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, ω2
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
\(\frac{z^3-1} {z^3+1} =0\)
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, …
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,ω2
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-e2z=0
e2z = e2inπ
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
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
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.
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
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 = 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
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-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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