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

View Answer

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, ω^{2}

b) f(z) has no zeroes

c) ±1, ±2, ±3

d) ±i

View Answer

Explanation: The zeroes of f(z) are explained by f(z) = 0

\(\frac{z^3-1} {z^3+1} =0\)

z

^{3}-1 = 0

z = \( {(1)} ^ {\frac {1}{3}} \) = 1,ω,ω

^{2}(cube rots of unity)

Thus, the zeroes of \(\frac{z^3-1} {z^3+1} \) are 1,ω,ω

^{2}

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, …

View Answer

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,ω^{2}

c) ±1, ±2, ±3, …

d) ±i

View Answer

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, …

View Answer

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-e

^{2z}=0

e

^{2z}= e

^{2inπ}

2z=2inπ

z=inπ

z=0, ±iπ,±i2π, …

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

View Answer

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 …

View Answer

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

View Answer

Explanation: Given that f (z) = \(\frac{z^4-1} {z^3+1} \) The zeroes are given by f (z) = 0

z

^{4}-1=0

z

^{4}=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

View Answer

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

View Answer

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

View Answer

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

View Answer

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

View Answer

Explanation: The poles of f(z) are obtained by equating the denominator to zero.

z

^{2}+4=0

z

^{2}=-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=z_{0} in R, then z_{0} is called a zero of f(z)

a) True

b) False

View Answer

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 = z

_{0}in R, then z

_{0}is called a zero of f(z). If f(z

_{0}) = 0 and f’ (z

_{0}) \(\neq \)0, then z = z

_{0}is called a simple zero of f(z) or a zero of the first order.

15. If f (z_{0}) = … = f^{n-1}(z_{0}) = 0 and f^{n}(z_{0}) \(\neq \) 0, then z_{0} is called a zero of order n.

a) True

b) False

View Answer

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-z

^{0})

^{n}\(\emptyset\)(z) where \(\emptyset\)(z) is analytic and \(\emptyset\)(z

_{0})\(\neq\) 0. This can be explained by zero of order n.

**Sanfoundry Global Education & Learning Series – Complex Integration.**

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