Singularities and Residues Questions and Answers

Q: The residue of \(f(z)\) at its pole is equal to 1. Then the value of \(\alpha\) is

The value is \(\alpha = 3\).\(f(z) = \frac{z^{2} + \alpha z}{(z + 1)^{2}}\) has a double pole at \(z = -1\) (the numerator is \(1 - \alpha \ne 0\) there unless \(\alpha = 1\)). For a pole of order 2, the residue is the derivative of \((z + 1)^{2}f(z)\) at the pole:\[ \begin{aligned} \operatorname{Res}_{z=-1} f &= \frac{d}{dz}(z^{2} + \alpha z)\Big|_{z=-1} \\ &= -2 + \alpha = 1 \Rightarrow \alpha = 3 \end{aligned} \]\(\alpha = 1\) would cancel a factor \((z + 1)\) and leave a simple pole with residue -1; \(\alpha = -1\) gives residue -3 and \(\alpha = 2\) gives 0. Only \(\alpha = 3\) gives residue 1.

Q: Let \(\Gamma\) be the circle given by \(z = 4e^{i\theta}\), where \(\theta\) varies from 0 to \(2\pi\). Then\[ \oint_{\Gamma} \frac{e^{z}}{z^{2} - 2z}\, dz = \]

The integral equals \(\pi i(e^{2} - 1)\).The integrand \(\frac{e^{z}}{z(z-2)}\) has simple poles at \(z = 0\) and \(z = 2\), and both lie inside the circle \(|z| = 4\). By the residue theorem the integral is \(2\pi i\) times the sum of the residues:\[ \begin{aligned} \mathrm{Res}_{z=0} &= \frac{e^{0}}{0 - 2} = -\frac{1}{2} \\ \mathrm{Res}_{z=2} &= \frac{e^{2}}{2} \\ \oint_{\Gamma} &= 2\pi i \cdot \frac{e^{2} - 1}{2} = \pi i(e^{2} - 1) \end{aligned} \]\(2\pi i(e^{2} - 1)\) forgets the factor \(\frac{1}{2}\) in each residue. The options with \(1 - e^{2}\) have the sign reversed, which is what you get by taking the residue at 0 as \(+\frac{1}{2}\) and at 2 as \(-\frac{e^{2}}{2}\).

Q: The value of the integral\[ \int_{C} \frac{z^{100}}{z^{101} + 1}\,dz \]where C is the circle of radius 2 centred at the origin taken in the anti-clockwise direction is

The integral equals \(2\pi i\).Let \(p(z) = z^{101} + 1\). Then \(p'(z) = 101z^{100}\), so the integrand is \(\frac{1}{101}\frac{p'(z)}{p(z)}\).All 101 roots of p lie on the unit circle, inside C of radius 2. By the argument principle the contour integral of p'/p counts them:\[ \begin{aligned} \oint_{C} \frac{z^{100}}{z^{101}+1}\,dz &= \frac{1}{101} \cdot 2\pi i \cdot 101 \\ &= 2\pi i \end{aligned} \]As a check, on a large circle the integrand behaves like 1/z, whose integral is \(2\pi i\). The value 0 would need no poles inside, \(-2\pi i\) has the wrong orientation, and \(2\pi\) drops the factor i.

Q: Let \(C\) be the ellipse \(\{z \in \mathbb{C} : |z - 2| + |z + 2| = 8\}\) traversed counter-clockwise. The value of the contour integral\[ \oint_{C} \frac{z^{2}\,dz}{z^{2} - 2z + 2} \]is equal to

The integral equals \(4\pi i\).The poles are the roots of \(z^{2} - 2z + 2 = 0\), namely \(z = 1 \pm i\). For \(z = 1 + i\), \(|z - 2| + |z + 2| = \sqrt{2} + \sqrt{10} \approx 4.58 \lt 8\), so both poles (they are symmetric) lie inside the ellipse.Write the integrand as \(1 + \frac{2z - 2}{z^{2} - 2z + 2}\). The constant integrates to zero, and the second term is \(q'/q\) for \(q = z^{2} - 2z + 2\), whose integral is \(2\pi i\) times the number of zeros of \(q\) inside \(C\):\[ \oint_{C} = 0 + 2\pi i \times 2 = 4\pi i. \]\(2\pi i\) counts only one pole, and 0 would hold only if both poles were outside the ellipse.

Q: The value of the contour integral\[ \int_{\Gamma} \frac{(1-z)(3\cos z+5\sin z)}{1-z^{101}}dz, \]where \(\Gamma = \left\{ z \in \mathbb{C} : |z| = \frac{7}{9} \right\}\), oriented in the counter-clockwise direction, is equal to

The integral equals 0.The integrand can only fail to be analytic where \(1 - z^{101} = 0\), that is at the 101st roots of unity, all of which lie on the unit circle \(|z| = 1\). The contour \(|z| = \frac{7}{9}\) lies strictly inside the unit circle, so the integrand is analytic on and inside \(\Gamma\).By Cauchy's theorem the integral is 0. The factor \(1 - z\) cancels the root at \(z = 1\), but that does not matter here since no root is enclosed. Values such as \(\pm 2\pi i\) or \(4\pi i\) would need poles inside the contour with nonzero residues, and there are none.

Q: Which of the following complex functions has a pole at \(z = 0\) ?

The function with a pole at 0 is \(f(z) = \dfrac{1 + z + 2z^{3}}{z^{4} - z^{7}}\).Factor the denominator: \(z^{4} - z^{7} = z^{4}(1 - z^{3})\). Near 0 the factor \(1 - z^{3}\) is close to 1 and the numerator is close to 1, so\[ \begin{aligned} f(z) &= \frac{1}{z^{4}} \\ &\quad \cdot \frac{1 + z + 2z^{3}}{1 - z^{3}} \end{aligned} \]with the second factor analytic and equal to 1 at 0. That is a pole of order 4.\(e^{1/z^{3}}\) and \(\sin(1/z)\) have Laurent series with infinitely many negative powers of \(z\), so 0 is an essential singularity of each, not a pole. \(z^{3} + 7z + 1\) is a polynomial, analytic everywhere, with no singularity at all.

Q: For which of the following functions does the function take complex values arbitrarily close to any complex number, inside any arbitrary neighbourhood of 0 ?

The function is \(\sin(1/z)\).The Casorati-Weierstrass theorem says that near an essential singularity a function takes values arbitrarily close to every complex number in every neighbourhood. \(\sin(1/z) = \sum (-1)^{k} z^{-(2k+1)}/(2k + 1)!\) has infinitely many negative powers in its Laurent series at 0, so 0 is an essential singularity.\(e^{z}\) and \(\cosh z + \sinh^{2} z\) are entire, so they are bounded near 0 and stay close to their value at 0. \(z^{3} + z^{2} + 1/z\) has a simple pole at 0, so it tends to \(\infty\) there and cannot come close to every value.

Q: Which of the following complex functions has an infinite number of poles ?

The function with infinitely many poles is \(\sec z\).\(\sec z = 1/\cos z\) has a simple pole wherever \(\cos z = 0\), that is at \(z = \pi/2 + k\pi\) for every integer \(k\): infinitely many isolated poles.\(e^{z}\) is entire with no poles. \(\cos(1/z)\) has a single singularity, at 0, and it is essential, not a pole. \(1/(z^{5} + z + 1)\) has poles only at the five roots of the denominator, a finite number.

Q: Let \(f(z) = \mathrm{cosec}\left(\dfrac{1}{z - 1}\right)\) for \(z \ne 1\), \(z \in \mathbf{C}\). Then at the point \(z = 1\), \(f\) has :

At \(z = 1\) the function has a non-isolated singularity.\(\mathrm{cosec}\,w = 1/\sin w\) has poles where \(\sin w = 0\), at \(w = k\pi\). With \(w = 1/(z - 1)\), the poles of \(f\) are at\[ z_{k} = 1 + \frac{1}{k\pi},\quad k = \pm 1, \pm 2, \ldots \]These accumulate at \(z = 1\). Every punctured disc around 1 contains poles, so \(f\) is not analytic on any punctured neighbourhood of 1, and the singularity there is not isolated.Removable singularities, poles and essential singularities are all isolated by definition, so options A, B and C do not apply.

Q: The value of the integral\[ \frac{1}{2\pi i}\int_{|z| = 1} z^{3}\cos(\tfrac{1}{z})\,dz \]is :

The value is 1/24.The integral divided by \(2\pi i\) is the residue at 0, the coefficient of \(1/z\) in the Laurent series:\[ \begin{aligned} z^{3}\cos\tfrac{1}{z} &= z^{3}\left(1 - \tfrac{1}{2z^{2}} + \tfrac{1}{24z^{4}} - \cdots\right) \\ &= z^{3} - \tfrac{z}{2} + \tfrac{1}{24z} - \cdots \end{aligned} \]The coefficient of \(1/z\) is 1/24. The value 0 would hold if the function were analytic inside the circle, but it has an essential singularity at 0. -1/6 is the coefficient that \(\sin(1/z)\) would give.

Direction for [ Question No: 1 To 1 ] :

Consider the functions \(f(z) = \frac{z^{2} + \alpha z}{(z + 1)^{2}}\) and \(g(z) = \sinh\left(z - \frac{\pi}{2\alpha}\right), \alpha \neq 0\).

The residue of \(f(z)\) at its pole is equal to 1. Then the value of \(\alpha\) is

A.-1
B.1
C.2
D.3

Exam Year(s)

GATE Mathematics - 2012,

Answer: D

The value is \(\alpha = 3\).

\(f(z) = \frac{z^{2} + \alpha z}{(z + 1)^{2}}\) has a double pole at \(z = -1\) (the numerator is \(1 - \alpha \ne 0\) there unless \(\alpha = 1\)). For a pole of order 2, the residue is the derivative of \((z + 1)^{2}f(z)\) at the pole:

\[ \begin{aligned} \operatorname{Res}_{z=-1} f &= \frac{d}{dz}(z^{2} + \alpha z)\Big|_{z=-1} \\ &= -2 + \alpha = 1 \Rightarrow \alpha = 3 \end{aligned} \]

\(\alpha = 1\) would cancel a factor \((z + 1)\) and leave a simple pole with residue -1; \(\alpha = -1\) gives residue -3 and \(\alpha = 2\) gives 0. Only \(\alpha = 3\) gives residue 1.

Discuss in Forum

Let \(\Gamma\) be the circle given by \(z = 4e^{i\theta}\), where \(\theta\) varies from 0 to \(2\pi\). Then

\[ \oint_{\Gamma} \frac{e^{z}}{z^{2} - 2z}\, dz = \]

A.\(2\pi i(e^{2} - 1)\)
B.\(\pi i(1 - e^{2})\)
C.\(\pi i(e^{2} - 1)\)
D.\(2\pi i(1 - e^{2})\)

Exam Year(s)

GATE Mathematics - 2018,

Answer: C

The integral equals \(\pi i(e^{2} - 1)\).

The integrand \(\frac{e^{z}}{z(z-2)}\) has simple poles at \(z = 0\) and \(z = 2\), and both lie inside the circle \(|z| = 4\). By the residue theorem the integral is \(2\pi i\) times the sum of the residues:

\[ \begin{aligned} \mathrm{Res}_{z=0} &= \frac{e^{0}}{0 - 2} = -\frac{1}{2} \\ \mathrm{Res}_{z=2} &= \frac{e^{2}}{2} \\ \oint_{\Gamma} &= 2\pi i \cdot \frac{e^{2} - 1}{2} = \pi i(e^{2} - 1) \end{aligned} \]

\(2\pi i(e^{2} - 1)\) forgets the factor \(\frac{1}{2}\) in each residue. The options with \(1 - e^{2}\) have the sign reversed, which is what you get by taking the residue at 0 as \(+\frac{1}{2}\) and at 2 as \(-\frac{e^{2}}{2}\).

Discuss in Forum

The value of the integral

\[ \int_{C} \frac{z^{100}}{z^{101} + 1}\,dz \]

where C is the circle of radius 2 centred at the origin taken in the anti-clockwise direction is

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

Exam Year(s)

GATE Mathematics - 2022,

Answer: D

The integral equals \(2\pi i\).

Let \(p(z) = z^{101} + 1\). Then \(p'(z) = 101z^{100}\), so the integrand is \(\frac{1}{101}\frac{p'(z)}{p(z)}\).

All 101 roots of p lie on the unit circle, inside C of radius 2. By the argument principle the contour integral of p'/p counts them:

\[ \begin{aligned} \oint_{C} \frac{z^{100}}{z^{101}+1}\,dz &= \frac{1}{101} \cdot 2\pi i \cdot 101 \\ &= 2\pi i \end{aligned} \]

As a check, on a large circle the integrand behaves like 1/z, whose integral is \(2\pi i\). The value 0 would need no poles inside, \(-2\pi i\) has the wrong orientation, and \(2\pi\) drops the factor i.

Discuss in Forum

Let \(C\) be the ellipse \(\{z \in \mathbb{C} : |z - 2| + |z + 2| = 8\}\) traversed counter-clockwise. The value of the contour integral

\[ \oint_{C} \frac{z^{2}\,dz}{z^{2} - 2z + 2} \]

is equal to

A.0
B.\(2\pi i\)
C.\(4\pi i\)
D.\(-\pi i\)

Exam Year(s)

GATE Mathematics - 2024,

Answer: C

The integral equals \(4\pi i\).

The poles are the roots of \(z^{2} - 2z + 2 = 0\), namely \(z = 1 \pm i\). For \(z = 1 + i\), \(|z - 2| + |z + 2| = \sqrt{2} + \sqrt{10} \approx 4.58 \lt 8\), so both poles (they are symmetric) lie inside the ellipse.

Write the integrand as \(1 + \frac{2z - 2}{z^{2} - 2z + 2}\). The constant integrates to zero, and the second term is \(q'/q\) for \(q = z^{2} - 2z + 2\), whose integral is \(2\pi i\) times the number of zeros of \(q\) inside \(C\):

\[ \oint_{C} = 0 + 2\pi i \times 2 = 4\pi i. \]

\(2\pi i\) counts only one pole, and 0 would hold only if both poles were outside the ellipse.

Discuss in Forum

The value of the contour integral

\[ \int_{\Gamma} \frac{(1-z)(3\cos z+5\sin z)}{1-z^{101}}dz, \]

where \(\Gamma = \left\{ z \in \mathbb{C} : |z| = \frac{7}{9} \right\}\), oriented in the counter-clockwise direction, is equal to

A.\(2\pi i\)
B.\(-2\pi i\)
C.0
D.\(4\pi i\)

Exam Year(s)

GATE Mathematics - 2026,

Answer: C

The integral equals 0.

The integrand can only fail to be analytic where \(1 - z^{101} = 0\), that is at the 101st roots of unity, all of which lie on the unit circle \(|z| = 1\). The contour \(|z| = \frac{7}{9}\) lies strictly inside the unit circle, so the integrand is analytic on and inside \(\Gamma\).

By Cauchy's theorem the integral is 0. The factor \(1 - z\) cancels the root at \(z = 1\), but that does not matter here since no root is enclosed. Values such as \(\pm 2\pi i\) or \(4\pi i\) would need poles inside the contour with nonzero residues, and there are none.

Discuss in Forum

Which of the following complex functions has a pole at \(z = 0\) ?

A.\(f(z) = e^{1/z^{3}}\)
B.\(f(z) = \sin\left(\dfrac{1}{z}\right)\)
C.\(f(z) = \dfrac{1 + z + 2z^{3}}{z^{4} - z^{7}}\)
D.\(f(z) = z^{3} + 7z + 1\)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: C

The function with a pole at 0 is \(f(z) = \dfrac{1 + z + 2z^{3}}{z^{4} - z^{7}}\).

Factor the denominator: \(z^{4} - z^{7} = z^{4}(1 - z^{3})\). Near 0 the factor \(1 - z^{3}\) is close to 1 and the numerator is close to 1, so

\[ \begin{aligned} f(z) &= \frac{1}{z^{4}} \\ &\quad \cdot \frac{1 + z + 2z^{3}}{1 - z^{3}} \end{aligned} \]

with the second factor analytic and equal to 1 at 0. That is a pole of order 4.

\(e^{1/z^{3}}\) and \(\sin(1/z)\) have Laurent series with infinitely many negative powers of \(z\), so 0 is an essential singularity of each, not a pole. \(z^{3} + 7z + 1\) is a polynomial, analytic everywhere, with no singularity at all.

Discuss in Forum

For which of the following functions does the function take complex values arbitrarily close to any complex number, inside any arbitrary neighbourhood of 0 ?

A.\(f(z) = e^{z}\)
B.\(f(z) = \sin\left(\dfrac{1}{z}\right)\)
C.\(f(z) = z^{3} + z^{2} + \dfrac{1}{z}\)
D.\(f(z) = \cosh z + \sinh^{2} z\)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: B

The function is \(\sin(1/z)\).

The Casorati-Weierstrass theorem says that near an essential singularity a function takes values arbitrarily close to every complex number in every neighbourhood. \(\sin(1/z) = \sum (-1)^{k} z^{-(2k+1)}/(2k + 1)!\) has infinitely many negative powers in its Laurent series at 0, so 0 is an essential singularity.

\(e^{z}\) and \(\cosh z + \sinh^{2} z\) are entire, so they are bounded near 0 and stay close to their value at 0. \(z^{3} + z^{2} + 1/z\) has a simple pole at 0, so it tends to \(\infty\) there and cannot come close to every value.

Discuss in Forum

Which of the following complex functions has an infinite number of poles ?

A.\(e^{z}\)
B.\(\sec z\)
C.\(\cos\left(\dfrac{1}{z}\right)\)
D.\(\dfrac{1}{z^{5} + z + 1}\)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: B

The function with infinitely many poles is \(\sec z\).

\(\sec z = 1/\cos z\) has a simple pole wherever \(\cos z = 0\), that is at \(z = \pi/2 + k\pi\) for every integer \(k\): infinitely many isolated poles.

\(e^{z}\) is entire with no poles. \(\cos(1/z)\) has a single singularity, at 0, and it is essential, not a pole. \(1/(z^{5} + z + 1)\) has poles only at the five roots of the denominator, a finite number.

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Let \(f(z) = \mathrm{cosec}\left(\dfrac{1}{z - 1}\right)\) for \(z \ne 1\), \(z \in \mathbf{C}\). Then at the point \(z = 1\), \(f\) has :

A.no singularity
B.a removable singularity
C.a pole
D.a non-isolated singularity

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: D

At \(z = 1\) the function has a non-isolated singularity.

\(\mathrm{cosec}\,w = 1/\sin w\) has poles where \(\sin w = 0\), at \(w = k\pi\). With \(w = 1/(z - 1)\), the poles of \(f\) are at

\[ z_{k} = 1 + \frac{1}{k\pi},\quad k = \pm 1, \pm 2, \ldots \]

These accumulate at \(z = 1\). Every punctured disc around 1 contains poles, so \(f\) is not analytic on any punctured neighbourhood of 1, and the singularity there is not isolated.

Removable singularities, poles and essential singularities are all isolated by definition, so options A, B and C do not apply.

Discuss in Forum

10.

The value of the integral

\[ \frac{1}{2\pi i}\int_{|z| = 1} z^{3}\cos(\tfrac{1}{z})\,dz \]

is :

A.0
B.1
C.\(\dfrac{-1}{6}\)
D.\(\dfrac{1}{24}\)

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: D

The value is 1/24.

The integral divided by \(2\pi i\) is the residue at 0, the coefficient of \(1/z\) in the Laurent series:

\[ \begin{aligned} z^{3}\cos\tfrac{1}{z} &= z^{3}\left(1 - \tfrac{1}{2z^{2}} + \tfrac{1}{24z^{4}} - \cdots\right) \\ &= z^{3} - \tfrac{z}{2} + \tfrac{1}{24z} - \cdots \end{aligned} \]

The coefficient of \(1/z\) is 1/24. The value 0 would hold if the function were analytic inside the circle, but it has an essential singularity at 0. -1/6 is the coefficient that \(\sin(1/z)\) would give.

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