Complex Analysis and Residues Questions and Answers

Q: The value of \(\oint z\exp\left(\frac{1}{z}\right)dz\), where the closed contour is a unit circle traversed anticlockwise is :

The value of the integral is \(\pi i\).The only singularity inside the unit circle is the essential singularity at z = 0, so expand in a Laurent series:\[ \begin{aligned} z\,e^{1/z} &= z\left(1 + \frac{1}{z} + \frac{1}{2!z^{2}} + \cdots\right) \\ &= z + 1 + \frac{1}{2z} + \cdots \end{aligned} \]The residue is the coefficient of \(1/z\), which is 1/2, and the integral is \(2\pi i \times \frac{1}{2} = \pi i\).\(2\pi i\) is what you get by taking the residue as 1, the coefficient of \(1/z\) in \(e^{1/z}\) alone, forgetting the extra factor z. Zero would follow only if the integrand were analytic inside the circle, and \(\pi\) loses the factor i.

Q: The value of the integral \(\int \frac{\sin z}{z}\,dz\) over the contour C, C given by |z| = 1 and traversed counter-clockwise, is :

The integral is 0.\(\frac{\sin z}{z}\) looks singular at \(z = 0\), but its Taylor series is \(1 - \frac{z^{2}}{3!} + \frac{z^{4}}{5!} - \cdots\), so the singularity is removable and the function is analytic inside and on \(|z| = 1\). The residue at 0 is the coefficient of \(\frac{1}{z}\), which is zero.By Cauchy's theorem, the integral of an analytic function round a closed contour is 0. \(2\pi i\) would be right for \(\frac{\cos z}{z}\) or \(\frac{e^{z}}{z}\), whose residue at 0 is 1. The options with \(\sqrt{3}\) do not arise from any residue here.

Q: It is given that the residue of the complex function \(\frac{e^{1/z}}{z^{n}}\) at an isolated singular point \(z = 0\) is \(\frac{1}{9!}\). The value of n must be equal to :

The value is \(n = -8\).Expand the essential singularity in a Laurent series and divide by \(z^{n}\):\[ \begin{aligned} \frac{e^{1/z}}{z^{n}} &= \sum_{k=0}^{\infty} \frac{z^{-k}}{k!}\cdot z^{-n} \\ &= \sum_{k=0}^{\infty} \frac{z^{-(k+n)}}{k!} \end{aligned} \]The residue is the coefficient of \(z^{-1}\), which needs \(k + n = 1\), so the residue is \(1/(1-n)!\). Setting \(1/(1-n)! = 1/9!\) gives \(1 - n = 9\), so \(n = -8\).With \(n = 1\) the residue would be \(1/0! = 1\); with \(n = -1\) it is \(1/2!\); and \(n = -9\) gives \(1/10!\). Only \(n = -8\) produces \(1/9!\).

Q: The function \(f(x) = \exp(2x)\sin(2x)\) can be expanded as the Taylor series :

The series is \(\sum_{n=0}^{\infty} 2^{3n/2}\sin\left(\frac{n\pi}{4}\right)\frac{x^{n}}{n!}\).Write the function as the imaginary part of one exponential:\[ \begin{aligned} e^{2x}\sin 2x &= \mathrm{Im}\, e^{(2+2i)x} \\ 2 + 2i &= 2\sqrt{2}\, e^{i\pi/4} \\ (2+2i)^{n} &= 2^{3n/2} e^{in\pi/4} \end{aligned} \]since \((2\sqrt{2})^{n} = (2^{3/2})^{n}\). Expanding \(e^{(2+2i)x} = \sum (2+2i)^{n}x^{n}/n!\) and taking the imaginary part gives the coefficient \(2^{3n/2}\sin(n\pi/4)/n!\).Check: the \(x\) coefficient of \(e^{2x}\sin 2x\) is 2, and \(2^{3/2}\sin(\pi/4) = 2\). The option with \(2^{n/2}\) gives \(\sqrt{2}\sin(\pi/4) = 1\), the one with \(2^{n}\) gives \(\sqrt{2}\), and the one with a fixed \(2^{3/2}\) is not a power of \(n\) at all.

Q: The Cauchy principal value of the integral \(\int_{-\infty}^{\infty} dx \frac{e^{2ix}\sin x}{x}\) is :

The principal value is 0.Write \(\sin x = (e^{ix} - e^{-ix})/2i\), so the integrand splits into two oscillating terms divided by \(x\):\[ I = \frac{1}{2i}\mathrm{P}\int\frac{e^{3ix} - e^{ix}}{x}dx \]For any \(k \gt 0\), closing the contour in the upper half plane and indenting around the pole at the origin gives \(\mathrm{P}\int e^{ikx}/x\,dx = i\pi\), independent of \(k\). The two terms are therefore equal and cancel:\[ I = \frac{1}{2i}(i\pi - i\pi) = 0 \]The options \(\pi i\) and \(2\pi i\) are what one term alone (or twice one term) would give, which forgets the subtraction; 1 has no source in the calculation.

11.

The value of \(\oint z\exp\left(\frac{1}{z}\right)dz\), where the closed contour is a unit circle traversed anticlockwise is :

A.\(\pi i\)
B.\(2\pi i\)
C.\(\pi\)
D.0

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: A

The value of the integral is \(\pi i\).

The only singularity inside the unit circle is the essential singularity at z = 0, so expand in a Laurent series:

\[ \begin{aligned} z\,e^{1/z} &= z\left(1 + \frac{1}{z} + \frac{1}{2!z^{2}} + \cdots\right) \\ &= z + 1 + \frac{1}{2z} + \cdots \end{aligned} \]

The residue is the coefficient of \(1/z\), which is 1/2, and the integral is \(2\pi i \times \frac{1}{2} = \pi i\).

\(2\pi i\) is what you get by taking the residue as 1, the coefficient of \(1/z\) in \(e^{1/z}\) alone, forgetting the extra factor z. Zero would follow only if the integrand were analytic inside the circle, and \(\pi\) loses the factor i.

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12.

The value of the integral \(\int \frac{\sin z}{z}\,dz\) over the contour C, C given by |z| = 1 and traversed counter-clockwise, is :

A.\(2\pi i\)
B.\(\frac{2\pi i}{\sqrt{3}}\)
C.0
D.\(\frac{\pi}{\sqrt{3}}\)

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: C

The integral is 0.

\(\frac{\sin z}{z}\) looks singular at \(z = 0\), but its Taylor series is \(1 - \frac{z^{2}}{3!} + \frac{z^{4}}{5!} - \cdots\), so the singularity is removable and the function is analytic inside and on \(|z| = 1\). The residue at 0 is the coefficient of \(\frac{1}{z}\), which is zero.

By Cauchy's theorem, the integral of an analytic function round a closed contour is 0. \(2\pi i\) would be right for \(\frac{\cos z}{z}\) or \(\frac{e^{z}}{z}\), whose residue at 0 is 1. The options with \(\sqrt{3}\) do not arise from any residue here.

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13.

It is given that the residue of the complex function \(\frac{e^{1/z}}{z^{n}}\) at an isolated singular point \(z = 0\) is \(\frac{1}{9!}\). The value of n must be equal to :

A.\(n = 1\)
B.\(n = -9\)
C.\(n = -1\)
D.\(n = -8\)

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: D

The value is \(n = -8\).

Expand the essential singularity in a Laurent series and divide by \(z^{n}\):

\[ \begin{aligned} \frac{e^{1/z}}{z^{n}} &= \sum_{k=0}^{\infty} \frac{z^{-k}}{k!}\cdot z^{-n} \\ &= \sum_{k=0}^{\infty} \frac{z^{-(k+n)}}{k!} \end{aligned} \]

The residue is the coefficient of \(z^{-1}\), which needs \(k + n = 1\), so the residue is \(1/(1-n)!\). Setting \(1/(1-n)! = 1/9!\) gives \(1 - n = 9\), so \(n = -8\).

With \(n = 1\) the residue would be \(1/0! = 1\); with \(n = -1\) it is \(1/2!\); and \(n = -9\) gives \(1/10!\). Only \(n = -8\) produces \(1/9!\).

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14.

The function \(f(x) = \exp(2x)\sin(2x)\) can be expanded as the Taylor series :

A.\(\sum_{n=0}^{\infty} 2^{\frac{3n}{2}} \sin(\frac{n\pi}{4})\frac{x^{n}}{n!}\)
B.\(\sum_{n=0}^{\infty} 2^{\frac{n}{2}} \sin(\frac{n\pi}{4})\frac{x^{n}}{n!}\)
C.\(\sum_{n=0}^{\infty} 2^{\frac{3}{2}} \sin(\frac{n\pi}{4})\frac{x^{n}}{n!}\)
D.\(\sum_{n=0}^{\infty} 2^{n} \sin(\frac{n\pi}{4})\frac{x^{n}}{n!}\)

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: A

The series is \(\sum_{n=0}^{\infty} 2^{3n/2}\sin\left(\frac{n\pi}{4}\right)\frac{x^{n}}{n!}\).

Write the function as the imaginary part of one exponential:

\[ \begin{aligned} e^{2x}\sin 2x &= \mathrm{Im}\, e^{(2+2i)x} \\ 2 + 2i &= 2\sqrt{2}\, e^{i\pi/4} \\ (2+2i)^{n} &= 2^{3n/2} e^{in\pi/4} \end{aligned} \]

since \((2\sqrt{2})^{n} = (2^{3/2})^{n}\). Expanding \(e^{(2+2i)x} = \sum (2+2i)^{n}x^{n}/n!\) and taking the imaginary part gives the coefficient \(2^{3n/2}\sin(n\pi/4)/n!\).

Check: the \(x\) coefficient of \(e^{2x}\sin 2x\) is 2, and \(2^{3/2}\sin(\pi/4) = 2\). The option with \(2^{n/2}\) gives \(\sqrt{2}\sin(\pi/4) = 1\), the one with \(2^{n}\) gives \(\sqrt{2}\), and the one with a fixed \(2^{3/2}\) is not a power of \(n\) at all.

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15.

The Cauchy principal value of the integral \(\int_{-\infty}^{\infty} dx \frac{e^{2ix}\sin x}{x}\) is :

A.\(2\pi i\)
B.\(\pi i\)
C.0
D.1

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: C

The principal value is 0.

Write \(\sin x = (e^{ix} - e^{-ix})/2i\), so the integrand splits into two oscillating terms divided by \(x\):

\[ I = \frac{1}{2i}\mathrm{P}\int\frac{e^{3ix} - e^{ix}}{x}dx \]

For any \(k \gt 0\), closing the contour in the upper half plane and indenting around the pole at the origin gives \(\mathrm{P}\int e^{ikx}/x\,dx = i\pi\), independent of \(k\). The two terms are therefore equal and cancel:

\[ I = \frac{1}{2i}(i\pi - i\pi) = 0 \]

The options \(\pi i\) and \(2\pi i\) are what one term alone (or twice one term) would give, which forgets the subtraction; 1 has no source in the calculation.

Discuss in Forum

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