region R, it is zero at a point z = z0 in R, then z0 is called a zero of f(z). If f(z0) = 0 and f’ (z0) $$\neq$$0, then z = z0 is called a simple zero of f(z) or a zero of the first order.

15. If f (z0) = … = fn-1(z0) = 0 and fn(z0) $$\neq$$ 0, then z0 is called a zero of order n.
a) True
b) False

Explanation: An analytic function f(z) is said to have a zero of order n, if f (z) can be expressed as f(z)= (z-z0)n $$\emptyset$$(z) where $$\emptyset$$(z) is analytic and $$\emptyset$$(z0)$$\neq$$ 0. This can be explained by zero of order n.

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