# Complex Analysis Questions and Answers – Residue

This set of Complex Analysis Multiple Choice Questions & Answers (MCQs) focuses on “Residue”.

1. Which of the following is the residue of $$\frac{1 – e^{2z}}{z^{4}}$$ at z = 0 ?
a) -4/3
b) 2/3
c) -2/3
d) 1/3

Explanation:
Given:f(z) = $$\frac{1 – e^{2z}}{z^{4}}$$
$$= \ \frac{1 – \left\lbrack 1 + \frac{2z}{1!} + \frac{(2z)^{2}}{2!} + \frac{(2z)^{3}}{3!} + \ \ldots \right\rbrack}{z^{4}}$$
$$= \frac{- \left\lbrack \frac{2}{1!} + \frac{4z}{2!} + \frac{8z^{2}}{3!} + \frac{16z^{3}}{4!} + \ldots \right\rbrack}{z^{3}}$$
Here, z = 0 is a pole of order 3
b1 = $$\frac{1}{(m – 1)!}\lim_{z \rightarrow z0}{\frac{d^{m – 1}} {\text{dz}^{m – 1}}\ \left\lbrack (z – z0)^{m}f(z) \right\rbrack\ }$$
Res f(z) at (z = 0) = $$\frac{1}{2!}\ \lim_{z \rightarrow 0}{\frac{d^{2}}{\text{dz}^{2}}\lbrack z^{3}f(z)\rbrack}$$
$$= \frac{1}{2}\ \lim_{z \rightarrow 0}{\frac{d^{2}}{\text{dz}^{2}}\left\lbrack – \left\lbrack \frac{2}{1} + 2z + \frac{4}{3}z^{2} + \frac{2}{3}z^{3} + \ldots \right\rbrack \right\rbrack}\$$
$$= \ \frac{1}{2}\ \lim_{z \rightarrow 0}{\frac{d}{\text{dz}}\left\lbrack – \left\lbrack 2 + \frac{8}{3}z + \frac{6}{3}z^{2}\ldots \right\rbrack \right\rbrack}$$
$$= \frac{1}{2}\lim_{z \rightarrow 0}{\left\lbrack – \left( \frac{8}{3} + \frac{12}{3}z + \ldots \right) \right\rbrack = \ \frac{1}{2}\left( \frac{- 8}{3} \right) = \ \frac{- 4}{3}}$$

2. Which of the following is the residue of f(z) = tan tan z at its singularity?
a) -1
b) 0
c) 1
d) 2

Explanation:
z= ±z/2, ±3π/2, ±5π/2, …are poles of order 1.
At each singularity,
Res {f(z)} = -1.

3. Which of the following is the residue of f(z) = tan tanz at z= π/2?
a) -1
b) 0
c) 1
d) 2

Explanation:
Res f(z) at $$\left( z = \frac{\pi}{2} \right) = \ \ \lim_{z \rightarrow \frac{\pi}{2}}{\left( z – \frac{\pi}{2} \right)\tan z = \ \lim_{z \rightarrow \frac{\pi}{2}}{\frac{\left( z – \frac{\pi}{2} \right)}{\cot z}\ \left\lbrack \frac{0}{0}\text{form} \right\rbrack}}$$

$$= \ \lim_{z \rightarrow \frac{\pi}{2}} {\frac {1} {- \text{cosec}^ {2}z} = – 1}$$ [L Hospital rule]

4. Which of the following is the residue of $$f(z) = \ \frac{1 – e^{- z}}{z^{3}}\$$ at $$z = 0$$?
a) -1/2
b) -1
c) 1
d) 1/2

Explanation:Given: f(z) = $$\frac{1 – e^{- z}}{z^{3}}$$
$$= \frac{1 – \left\lbrack 1 – \frac{z}{1!} + \frac{z^{2}}{2!} – \frac{z^{3}}{3!} + \ldots \right\rbrack}{z^{3}} = \ \frac{1 – \frac{z}{2!} + \frac{z^{2}}{3!} – \frac{z^{3}}{4!} + \ldots}{z^{2}}$$
Here, z=0 is a pole of order 2.

$$b1 = \frac{1}{(m – 1)!}\lim_{z \rightarrow z0}{\frac{d^{m – 1}}{\text{dz}^{m – 1}}\ \left\lbrack (z – z0)^{m}f(z) \right\rbrack\ }$$
Res f(z) at (z=0) $$= \frac{1}{1!}\lim_{z \rightarrow 0}{\frac{d}{\text{dz}}\left\lbrack z^{2}f(z) \right\rbrack = \ \lim_{z \rightarrow 0}{\frac{d}{\text{dz}}\left\lbrack 1 – \frac{z}{2!} + \frac{z^{2}}{3!} – \frac{z^{3}}{4!} + \ldots \right\rbrack}}$$ $$= lim_{z \rightarrow 0\ }{\left\lbrack \frac{- 1}{2!} + \ \frac{2z}{3!} – \frac{3z^{2}}{4!} + \ldots \right\rbrack = \ \frac{- 1}{2!} = \ \frac{- 1}{2}}$$

5. Which of the following is the residue of$$f(z) = \frac{z} {{(z – 1)} ^ {2}}$$ at its pole?
a) 1
b) 0
c) -1
d) 2

Explanation:
Given:f(z) = $$\frac{z}{{(z – 1)}^{2}}$$
Here, z=1 is a pole of order 2.
Re[z=z0] $$= \frac{1}{(m – 1)!}\lim_{z \rightarrow z0}{\frac{d^{m – 1}}{\text{dz}^{m – 1}}\ \left\lbrack (z – z0)^{m}f(z) \right\rbrack\ }$$
Here, m=2
Re[z=1] =
$$\lim_{z \rightarrow 1}{\frac{1}{1!}\frac{d}{\text{dz}}\left\lbrack (z – 1)^{2}\frac{z}{(z – 1)^{2}} \right\rbrack = \ \lim_{z \rightarrow 1}{\frac{d}{\text{dz}}\lbrack z\rbrack = \ \lim_{z \rightarrow 1}{(1) = 1}}}$$
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6. Which of the following is the residue of $$f(z) = \frac{z^{2}}{(z – 2){(z + 1)}^{2}}\ at\ z = 2$$?
a) 4/9
b) 1/3
c) 1/5
d) 9/4

Explanation:
f(z) $$= \frac{z^{2}}{(z – 2){(z + 1)}^{2}}\$$
Here,z=2 is a simple pole
Res f(z) at (z=2) $$\lim_{z \rightarrow 2}{(z – 2)\frac{z^{2}}{(z – 2){(z + 1)}^{2}} = \ \frac{4}{9}}$$

7. Which of the following is the residue of $$f(z) = \frac{(z + 1)}{(z – 1)(z + 2)}$$ at z = 1?
a) 2/3
b) 1/3
c) 1/2
d) 2/5

Explanation:
Given:f(z) = $$\frac{z + 1}{(z – 1)(z + 2)}$$
Here z=1 is a simple pole.
Res f(z) at (z = 1) = $$\lim_{z \rightarrow 1} {(z – 1) \frac {z + 1} {(z – 1) (z + 2)}}$$
$$= \lim_{z \rightarrow 1} {\frac {z + 1} {z + 2} = \ \frac {2}{3}}$$

8. Which of the following is the residue of $$f(z) = \frac{e^{2z}} {{(z + 1)} ^ {2}}$$ at its pole?
a) 2 e-2
b) e-3
c) e4
d) 3e-3

Explanation:
Given:f(z) = $$\frac{e^{2z}} {{(z + 1)} ^ {2}}$$ Here, z = – 1 is a pole of order 2
We know that, Res (z = z0) = $$\frac {1} {(m – 1)!} \lim_{z \rightarrow z0} {\frac {d^ {m – 1}} {\text{dz}^ {m – 1}} \left\lbrack (z – z0) ^{m}f(z) \right\rbrack\ }$$
Here, m = 2
Res = [z = – 1] $$= \ \lim_{z \rightarrow – 1} {\frac {1} {1!} \frac {d}{\text{dz}}\left\lbrack (z + 1) ^ {2} \frac{e^{2z}} {(z + 1) ^ {2}} \right\rbrack}$$
$$= \lim_{z \rightarrow – 1} {\frac{d}{\text{dz}}\left\lbrack e^{2z} \right\rbrack = \ \lim_{z \rightarrow – 1} {2e^{2z} = 2e^ {- 2}}}$$

9. Which of the following is the residue of cot z at z = 0?
a) 1
b) 0
c) -1
d) 2

Explanation:
Let f(z) = cot cot z = $$\frac {cos z} {sin z} = \frac{∅(z)}{φ(z)}$$
Here, z = 0 is a simple pole
∅(z) = cos cos z
∅(0) = 1

φ(z) = sin sin z
φ'(z) = cos cos z
φ'(0) = 1
Res [f(z),0]= $$\frac{∅(0)}{φ(0)} = \frac{1}{1}$$ = 1

10. Which of the following is the residue at $$z = 0$$ of the function$$f(z) = \ e^{\frac{1}{z}}$$?
a) 1
b) 0
c) 2
d) 3

Explanation: The residues are the co-efficient of $$\frac{1}{z}$$ in the Laurent’s expansions of f(z) about z = 0.
$$e^{\frac{1}{z}} = 1 + \frac {\left (\frac{1}{z} \right)} {1!} + \ \frac {\left (\frac{1}{z} \right) ^ {2}} {2!} + \ldots = 1 + \frac {1} {1!} \left (\frac{1}{z} \right) + \ \frac {1} {2!} \left (\frac {1} {z^ {2}} \right) + \ \frac {1} {3!} \left (\frac {1} {z^ {3}} \right) + \ldots$$
Res[f(Z),0] = Co-efficient of $$\frac{1}{z}$$ in Laurent’s expansion.
Res[f(z),0] $$\frac {1} {1!}$$ = 1 by definition of residue.

11. Which of the following is the residue atz = 0 of the function f(z) = $$\frac {\sin z} {z^ {4}}$$?
a) $$\frac {- 1}{6}$$
b) $$\frac {1}{3}$$
c) $$\frac {1}{4}$$
d) $$\frac {1}{5}$$

Explanation: $$\frac {sin z} {{z}^4} [z-\frac{{z}^3} {3!} +\frac{{z}^5} {5!} …] = \frac{1} {{z}^3}-\frac{1} {3!} \frac {1}{z} + \frac{z} {5!}-….$$
Res[f(z),0] = Co-efficient of $$\frac {1} {z}$$ in Laurent’s Expansion…
Res[f(z),0] = $$\frac {-1} {3!} = \frac {-1} {6}$$ by definition of residue.

12. Which of the following is the residue at z = 0 of the function f(z) = z cos $$\frac{1}{z}$$?
a) $$\frac {- 1}{2}$$
b) $$\frac {1}{3}$$
c)$$\ \frac {1}{4}$$
d) $$\frac {- 1}{4}$$

Explanation: z cos$${\frac{1}{z} = z\left\lbrack 1 – \frac{1} {2!}\frac{1}{z^{2}} + \frac{1}{4!}\frac{1} {z^ {4}} – \ldots \right\rbrack = z – \frac{1} {2!}\frac {1}{z} + \frac{1} {4!}\frac {1} {z^{3}} – \ldots}$$
Res[f(z),0] = co-efficient of 1/z in Laurent’s expansion.
Res[f(z),0] = -1/2!= -1/2

13. Which of the following is the residue of f(z) = $$\frac {1} {z^{2}e^{z}}$$?
a) -1
b) 0
c) 1
d) -2

Explanation: f(z) = $$\frac {e^ {- z}} {z^ {2}} = \frac {1 – z + \frac {z^ {2}} {2!} – \frac {z^ {3}} {3!} + \ldots}{z^ {2}}$$
$$\ \ \ \ \ \ \ \ \ \ = \ \frac {1} {z^ {2}} – \frac{1}{z} + \frac {1} {2!} – \frac{z} {3!} + \ldots$$
Res[f(Z),0] = Co-efficient of $$\frac{1}{z}$$ in Laurent’s expansion.
Res[f(Z),0] = -1

14. Which of the following is the residue of $$\text {z} e^{\frac{2}{z}}$$ at z = 0?
a) 4
b) 3
c) 2
d) 1

Explanation: The residues are the co – efficient of $$\frac{1}{z}$$ in the Laurent’s expansion of f(z) about z=0
$$z e^{\frac{2}{z}} = z\left (1 +\frac{2}{z} + (\frac{2}{z}) ^2 + (\frac{2}{z}) ^3 + …\right)$$
$$\ \ \ \ \ \ =z+2+ \frac {4} {{z}^2} (z)+\frac {8} {{z}^3} (z)… = z+2+\frac{4}{z} +\frac {8} {{z}^2} +…$$
Res[f(z),0] = Co-efficient of $$\frac{1}{z}$$ in Laurent’s expansion
Res[f(z),0] = 4 by definition of residue.

15. Which of the following is the residue of the function f(z) = $$\frac {z sin z} {{(z – \pi)} ^ {3}}$$ atz=π?
A) -1
b) 0
c) 1
d) 2

Explanation: Given: f(z) = $$\frac {z\sin z} {{(z – \pi)} ^ {3}}$$
Here, z = π is a pole of order 3
$$Res[f(z), \pi] = \lim_{z\rightarrow \pi}\frac{1}{2!}\frac {{d}^ 2} {d{z}^2}[(z-\pi)^3f(z)] = \lim_{z\rightarrow \pi}\frac{1}{2}\frac{{d}^2}{d{z}^2}[(z-\pi)^3 \frac{zsinz}{(z-\pi)^3}]$$ $$= \lim_{z \rightarrow \pi}\ \frac{1}{2}\frac{d^{2}}{dz^{2}}\lbrack z\sin{z\rbrack} = \ \lim_{z \rightarrow \pi}\ \frac{1}{2}\frac{d}{\text{dz}}\lbrack z\cos{z + \sin{z\rbrack}}$$ $$= \lim_{z \rightarrow \pi}\ \frac{1}{2}\lbrack z( – \sin{z) + \cos{z + \cos{z\rbrack = \ }}}\lim_{z \rightarrow \pi}\ \frac{1}{2}\lbrack – z\sin{z + 2\cos{z\rbrack}}$$ $$= \frac {1}{2}\left\lbrack 0 + 2 (- 1) \right\rbrack = \ – 1$$

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