Complex Analysis Questions and Answers – Residue Theorem

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

1. Which of the following is the bilinear transformation that maps the points z=0,1, i onto the points w=i,0, ∞ respectively?
a) $$\frac {z + 1} {z – i}$$
b) $$\frac {z – i} {z + 1}$$
c) $$\frac {z – 1} {z + 1}$$
d) $$\frac {z – i}{z + i}$$

Explanation: Given: z1=0, z2= -1, z3 = i
w1=i, w2=0, w3= ∞
Let the required transformation be
$$\frac{\left(w – w_{1} \right)(w_{2} – w_{3})}{\left( w – w_{3} \right)(w_{2} – w_{1})} = \frac{\left( z – z_{1} \right)(z_{2} – z_{3})}{\left( z – z_{3} \right)(z_{2} – z_{1})}$$
$$\frac {w – w_{1}} {w_{2} – w_1} = \frac {\left (z – z_{1} \right) (z_{2} – z_{3})} {\left (z – z_{3} \right) (z_{2} – z_{1})}$$
$$\frac {w – i}{0 – i} = \frac {(z – 0) (- 1 – i)} {(z – i)( – 1 – 0)}$$
$$\frac {w – i}{- i} = \frac{z} {(z – i)} (i + 1)$$
$$w – i = \frac{z} {z – i}( – i + 1)$$
$$w = \frac{z} {z – i}( – i + 1) + i = \frac {- iz + z + iz + i}{(z – i)} = \frac {z + 1} {z – i}$$

2. Which of the following is the bilinear transformation that maps the points ∞, i, 0 onto the points 0, i, ∞ respectively?
respectively?
a) -1/z
b) 1/z2
c) 1/z
d) -1/z2

Explanation:
Given: z1=∞, z2= i, z3=0
w1=0, w2=i, w3=∞
Let the required transformation be
$$\frac{\left( w – w_{1} \right)(w_{2} – w_{3})}{\left( w – w_{3} \right)(w_{2} – w_{1})} = \frac{\left( z – z_{1} \right)(z_{2} – z_{3})}{\left( z – z_{3} \right)(z_{2} – z_{1})}$$
$$\frac {w – w_{1}} {w_{2} – w_{1}} = \frac{z_{2} – z_{3}} {z – z_{3}}$$
$$\frac {w – 0} {i – 0} = \frac {i – 0} {z – 0}$$
$$w = \frac{-1} {z} = \frac {0\ z + (- 1)} {(1)z + 0}$$

3. Which of the following is the bilinear transformation that maps the points 1, i, -1 onto the points 0, 1, ∞ respectively?
A) $$\frac {(- i)z + i}{(1)z + 1}$$
b) 0
c) 1
d) -1

Explanation:
Given: z1=1, z2= i, z3=-1
w1=0, w2=1, w3=∞
Let the required transformation be
$$\frac{\left( w – w_{1} \right)(w_{2} – w_{3})}{\left( w – w_{3} \right)(w_{2} – w_{1})} = \frac{\left( z – z_{1} \right)(z_{2} – z_{3})}{\left( z – z_{3} \right)(z_{2} – z_{1})}$$
$$\frac {\left (w – w_{1} \right)} {\left(w_{2} – w_{1} \right)} = \frac {\left (z – z_{1} \right) (z_{2} – z_{3})} {\left (z – z_{3} \right) (z_{2} – z_{1})}$$
$$\frac {w – 0} {1 – 0} = \frac {(z – 1) (i + 1)} {(z + 1) (i – 1)}$$
$$w = \frac {(z – 1) (i + 1)} {(z + 1) (i – 1)}$$
$$w = \frac {z – 1} {z + 1} \lbrack – i\rbrack$$
$$w = \frac {(- i)z + i}{(1)z + 1}$$

4. Which of the following is obtained by evaluating $$\int_C \frac {(z – 1)} {(z – 1) ^ {2} (z – 2)}$$ dz, where C is the Circle $$|z – i|$$ = 2?
a)-2πi
b) 2πi
c) 4πi
d)-4πi

Explanation: Let f(z) = $$\frac {(z – 1)} {(z – 1) ^ {2} (z – 2)} = \frac {1} {(z – 1) (z – 2)}$$
Singular points of the function f(z) are got by equating the denominator to zero, we get (z-1) (z-2) = 0.
z = 1 is a simple pole and lies inside C.
z = 2 is a simple pole and lies outside C.
C is the circle |z-i| = 2 with centre(0,1) and radius r=2
Res[f(z),1] = limz->1(z-1) f(z)
limz->1(z-1) $$\frac {1} {(z-2) (z-1)} = \frac {1} {1-2} = -1$$
By Cauchy’s residue theorem,
Cf(z)dz = 2πi [sum of the residue]
2πi[-1] = -2πi

5. Which of the following is obtained by evaluating$$\int_C \frac{dz} {z sin sinz }$$, where C is |z| =1?
a) 0
b) 1
c)-1
d) 2

Explanation: Let f(z)$$= \frac {1} {z sin}$$ The singularity of f(z) is given by
z sinz = 0
$$z\left\lbrack z – \frac {z^ {3}} {3!} + \frac {z^ {5}} {5!} – \ldots \right\rbrack = 0$$
$$z^ {2}\left\lbrack 1 – \frac {z^ {2}} {3!} + \ \frac {z^ {4}} {5!} – \ldots \right\rbrack = 0$$
z = 0 is a pole of order 2
z = 0 lies inside C
Res[f(z),0] =lim_{z->0} $$\frac {1} {1!} \frac {d}{\text{dz}} \left\lbrack \frac {z^ {2}} {zsin} \right\rbrack = lim_{z->0}\frac{d}{dz}\left\lbrack \frac{z} {zsin} \right\rbrack$$
$$= lim_{z->0}\left\lbrack \frac {sinz(1) – zcosz}{sin^ {2}z} \right\rbrack = lim_{z->0} \left\lbrack \frac {sinz – zcosz}{sin^ {2}z} \right\rbrack$$
$$= \frac {0 – 0}{0} \rightarrow \left\lbrack \frac {0}{0} form \right\rbrack$$
$$= lim_{z->0}\frac {cosz – \lbrack z (-sinz) + cosz(1) \rbrack}{2sin z cos z} = \frac {zsinz} {2sinz cosz}$$
$$= lim_{z->0}\frac{z} {2 cosz} = \frac {0}{2}$$ = 0
By Cauchy’s Residue Theorem,
∫f(z)dz = 2πi(0) = 0
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6. Which of the following is obtained by evaluating ∫C sin⁡(1/z)dz sin ($$\frac{1}{z}$$)dz, where C is any circle with centre origin?
a) 2πi
b) 4πi
c) 8πi
d) 3πi

Explanation: Let f(z) = sin$$(\frac{1}{z})$$
The only singular point is z = 0 which is inside C.
sin $$(\frac{1}{z}) = \frac{1}{z} – \frac {1} {3! z^ {3}} +$$ ….
Res[f(z),0] = co – efficient of $$\frac{1}{z}$$ = 1
By Cauchy’s residue theorem,
$$\int_C$$ f(z)dz = 2πi [sum of residues]
$$\int_C$$ sinsin($$\frac{1}{z}$$)dz=2πi[1] =2πi

7. Which of the following is obtained by evaluating$$\int_C \frac {tan \frac{z} {2}} {{(z – 1 – i)} ^ {2}} {dz}$$, where C is the boundary of square, whose sides along the lines x = ±2 and y = ±2?
a) πi sec2 $$(\frac {1 + i}{2})$$
b) 2πi
c) πi
d) 4πi

Explanation: Let f(z) = $$\frac {tan tan \frac{z}{2}} {{\lbrack z – (1 + i)\rbrack}^{2}}$$
The poles are given by
[z-(1+i)] 2=0
z = 1 + i is a pole of order 2
The pole z = 1 + i lies inside C
Res[f(z), 1 + i] = $$\frac {1} {(2 – 1)!} \frac {d}{dz} [ [z – (1 + i)] ^ {2}f(z)]$$
$$= \frac {1} {1!} \frac {d}{dz}\left\lbrack \left\lbrack z – (1 + i) \right\rbrack^ {2} \frac {tan tan \frac{z}{2}} {\left\lbrack z – (1 + i) \right\rbrack^ {2}} \right\rbrack\$$
$$\left\lbrack \sec^ {2} \frac{z}{2} \right\rbrack\frac{1}{2} = \frac {1}{2} sec^ {2} \left (\frac {1 + i}{2} \right)$$
By Cauchy’s residue theorem,
$$\int f(z)dz=2 \pi$$ [sum of the residues]
2πi $$\left\lbrack \frac {1}{2} \sec^ {2} \left (\frac {1 + i}{2} \right) \right\rbrack = πi sec^ {2} \left (\frac {1 + i}{2} \right)$$

8. Which of the following is obtained by evaluating $$\int \frac{tantan\frac{z}{2}} {(z-a) ^2}$$ and C is the boundary of the square whose sides lie along x = ±2 and y = ±2?
a) πi sec2$$\frac{a}{2}$$
b) πi sin2$$\frac{a}{2}$$
c) 2πi
d) 4πi

Explanation: $$\frac {tan tan\frac{z}{2}} {{(z – a)} ^ {2}}$$
The poles are given by (z – a) 2= 0
z = a is a pole of order 2.
z = a lies inside C.
Res[f(z), a] = $$\frac {1} {(2 – 1)!} \frac {d}{dz} (z – a) ^ {2}$$ f(z)
$$= \frac {1} {1!} \frac {d}{dz} [{(z – a) ^ 2} \frac {tan tan \frac{z}{2}} {(z – a) ^ {2}}]$$
$$= \frac{d}{dz} [tan tan \frac{z}{2}] = sec^ {2} \frac{z}{2} \frac {1}{2} = \frac {1}{2} \sec^ {2} \frac{a}{2}$$
By Cauchy’s Residue Theorem,
$$\int$$ f(z)dz = 2πi ($$\sum$$ of the residue)
$$= 2πi \frac {1}{2} sec^ {2} \frac{a}{2} = πi \sec^ {2} \frac{a}{2}$$

9. Which of the following is obtained by evaluating $$\frac {z dz} {{(z^ {2} + 1)} ^ {2}},$$ where C is the circle |z – i| = 1?
a) 0
b) 1
c) -1
d) 2

Explanation: Let f(z) = $$\frac {z dz} {{(z^ {2} + 1)} ^ {2}} = \frac{z} {{[(z + i)(z – i)]} ^ {2}}$$
z = i is a pole of order 2, lies inside $$|z – i|$$ = 1
z = – i is a pole of order 2, lies outside $$|z – i|$$ = 1
Given: $$|z – i|$$ = 1
Here, Centre is (0,1) and radius is 1
Res[f(z),i]= $$\frac {1} {1!} \frac {d}{dz} [\frac {(z – i)^{2}z} {(z + i)^{2} (z – i)^{2}}]$$
$$= \frac {d}{dz} \frac{z}{(z+i)^{2}}$$
$$= \frac {(z + i)^{2}(1) – z2(z + i)} {{(z + i)} ^ {4}}$$
$$= \frac {z + i – 2z} {{(z + i)} ^ {3}}$$
$$= \frac {i – z} {{(z +i)} ^ {3}}$$ = 0
By Cauchy’s Residue Theorem,
$$\int$$ f(z) dz = 2πi $$\sum$$ of all residues
2πi[0] = 0

10. Which of the following is the residue of the function f(z) = $$\frac {- 1} {z – 1} – 2\left\lbrack 1 + (z – 1) + \ (z – 1) ^ {2} + \ldots \right\rbrack$$ at z = 1?
a) 0
b) -1
c) 1
d) 2

Explanation: The residue of f(z)at z = 1 is equal to the co – efficient of $$\frac {1} {z – 1}$$in the Laurent’s series of f(z) about z = 1.
The co – efficient of $$\frac {1} {z – 1}$$ in the Laurent’s series of f(z) about z = 1 is equal to – 1.

11. Which of the following is the residue of $$z^ {2} \sin\sin\ \frac{1}{z}$$ at$$z = 0$$?
a) $$\frac {- 1}{5}$$
b) $$\frac {1}{6}$$
c) $$\frac {- 1}{6}$$
d) $$\frac {1}{4}$$

Explanation: Let f(z) = z^ {2} sin sin $$\frac{1}{z}$$
f(z) = $$z^ {2}\left\lbrack \frac {\left(\frac{1}{z} \right)} {1!} – \frac {\left (\frac{1}{z} \right) ^ {3}} {3!} + \ldots \right\rbrack$$
f(z) = $$\frac{z} {1!} – \frac {1} {6z} + \ldots$$
Res [f(z),0] = co – efficient of $$\frac{1}{z}$$in the Laurent’s expansion
$$= \frac {- 1}{6}$$

12. Which of the following is the residue of $$cosec^ {2}$$ z at z = 0?
a) 0
b) 1
c)-1
d) 2

Explanation: f(z) = cosec2z = $$\frac {1} {\sin^ {2}z} = {(sinsinz)} ^ {- 2}$$
$$= \left\lbrack z – \ \frac {z^ {3}} {3!} + \frac {z^ {5}} {5!} – \ldots \right\rbrack^ {- 2} = \ z^ {- 2}\left\lbrack 1 – \frac {z^ {3}} {3!} + \frac {z^ {4}} {5!} – \ldots \right\rbrack ^ {- 2}$$
$$= \frac {1} {z^ {2}}\left\lbrack 1 – \left (\frac {z^ {2}} {3!} – \frac {z^ {4}} {5!} + \ldots \right) \right\rbrack^ {- 2}$$
$$= \frac {1} {z^ {2}}\left\lbrack 1 + 2\left (\frac {z^ {2}} {3!} – \frac {z^ {4}} {5!} + \ldots \right) + \ldots \right\rbrack$$
$$= \frac {1} {z^ {2}} + 2\left (\frac {1} {3!} – \frac {z^ {2}} {5!} + \ldots \right) + \ldots$$
Res[f(z),0] = Co – efficient of $$\frac{1}{z}$$ in the Laurent’s series expansion = 0

13. Which of the following is the residue of the function f(z) = $$\frac {4} {z^ {3} (z – 2)}$$ at a simple pole?
a) $$\frac {1}{2}$$
b) $$\frac {1}{3}$$
c) $$\frac {1}{4}$$
d) $$\frac {1}{5}$$

Explanation: Here, z = 2 is a simple pole.
Res f(z) at (z = 2) = z – 2) $$\frac {4} {z^ {3} (z – 2)}$$
$$= \frac {4} {z^ {3}} = \frac {4}{8} = \frac {1}{2}$$

14. Which of the following is the residue of $$\frac {4} {z^ {4} (z – 3)}$$at a simple pole?
a) $$\frac {4}{81}$$
b) $$\frac {4}{9}$$
c) $$\frac {9}{4}$$
d) $$\frac {81}{4}$$

Explanation: Here, z = 3 is a simple pole.
Res f(z) at (z = 3) = (z – 3) $$\frac {4} {z^ {4} (z – 3)}$$
$$= \frac {4} {z^ {4}}$$
$$= \frac {4} {3^ {4}} = \frac {4}{81}$$

15. Which of the following is the residue at the pole of the function f(z) = $$\frac {2z + 3} {{(z + 2)} ^ {2}}$$?
A) 2
b) 3
c) 4
d) 5

Explanation: Let f(z) =$$\frac {2z + 3} {{(z + 2)} ^ {2}} =\frac {2(z + 2) – 1} {{(z + 2)} ^ {2}} =\frac {2} {z + 2} -\frac {1} {{(z + 2)} ^ {2}}$$
which is the principal part of f(z)
Residue at the pole z = – 2 is the co – efficient of $$\frac {1} {z + 2}$$ which is equal to 2.

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