Pravegaa PYQ Series · Volume 1: Mathematical Physics

MP05: Contour Integration and Residues

38 previous-year questions with complete solutions, from IIT JAM, GATE, CSIR NET-JRF, JEST and TIFR, arranged from IIT JAM and GATE through to TIFR and from the most recent paper backwards. Click Show solution under any question.

IIT JAM: 1
GATE: 7
CSIR NET: 13
JEST: 10
TIFR GS: 7

MP05.1IIT JAM 2019, Q52NAT | 2 marks

The value of \(\left|\displaystyle\int_0^{3+i}\left(\bar{z}\right)^2dz\right|^2\), along the line \(3y=x\), where \(z=x+iy\) is

(Round off to 1 decimal places)

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On the line \(x=3y\), \(z=(3+i)y\) with \(y\) from \(0\) to \(1\), so \[ \bar{z}=(3-i)y,\qquad dz=(3+i)\,dy . \] \[ \int_0^{3+i}\left(\bar{z}\right)^2dz=\int_0^1(3-i)^2(3+i)\,y^2\,dy=(3-i)\,(3-i)(3+i)\int_0^1y^2dy=\frac{10(3-i)}{3}. \] \[ \left|\frac{10(3-i)}{3}\right|^2=\frac{100\times10}{9}=\frac{1000}{9} \;\Rightarrow\;\boxed{111.1} \]

Answer: 111.1 (range 111.0 to 111.2)

MP05.2GATE 2025, Q21MCQ | 1 mark

Consider the function \(f(z)=\dfrac{1}{z^2(z-2)^3}\) of a complex variable \(z\). The residues of the function at \(z=0\) and \(z=2\), respectively, are

\(-\dfrac38\) and \(\dfrac38\)
\(\dfrac38\) and \(-\dfrac{3}{16}\)
\(-\dfrac{3}{16}\) and \(\dfrac{3}{16}\)
\(-\dfrac38\) and \(\dfrac{3}{16}\)

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At \(z=0\) there is a pole of order 2: \[\operatorname{Res}(f,0)=\lim_{z\to0}\frac{d}{dz}\left[\frac{1}{(z-2)^3}\right]=\lim_{z\to0}\frac{-3}{(z-2)^4}=-\frac{3}{16}.\] At \(z=2\) there is a pole of order 3: \[\operatorname{Res}(f,2)=\frac{1}{2!}\lim_{z\to2}\frac{d^2}{dz^2}\left[\frac{1}{z^2}\right]=\frac12\lim_{z\to2}\frac{6}{z^4}=\frac12\cdot\frac{6}{16}=\frac{3}{16}.\] (Check: the sum of residues is zero, as it must be since \(f\sim z^{-5}\) at infinity.) \[\boxed{-\dfrac{3}{16}\ \text{and}\ \dfrac{3}{16}}\]

Answer: (C) \(-\frac{3}{16}\) and \(\frac{3}{16}\)

MP05.3GATE 2025, Q55MSQ | 2 marks

Consider the integral \[I=\frac{1}{2\pi i}\oint\frac{z^4-1}{\left(z-\frac ab\right)\left(z-\frac ba\right)}\,dz\] where \(z\) is a complex variable and \(a,b\) are positive real numbers. The integral is taken over a unit circle with center at the origin. Which of the following option(s) is/are correct?

\(I=\dfrac58\) when \(a=1,b=2\)
\(I=\dfrac{10}{3}\) when \(a=1,b=3\)
\(I=\dfrac58\) when \(a=2,b=1\)
\(I=\dfrac58\) when \(a=3,b=2\)

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The poles are at \(z=a/b\) and \(z=b/a\), which are reciprocals of each other. For \(a\neq b\), exactly one, \(z_0=\min(a/b,b/a)<1\), lies inside the unit circle, and it is a simple pole: \[I=\operatorname{Res}(f,z_0)=\frac{z_0^4-1}{z_0-1/z_0}.\]

(A) \(a=1,b=2\): \(z_0=\frac12\), \(I=\dfrac{\frac1{16}-1}{\frac12-2}=\dfrac{-15/16}{-3/2}=\dfrac58\). Correct.
(B) \(a=1,b=3\): \(z_0=\frac13\), \(I=\dfrac{\frac1{81}-1}{\frac13-3}=\dfrac{-80/81}{-8/3}=\dfrac{10}{27}\neq\dfrac{10}{3}\). Incorrect.
(C) \(a=2,b=1\): the poles are again \(2\) and \(\frac12\), so \(z_0=\frac12\) and \(I=\dfrac58\). Correct.
(D) \(a=3,b=2\): \(z_0=\frac23\), \(I=\dfrac{\frac{16}{81}-1}{\frac23-\frac32}=\dfrac{-65/81}{-5/6}=\dfrac{26}{27}\neq\dfrac58\). Incorrect.

\[\boxed{\text{(A), (C)}}\]

Answer: (A), (C)

MP05.4GATE 2023, Q56MSQ | 2 marks

Consider the complex function \[ f(z)=\frac{z^2\sin z}{(z-\pi)^4} \] At \(Z=\pi\), which of the following options is(are) CORRECT?

The order of the pole is 4
The order of the pole is 3
The residue at the pole is \(\dfrac\pi6\)
The residue at the pole is \(\dfrac{2\pi}{3}\)

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Put \(z-\pi=t\); then \(\sin z=\sin(t+\pi)=-\sin t\) and \[ f=\frac{-(t+\pi)^2\sin t}{t^4}=\frac{-\left(t^2+2\pi t+\pi^2\right)\left(t-\dfrac{t^3}{3!}+\cdots\right)}{t^4}. \] The numerator starts at order \(t\) (coefficient \(-\pi^2\neq0\)), so \(f\sim-\pi^2/t^3\): the pole is of order 3, not 4.

Residue \(=\) coefficient of \(t^{-1}\) \(=\) coefficient of \(t^3\) in \(-(t^2+2\pi t+\pi^2)(t-t^3/6+\cdots)\): \[ \text{Res}=-\left(1-\frac{\pi^2}{6}\right)=\frac{1}{6}\left(\pi^2-6\right)\approx0.645 , \] which matches neither \(\pi/6\) nor \(2\pi/3\). Only (B) is correct.

\(\boxed{\text{(B)}}\)

Answer: (B)

MP05.5GATE 2021, Q60NAT | 2 marks

A contour integral is defined as \[I_n=\oint_C\frac{dz}{(z-n)^2+\pi^2}\] where \(n\) is a positive integer and \(C\) is the closed contour, as shown in the figure, consisting of the line from \(-100\) to \(100\) and the semicircle traversed in the counter-clockwise sense.The value of \(\sum_{n=1}^5I_n\) (in integer) is

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The integrand has simple poles where \((z-n)^2=-\pi^2\): \[z=n+i\pi\quad\text{and}\quad z=n-i\pi.\] For \(n=1,\dots,5\), the pole \(z=n+i\pi\) lies in the upper half plane well inside the semicircle of radius 100, while \(z=n-i\pi\) lies outside \(C\). The residue at \(z=n+i\pi\) is \[\operatorname{Res}=\lim_{z\to n+i\pi}\frac{z-(n+i\pi)}{(z-n-i\pi)(z-n+i\pi)}=\frac{1}{2i\pi}.\] By the residue theorem, \[I_n=2\pi i\times\frac{1}{2\pi i}=1\quad\text{for every }n,\qquad\sum_{n=1}^5I_n=\boxed{5}\]

Answer: 5 (range 5 to 5)

MP05.6GATE 2020, Q32NAT | 1 mark

For a complex variable \(z\) and the contour \(c:|z|=1\) taken in the counter clockwise direction, \[ \frac{1}{2\pi i}\oint_C\left(z-\frac{2}{z}+\frac{3}{z^2}\right)dz=\underline{\hspace{2cm}} \]

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The only singularity inside \(|z|=1\) is the pole at \(z=0\). The integrand is already a Laurent series about \(z=0\); the residue is the coefficient of \(1/z\), which is \(b_1=-2\). (The terms \(z\) and \(3/z^2\) integrate to zero around a closed contour.) By the residue theorem, \[ \frac{1}{2\pi i}\oint_C\left(z-\frac{2}{z}+\frac{3}{z^2}\right)dz=\frac{1}{2\pi i}\cdot2\pi i\,(-2)=\boxed{-2} \]

Answer: \(-2\)

MP05.7GATE 2019, Q37MCQ | 2 marks

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Chapter 5 Contour Integration and Residues Previous Year Question with Solution

MP05: Contour Integration and Residues

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