Residue Theorem MCQ Quiz - Objective Question with Answer for Residue Theorem - Download Free PDF

Explanation -

In Complex Analysis, the Residue Theorem helps to compute complex line integrals. If a function has a simple pole at z₀, then the residue at that point is equal to the limit as z approaches z₀ of (z-z₀) times the function. This is essentially "stripping off" the (z-z₀) term in the function's Laurent Series Expansion.

Hence option (i) is correct.

Concept:

For simple poles at z = a, b, c…

Residue of $\{f\left(z\right){\}}_{z=a}=\underset{z\to a}{lim}\left\{\left(z-a\right)\times f\left(z\right)\right\}$

For multiple poles at z = a, a, a … n times

{Residue of f(z)}_z=a $=\frac{1}{\left(n-1\right)!}{\left\{\frac{d^{n-1}}{d_z^{n-1}}{\left(z-a\right)}^{n}f\left(z\right)\right\}}_{z=a}$

Calculation:

Given, $f\left(z\right)=\frac{1}{\left(z-4\right){\left(z+1\right)}^3}$

For a simple pole at z = 4

Residue of {f(z)}_z=4 $=\underset{z\to 4}{lim}\left\{\left(z-4\right).\frac{1}{\left(z-4\right){\left(z+1\right)}^3}\right\}$

$=\underset{z\to 4}{lim}\frac{1}{{\left(z+1\right)}^3}\Rightarrow \frac{1}{5^3}=\frac{1}{125}$

For multiple pole (n = 3) at z = -1

Residue will be  

$\Rightarrow \frac{1}{\left(3-1\right)!}\left\{\frac{d^2}{d{z}^2}{\left(z+1\right)}^3.\frac{1}{\left(z-4\right){\left(z+1\right)}^3}\right\}$

$\Rightarrow z_{\to -1}\frac{1\left(-1\right)\left(-2\right)}{2}\frac{1}{{\left(z-4\right)}^3}$

Residue Theorem:

If f(z) is analytic in a closed curve C except at a finite number of singular points within C, then

$\underset{C}{\int }f\left(z\right)dz=2\pi i\times \left[\mathrm{s}\mathrm{u}\mathrm{m}\mathrm{o}\mathrm{f}\mathrm{r}\mathrm{e}\mathrm{s}\mathrm{i}\mathrm{d}\mathrm{u}\mathrm{e}\mathrm{s}\mathrm{a}\mathrm{t}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{g}\mathrm{u}\mathrm{a}\mathrm{l}\mathrm{r}\mathrm{p}\mathrm{o}\mathrm{i}\mathrm{n}\mathrm{t}\mathrm{s}\mathrm{w}\mathrm{i}\mathrm{t}\mathrm{h}\mathrm{i}\mathrm{n}\mathrm{C}\right]$

Formula to find residue:

1. If f(z) has a simple pole at z = a, then

$Resf\left(a\right)=\underset{z\to a}{lim}\left[\left(z-a\right)f\left(z\right)\right]$

2. If f(z) has a pole of order n at z = a, then

$Resf\left(a\right)=\frac{1}{\left(n-1\right)!}{\left\{\frac{d^{n-1}}{d{z}^{n-1}}\left[{\left(z-a\right)}^{n}f\left(z\right)\right]\right\}}_{z=a}$

Important Points:

Cauchy’s Theorem:

If f(z) is an analytic function and f’(z) is continuous at each point within and on a closed curve C, then

$\underset{C}{\oint }f\left(z\right)dz=0$

Cauchy’s Integral Formula:

If f(z) is an analytic function within a closed curve and if a is any point within C, then

$f\left(a\right)=\frac{1}{2\pi i}\underset{C}{\oint }\frac{f\left(z\right)}{z-a}dz$

$f^n\left(a\right)=\frac{n!}{2\pi i}\underset{C}{\oint }\frac{f\left(z\right)}{{\left(z-a\right)}^{n+1}}dz$

Cauchy's Residue Theorem:

Residue of f(z):

Residue of f(z) is denoted as Res[f(z) : z = z₀]

z₀ is a simple pole of the function f(z)

If f(z) = p(z) / q(z)

Where, p(z), q(z) are polynomials

Then residue is,

Res[f(z) : z = z₀] = $\underset{z\to z_0}{lim}\left[\left(z-z_0\right)f\left(z\right)\right]$

If f(z) has a pole of order 'm' at z = z₀ then

Res [f(z) : z = z₀] $=\frac{1}{\left(m-1\right)!}\left\{\underset{z\to z_0}{lim}\left[\frac{d^{m-1}}{d{z}^{m-1}}{\left(z-z_0\right)}^{m}f\left(z\right)\right]\right\}$

Calculation:

Given,

$f\left(z\right)=\frac{1}{{\left(z^2+a^2\right)}^2}=\frac{1}{{\left(z+ia\right)}^2{\left(z-ia\right)}^2}$

Pole of f(z) has order "2"

Res [f(z) : z = ia] $=\frac{1}{\left(2-1\right)!}\left\{\underset{z\to ia}{lim}\left[\frac{d^{\left(2-1\right)}}{d{z}^{\left(2-1\right)}}{\left(z-ia\right)}^2\frac{1}{{\left(z+ia\right)}^2{\left(z-ia\right)}^2}\right]\right\}$

$=\underset{z\to ia}{lim}\left\{\frac{d}{dz}\frac{1}{{\left(z+ia\right)}^2}\right\}$

$=\underset{z\to ia}{lim}\frac{-2}{{\left(z+ia\right)}^3}$

$=\frac{-i}{4a^3}$

The residue of f(z) is $\frac{-i}{4a^3}$

Residue Theorem:

If f(z) is analytic in a closed curve C except at a finite number of singular points within C, then

$\underset{C}{\int }f\left(z\right)dz=2\pi i\times \left[\mathrm{s}\mathrm{u}\mathrm{m}\mathrm{o}\mathrm{f}\mathrm{r}\mathrm{e}\mathrm{s}\mathrm{i}\mathrm{d}\mathrm{u}\mathrm{e}\mathrm{s}\mathrm{a}\mathrm{t}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{g}\mathrm{u}\mathrm{a}\mathrm{l}\mathrm{r}\mathrm{p}\mathrm{o}\mathrm{i}\mathrm{n}\mathrm{t}\mathrm{s}\mathrm{w}\mathrm{i}\mathrm{t}\mathrm{h}\mathrm{i}\mathrm{n}\mathrm{C}\right]$

Formula to find residue:

1. If f(z) has a simple pole at z = a, then

$Resf\left(a\right)=\underset{z\to a}{lim}\left[\left(z-a\right)f\left(z\right)\right]$

2. If f(z) has a pole of order n at z = a, then

$Resf\left(a\right)=\frac{1}{\left(n-1\right)!}{\left\{\frac{d^{n-1}}{d{z}^{n-1}}\left[{\left(z-a\right)}^{n}f\left(z\right)\right]\right\}}_{z=a}$

Important Points:

Cauchy’s Theorem:

If f(z) is an analytic function and f’(z) is continuous at each point within and on a closed curve C, then

$\underset{C}{\oint }f\left(z\right)dz=0$

Cauchy’s Integral Formula:

If f(z) is an analytic function within a closed curve and if a is any point within C, then

$f\left(a\right)=\frac{1}{2\pi i}\underset{C}{\oint }\frac{f\left(z\right)}{z-a}dz$

$f^n\left(a\right)=\frac{n!}{2\pi i}\underset{C}{\oint }\frac{f\left(z\right)}{{\left(z-a\right)}^{n+1}}dz$

Concept:

Pole:

The value for which f(z) fails to exists i.e. the value at which the denominator of the function f(z) = 0.

When the order of a pole is 1, it is known as a simple pole.

Residue:

If f(z) has a simple pole at z = a, then

$Resf(a)=\underset{z\to a}{lim}\left(z-a\right)f\left(z\right)$

If f(z) has a pole of order n at z = a, then

$Res\left(atz=a\right)=\frac{1}{\left(n-1\right)!}{\left\{\frac{d^{n-1}}{d{z}^{n-1}}\left[{\left(z-a\right)}^{n}f\left(z\right)\right]\right\}}_{z=a}$

Calculate:

Given:

$f\left(z\right)=\frac{z^2}{z^2+a^2}$

For calculating pole:

z² + a² = 0

∴ (z + ia)(z - ai) = 0

∴ z = ai, -ai.

∴ z has simple pole at z = ai and -ai.

Residue:

If f(z) has a simple pole at z = a, then

$Resf(a)=\underset{z\to a}{lim}\left(z-a\right)f\left(z\right)$

For pole at z = ai

$Resf(ai)=\underset{z\to ai}{lim}\left(z-ai\right)\left(\frac{z^2}{z^2+a^2}\right)$

$Resf(ai)=\underset{z\to ai}{lim}\left(z-ai\right)\left(\frac{z^2}{(z-ai)(z+ai)}\right)$

$Resf(ai)=\frac{(ai{)}^2}{2ai}\Rightarrow \frac{ai}{2}$

For pole at z = -ai

$Resf(-ai)=\underset{z\to -ai}{lim}\left(z+ai\right)\left(\frac{z^2}{(z-ai)(z+ai)}\right)$

$Resf(-ai)=\frac{(-ai{)}^2}{-2ai}\Rightarrow -\frac{ai}{2}$

∴ z has a simple pole at z = ai and $\frac{ia}{2}$ is a residue at z = ia of f(z)

$\mathrm{R}\mathrm{e}\mathrm{s}\mathrm{f}{\left(\mathrm{z}\right)}_{\mathrm{z}=\mathrm{a}}=\frac{1}{\left(\mathrm{n}-1\right)!}\frac{{\mathrm{d}}^{\mathrm{n}-1}}{\mathrm{d}{\mathrm{z}}^{\mathrm{n}-1}}{\left(\mathrm{z}-\mathrm{a}\right)}^{\mathrm{n}}{\mathrm{f}\left(\mathrm{z}\right)]}_{\mathrm{z}=\mathrm{a}}$

Here we have n = 2 and a = 2

Thus Res $\mathrm{f}{\left(\mathrm{z}\right)}_{\mathrm{z}=2}=\frac{1}{\left(2-1\right)!}\frac{\mathrm{d}}{\mathrm{d}\mathrm{z}}{\left[{\left(\mathrm{z}-2\right)}^2\frac{1}{{\left(\mathrm{z}-2\right)}^2{\left(\mathrm{z}+2\right)}^2}\right]}_{\mathrm{z}=2}$

$\begin{array}{l}=\frac{\mathrm{d}}{\mathrm{d}\mathrm{z}}{\left[\frac{1}{{\left(\mathrm{z}+2\right)}^2}\right]}_{\mathrm{z}=2}={\left[-\frac{2}{{\left(\mathrm{z}+2\right)}^3}\right]}_{\mathrm{z}=2}\\ =-\frac{2}{64}=-\frac{1}{32}\end{array}$

Calculation:

Given: integral: $\frac{1}{\pi j}\oint \frac{dz}{Z^2-1}$

From residue theorem:

$\int f\left(z\right)dz=2\pi i\left[sumofresidues\right]$

Residue at z = 1

$\begin{array}{c}lt\\ z\to 1\end{array}\left(z-1\right)\frac{1}{\left(z-1\right)\left(z+1\right)}=\frac{1}{2}$

Residue at z = -1 (Here the contour is traveled in a clockwise sense, hence extra Negative sign will come)

$\begin{array}{c}lt\\ z\to -1\end{array}\left(z+1\right)\frac{-1}{\left(z+1\right)\left(z-1\right)}=\frac{1}{2}$

Sum of residues $=\frac{1}{2}+\frac{1}{2}=1$

Concept:

Pole – a point on which functional value is infinite.

If z = a is a simple pole of f(z) then the residue of f(z) at z = a is given by,
$Re{s}_{z=a}f\left(z\right)=\underset{z\to a}{lim}\left(z-a\right)f\left(z\right)$

Calculation:

Given:

$f\left(z\right)=\frac{2z+1}{z^2-z-2}$

We have to find residue at z = 2.

$f\left(z\right)=\frac{2z+1}{z^2-z-2}=\frac{2z+1}{(z-2)(z+1)}$

Value of f(z) is infinite at x = 2. hence z = 2 is a simple pole.

Residue of f(z) at z = 2, 

$Re{s}_{z=2}f\left(z\right)=\underset{z\to 2}{lim}\left(z-2\right)\times \frac{2z+1}{\left(z-2\right)\times \left(z+1\right)}$

$=\frac{2\times 2+1}{2+1}=\frac{5}{3}$

Concept:

For a given complex function with poles, the complex integral $I=\underset{C}{\oint }f\left(z\right)dz$ is given by

Residue theorem as;

$I=\underset{C}{\oint }f\left(z\right)dz=2\pi i\times \left\{SumofresidueofpolesinsideoronC\right\}$

Calculation:

Given C, |Z|= 3

|x + iy| = 3

$\sqrt{x^2+y^2}=3$

x² + y² = 9

$f\left(z\right)=\left(\frac{z^2-1}{z^2+1}\right)e^z$

Poles are defined by z² + 1 = 0

(z + i)(z - i) = 0

z = i, -i (simple poles lying inside of closed path)

Residue of f(z) at z = i $\Rightarrow \underset{z\to i}{lim}\left(z-i\right).\frac{z^2-1}{\left(z+i\right)\left(z-i\right)}{e}^z$

$\Rightarrow \frac{\left(i^2-1\right)e^i}{2i}\to -\frac{2e^i}{xi}=i{e}^i$

Residue of f(z) at z = -i $\Rightarrow \underset{z\to -i}{lim}\left(z+i\right).\frac{z^2-1}{\left(z+i\right)\left(z-i\right)}{e}^z$

$\Rightarrow \frac{\left(i-1\right)}{-2i}.e^{-i}\to -i{e}^{-i}$

Now, using Cauchy’s Residue Theorem,

$I=\underset{c}{\oint }\frac{z^2-1}{z^2+1}{e}^{z}dz=2\pi i\left(Sumofresidue\right)$

= 2πi (ieⁱ – ie^-i)

= (2πi) (i) (eⁱ – e^-i)

= 2πi² (2i sin (1))

Concept:

Cauchy’s Theorem:

If f(z) is an analytic function and f’(z) is continuous at each point within and on a closed curve C, then

$\underset{C}{\oint }f\left(z\right)dz=0$

Cauchy’s Integral Formula:

If f(z) is an analytic function within a closed curve and if a is any point within C, then

$f\left(a\right)=\frac{1}{2\pi i}\underset{C}{\oint }\frac{f\left(z\right)}{z-a}dz$

$f^n\left(a\right)=\frac{n!}{2\pi i}\underset{C}{\oint }\frac{f\left(z\right)}{{\left(z-a\right)}^{n+1}}dz$

Residue Theorem:

If f(z) is analytic in a closed curve C except at a finite number of singular points within C, then

$\underset{C}{\int }f\left(z\right)dz=2\pi i\times \left[\mathrm{s}\mathrm{u}\mathrm{m}\mathrm{o}\mathrm{f}\mathrm{r}\mathrm{e}\mathrm{s}\mathrm{i}\mathrm{d}\mathrm{u}\mathrm{e}\mathrm{s}\mathrm{a}\mathrm{t}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{g}\mathrm{u}\mathrm{a}\mathrm{l}\mathrm{r}\mathrm{p}\mathrm{o}\mathrm{i}\mathrm{n}\mathrm{t}\mathrm{s}\mathrm{w}\mathrm{i}\mathrm{t}\mathrm{h}\mathrm{i}\mathrm{n}\mathrm{C}\right]$

Formula to find residue:

1. If f(z) has a simple pole at z = a, then

$Resf\left(a\right)=\underset{z\to a}{lim}\left[\left(z-a\right)f\left(z\right)\right]$

2. If f(z) has a pole of order n at z = a, then

$Resf\left(a\right)=\frac{1}{\left(n-1\right)!}{\left\{\frac{d^{n-1}}{d{z}^{n-1}}\left[{\left(z-a\right)}^{n}f\left(z\right)\right]\right\}}_{z=a}$

Application:

${\int }_C\frac{\cos \pi z^2}{\left(z-1\right)\left(z-2\right)}dz$

Simple poles: z = 1, 2

Both the poles lie within the given region.

Residue at z = 1,

$\underset{z\to 1}{lim}\frac{\cos \pi z^2}{\left(z-1\right)\left(z-2\right)}\cdot \left(z-1\right)=\frac{\cos \pi }{\left(-1\right)}=1$

Residue at z = 2

$\underset{z\to 2}{lim}\frac{\cos \pi z^2}{\left(z-1\right)\left(z-2\right)}\left(z-2\right)=\cos 4\pi =1$

Value of integral = 2πi [1 + 1] = 4πi

Concept:

Cauchy’s Theorem:

If f(z) is an analytic function and f’(z) is continuous at each point within and on a closed curve C, then

$\underset{C}{\oint }f\left(z\right)dz=0$

Cauchy’s Integral Formula:

If f(z) is an analytic function within a closed curve and if a is any point within C, then

$f\left(a\right)=\frac{1}{2\pi i}\underset{C}{\oint }\frac{f\left(z\right)}{z-a}dz$

$f^n\left(a\right)=\frac{n!}{2\pi i}\underset{C}{\oint }\frac{f\left(z\right)}{{\left(z-a\right)}^{n+1}}dz$

Residue Theorem:

If f(z) is analytic in a closed curve C except at a finite number of singular points within C, then

$\underset{C}{\int }f\left(z\right)dz=2\pi i\times \left[\mathrm{s}\mathrm{u}\mathrm{m}\mathrm{o}\mathrm{f}\mathrm{r}\mathrm{e}\mathrm{s}\mathrm{i}\mathrm{d}\mathrm{u}\mathrm{e}\mathrm{s}\mathrm{a}\mathrm{t}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{g}\mathrm{u}\mathrm{a}\mathrm{l}\mathrm{r}\mathrm{p}\mathrm{o}\mathrm{i}\mathrm{n}\mathrm{t}\mathrm{s}\mathrm{w}\mathrm{i}\mathrm{t}\mathrm{h}\mathrm{i}\mathrm{n}\mathrm{C}\right]$

Formula to find residue:

1. If f(z) has a simple pole at z = a, then

$Resf\left(a\right)=\underset{z\to a}{lim}\left[\left(z-a\right)f\left(z\right)\right]$

2. If f(z) has a pole of order n at z = a, then

$Resf\left(a\right)=\frac{1}{\left(n-1\right)!}{\left\{\frac{d^{n-1}}{d{z}^{n-1}}\left[{\left(z-a\right)}^{n}f\left(z\right)\right]\right\}}_{z=a}$

Application:

Given function is $f\left(z\right)=\frac{z^2}{z^2-1}$

Poles: z = 1, -1

|z – 1| = 1

⇒ |x – 1 + iy| = 1

$\Rightarrow \sqrt{{\left(x-1\right)}^2+y^2}=1$

The given region is a circle with the centre at (1, 0) and the radius is 1.

Only pole z = 1, lies within the given region.

Residue at z = 1 is, $\underset{z\to 1}{lim}\frac{z^2}{\left(z+1\right)}=0.5$

The value of the integral = 2πi × 0.5 = πi

$f\left(z\right)=\frac{z^2}{{\left(z^2-3z+2\right)}^2}$

$=\frac{z^2}{{\left(z-1\right)}^2{\left(z-2\right)}^2}$

Poles of f(z) are z = 1, 2

Both Z = 1 and 2 are inside the curve |z| = 4

$\underset{C}{\int }f\left(z\right)dz=2\pi i$ [sum of residues at z = 1 and z = 2]

If f has a pole of order n at z = a, then Residue of f(z) at z = a is

$=\frac{1}{\left(n-1\right)!}\frac{d^{n-1}}{d{z}^{n-1}}{\left[{\left(z-a\right)}^{n}f\left(z\right)\right]}_{z=a}$

Residue at z = 1,

$\underset{z\to 1}{\mathrm{I}\mathrm{t}}\frac{d}{dz}\left[\frac{z^2.{\left(z-1\right)}^2}{{\left(z-1\right)}^2{\left(z-2\right)}^2}\right]$

$=\frac{{\left(1-2\right)}^2\left(2\right)-{\left(1\right)}^2\left(2\left(-1\right)\right)}{{\left(1-2\right)}^4}=4$

Residue at z = 2,

$=\underset{z\to 2}{\mathrm{l}\mathrm{t}}\frac{d}{dz}\left[\frac{z^2}{{\left(z-1\right)}^2}\right]$

$=\underset{z\to 2}{\mathrm{l}\mathrm{t}}\left[\frac{{\left(z-1\right)}^2\left(2z\right)-z^2\left(2\left(z-1\right)\right)}{{\left(z-1\right)}^4}\right]$

$=\frac{{\left(2-1\right)}^2-4\left(2\left(1\right)\right)}{{\left(2-1\right)}^4}=-4$

$\underset{C}{\int }f\left(z\right)dz=2\pi i\left(4-4\right)=0$

Concept:

Residue Theorem: 

If f(z) is analytic in a closed curve C except at a finite number of singular points within C, then

∫_cf(z) dz = 2πj × [sum of residues at the singular points within C]

Formula to find residue:

1. If f(z) has a simple pole at z = a, then

$Resf\left(a\right)=\underset{z\to a}{lim}\left[\left(z-a\right)f\left(z\right)\right]$

2. If f(z) has a pole of order n at z = a, then

$Resf\left(a\right)=\frac{1}{\left(n-1\right)!}{\left\{\frac{d^{n-1}}{d{z}^{n-1}}\left[{\left(z-a\right)}^{n}f\left(z\right)\right]\right\}}_{z=a}$

Application:

Given (-1 - j), (3 - j), (3 + j) and (-1 + j) are the vertices of a rectangle C in the complex plane

f(z) from the given data is,

$f(z)={\displaystyle \frac{1}{z^2(z-4)}}$

Poleas of f(z) is

z = 0 of order n = 2, lies in side the closed curve.

z = 4 of order n = 1, lies outside the closed curve.

∴ ${\displaystyle {\oint }_C{\displaystyle \frac{dz}{z^2(z-4)}}=2\pi jResf\left(0\right)}$ 

⇒ $Resf\left(0\right)=\frac{1}{\left(2-1\right)!}{\left\{\frac{d^{2-1}}{d{z}^{2-1}}\left[{\left(z\right)}^2{\displaystyle \frac{1}{z^2(z-4)}}\right]\right\}}_{z=0}$

$Resf(0)={\displaystyle \frac{d}{dz}}{\left[{\displaystyle \frac{1}{z-4}}\right]}_{z=0}=-{\displaystyle \frac{1}{16}}$

${\displaystyle {\oint }_C{\displaystyle \frac{dz}{z^2(z-4)}}=2\pi j\left({\displaystyle \frac{-1}{16}}\right)={\displaystyle \frac{-j\pi }{8}}}$

Additional Information

Cauchy’s Theorem:

If f(z) is an analytic function and f’(z) is continuous at each point within and on a closed curve C, then

$\underset{C}{\oint }f\left(z\right)dz=0$

Cauchy’s Integral Formula:

If f(z) is an analytic function within a closed curve and if a is any point within C, then

$f\left(a\right)=\frac{1}{2\pi i}\underset{C}{\oint }\frac{f\left(z\right)}{z-a}dz$

$f^n\left(a\right)=\frac{n!}{2\pi i}\underset{C}{\oint }\frac{f\left(z\right)}{{\left(z-a\right)}^{n+1}}dz$