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) \(\frac {z – i}{z + i}\)
View Answer
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
View Answer
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
View Answer
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
View Answer
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
View Answer
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
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
View Answer
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
View Answer
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
View Answer
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
View Answer
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
View Answer
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) \(\frac {1}{4} \)
View Answer
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
View Answer
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) \(\frac {1}{5} \)
View Answer
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) \(\frac {81}{4} \)
View Answer
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
View Answer
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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