Cauchy's Integral Theorem MCQ Quiz - Objective Question with Answer for Cauchy's Integral Theorem - Download Free PDF

Explanation:

Let γ be the positively oriented circle in the complex plane given by {z ∈ ℂ: |z – 1| = 1}.

Then ${\displaystyle \frac{1}{2\pi i}{\int }_{\gamma }\frac{dz}{z^3-1}}$ 

Concept - 

Cauchy Integral Formula - 

Let f(z) be analytic in a region D and let C be a closed curve in D. If a is any point in D, then 

${\displaystyle {\int }_C\frac{f(z)}{z-a}dz=2\pi i.f(a).\eta (\gamma :a)}$

here $\eta (\gamma :a)$ is a winding number. The winding number measures the number of times a path (counter-clockwise) winds around a point.

Explanation -

We have γ = {z ∈ ℂ: |z – 1| = 1}.

Now ${\displaystyle \frac{1}{2\pi i}{\int }_{\gamma }\frac{dz}{z^3-1}}$

= ${\displaystyle \frac{1}{2\pi i}{\int }_{\gamma }\frac{dz}{(z-1)(z^2+z+1)}}$

= ${\displaystyle \frac{1}{2\pi i}{\int }_{\gamma }\frac{1/(z^2+z+1)}{(z-1)}dz}$

Now using Cauchy Integral formula we get -

= ${\displaystyle \frac{1}{2\pi i}\times 2\pi i\times f(1)={\displaystyle \frac{1}{2\pi i}\times 2\pi i\times \frac{1}{3}=\frac{1}{3}}}$

Hence option(ii) is correct.

Concept:

Cauchy's integral formula: Let f(z) be a complex analytic function within and on a closed contour C inside a simply-connected domain, and if a is any point in the middle of C, then

f(a) = 2πi ${\int }_C\frac{f(z)}{z-a}dz$

Explanation:

${\oint }_{|1-z|=1}\frac{e^z}{z^2-1}dz$

Here C: |1 - z| = 1

The singularities of $\frac{e^z}{z^2-1}$ are z² = 1 i.e., z = ± 1 out of which z = 1 lies inside C.

So, f(z) = $\frac{e^z}{z+1}$

Then using Cauchy's integral formula

${\oint }_{|1-z|=1}\frac{e^z}{z^2-1}dz$ = 2πi(f(1)) = 2πi × $\frac{e}{2}$ = (πi)e

Option (2) is correct

Concept:

Cauchy's integral theorem: If f(z) is an analytic function in a simply connected region R, then ∫c f(z) dz = 0 for every closed contour c contained in R.

Cauchy's integral formula: Let f(z) be a complex analytic function within and on a closed contour C inside a simply connected domain, and if a is any point in the middle of C, then

$\frac{1}{2\pi i}{\int }_C\frac{f(z)}{(z-a)}dz$ = f(a)

Generalisation of Cauchy’s Integral Formula: If f(z) is an analytic function within and on a simple closed curve C and if a is any point within C, then

$\frac{n!}{2\pi i}{\int }_C\frac{f(z)}{(z-a{)}^{n+1}}dz$ = fⁿ(a)

Explanation:

C denote the unit circle centered at the origin in ℂ

So C: |z| = 1 ⇒ $z\overline{z}=1$ ⇒ $\overline{z}=\frac{1}{z}$

$\frac{1}{2\pi i}{\int }_C{|1+z+z^2|}^{2}dz$

= $\frac{1}{2\pi i}{\int }_C(1+z+z^2)(\overline{1+z+z^2})dz$

= $\frac{1}{2\pi i}{\int }_C(1+z+z^2)(1+\overline{z}+{\overline{z}}^2)dz$

= $\frac{1}{2\pi i}{\int }_C(1+z+z^2+\overline{z}+z\overline{z}+z^2\overline{z}+{\overline{z}}^2+z{\overline{z}}^2+z^2{\overline{z}}^2)dz$

=$\frac{1}{2\pi i}{\int }_C(1+z+z^2+\frac{1}{z}+z\cdot \frac{1}{z}+z^2\cdot \frac{1}{z}+\frac{1}{z^2}+z\cdot \frac{1}{z^2}+z^2\cdot \frac{1}{z^2})$dz 

= $\frac{1}{2\pi i}{\int }_C(1+z+z^2+1+z+1+\frac{1}{z^2}+\frac{1}{z}+\frac{1}{z})$dz

= $\frac{1}{2\pi i}{\int }_C(3+2z+z^2)dz$ + $\frac{1}{2\pi i}{\int }_C(\frac{1}{z^2}+\frac{2}{z})dz$

= 0 + $\frac{1}{2\pi i}{\int }_C\frac{1}{z^2}dz$ + $\frac{1}{2\pi i}{\int }_C\frac{2}{z}dz$(by Cauchy's integral theorem)

= f'(0) + g(0) where f(z) = 1 and g(z) = 2 (using Generalisation of Cauchy’s Integral Formula and Cauchy's integral formula) 

= 0 + 2 = 2

Option (3) is true

Concept:

Let ϕ(z) be a meromorphic function on a closed curve C. Let ϕ(z) has n number of zeros and p number of poles inside C then 

$\frac{1}{2\pi i}{\int }_C\frac{{\varphi }^{\prime }(z)}{\varphi (z)}dz$ = n - p

Explanation:

Let ϕ(z) = zf(z)

Then, number of zeros of ϕ(z) = number of zeros of f(z) + 1 = n_o + 1 (as z = 0 is also a zeros of zf(z))

and number of poles of ϕ(z) = number of poles of zf(z) = n_p

Hence $\frac{1}{2\pi i}{\int }_C\frac{{\varphi }^{\prime }(z)}{\varphi (z)}dz$ = no + 1 - np 

⇒ $\frac{1}{2\pi i}{\int }_C\frac{(zf{)}^{\prime }}{zf}dz=n_0-n_p+1.$ (putting the value of ϕ(z) = zf(z))

(1) is correct

Concept:

Cauchy's argument principle: If f(z) is a meromorphic function inside and on some closed contour C, and f has no zeros or poles on C, then

$\frac{1}{2\pi i}{\int }_C\frac{f^{\prime }(z)}{f(z)}dz$ = N - P, where N and P denote respectively the number of zeros and poles of f(z) inside the contour C, with each zero and pole counted as many times as its multiplicity and order, respectively

Explanation:

${\int }_C\frac{z^{100}}{z^{101}+1}dz$

= $\frac{1}{101}{\int }_C\frac{101z^{100}}{z^{101}+1}dz$

= $\frac{1}{101}{\int }_C\frac{f^{\prime }(z)}{f(z)}dz$ where f(z) = z¹⁰¹ + 1

f(z) is a polynomial of degree 101 so number of zeros of f(z) is 101

Hence using Cauchy's argument principle,

${\int }_C\frac{z^{100}}{z^{101}+1}dz$ = 2πi × $\frac{1}{101}$ × 101 = 2πi

Therefore option (4) is true

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:

${\int }_C\frac{z^2+1}{z^2-2z}dz$

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

The simple poles are: z = 0, 2

The given region is a unit circle.

The residue at z = 2 is zero as it lies outside the given region.

The reside at z = 0, is given by

$=\underset{z\to 0}{lim}z\frac{z^2+1}{z\left(z-2\right)}dz=-\frac{1}{2}$

Concept:

Cauchy Riemann equations:

$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y};\frac{\partial v}{\partial x}=-\frac{\partial u}{\partial y}$

Calculation:

z = x + iy

Let F(z) = u + iv

F(z) = iz + k Re(z) + i Im(z) = i (x + iy ) + kx + iy = i (x + y) + (kx – y)

u = (kx – y), v = (x + y)

$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}\Rightarrow k=1$

Concept:

Residue Theorem:

If f(z) is analytic inside and on a closed curve ‘C’ except at a finite number of singularities inside ‘C’

Then:

∮f(z) dz = 2πi (sum of residues)

Singularity: A point where the function f(z) fails to be analytic

Analysis:

Given $f\left(x\right)=\frac{\sin \left(x\right)}{x^2\left(x^2+4\right)}$ 

$f\left(x\right)=\frac{\sin \left(x\right)}{x^2\left(x+2i\right)\left(x-2i\right)}$ 

x = 0 pole of order 2

x = 2i, -2i

Given curve or region is |x - i| = 2

x = -2i lies outside |x - i| = 2

So the singular points are x = 0 & x = 2i

Residue at x = 0

$\underset{x\to 0}{lim}\frac{d}{dx}\left[\frac{\sin x}{\left(x^2+4\right)}\right]$

$=\underset{x\to 0}{lim}\left[\frac{\cos x\left(x^2+y\right)-\sin x\left(2x\right)}{{\left(x^2+4\right)}^2}\right]$
$=\underset{x\to 0}{lim}\frac{\cos x}{x^2+y}=\frac{1}{4}=0.25$

Residue at x = 2i

$\Rightarrow \underset{x\to 2i}{lim}\left(x-2i\right)\frac{\sin x}{x^2\left(x+2i\right)\left(x-2i\right)}$

$\Rightarrow \frac{\sin \left(2i\right)}{{\left(2i\right)}^2\left(4i\right)}$

$\Rightarrow \frac{\sin \left(2i\right)}{-16i}$

 ${\oint }_{C}f\left(x\right)dx={\oint }_C\frac{\sin \left(x\right)}{x^2\left(x^2+4\right)}dx=2\pi i$ (sum of residues)

$=2\pi i\left[\frac{1}{4}-\frac{\sin \left(2i\right)}{16i}\right]$

$=\frac{\pi i}{2}-\frac{\pi \sin \left(2i\right)}{8}$

$f\left(z\right)=\underset{C}{\oint }\frac{1}{{\left(z-1\right)}^3\left(z-3\right)}$

(z - 1)³ (z - 3) = 0

⇒ z = 1, z = 3

The function has a simple pole at z = 3 and has a multiple pole at z = 1

Both z = 1, and z = 3 are inside the region C

According to Cauchy’s Residue theorem

$\underset{c}{\int }f\left(z\right)dz=2\pi i\left[sumoftheresiduesatthepolesinside{}^{\prime }{C}^{\prime }\right]$

If z = a is a pole of order ‘m’, then residue of f(z) at z = a is,

$\frac{1}{\left(m-1\right)!}\begin{array}{c}lim\\ z\to a\end{array}\frac{d^{m-1}}{d{z}^{m-1}}\left[{\left(z-a\right)}^{m}f\left(z\right)\right]$

Residue at z = 3 is,

$=\begin{array}{c}lim\\ z\to 3\end{array}\left(z-3\right)\frac{1}{{\left(z-1\right)}^3\left(z-3\right)}$

$=\begin{array}{c}lim\\ z\to 3\end{array}\frac{1}{{\left(z-1\right)}^3}=\frac{1}{8}$

Residue at z = 1 is,

$=\frac{1}{\left(3-1\right)!}\begin{array}{c}lim\\ z\to 1\end{array}\frac{d^2}{d{z}^2}\left({\left(z-1\right)}^3\frac{1}{{\left(z-1\right)}^3\left(z-3\right)}\right)$

$=\frac{1}{2!}\begin{array}{c}lim\\ z\to 1\end{array}\frac{d^2}{d{z}^2}\left(\frac{1}{z-3}\right)$

$=\frac{1}{2}\begin{array}{c}lim\\ z\to 1\end{array}2{\left(z-3\right)}^{-3}$

$=\frac{1}{2}\times 2\times \frac{1}{{\left(-2\right)}^3}=-\frac{1}{8}$

Concept:

Cauchy’s Integral Formula  - If f(z) is analytic within and on the closed curve C and if z_o is any point inside C then

${\oint }_c\frac{f\left(z\right)}{z-z_o}=2\pi if(z_o)$

$f(z_o)=\frac{1}{2\pi i}{\oint }_c\frac{f\left(z\right)}{z-z_o}$

Cauchy’s Residue Theorem – If f(z) is analytic in a closed curve C except at a finite number of singular point lies inside C then,

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

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

$\mathrm{Re}sf(a)=\underset{z\to a}{\mathrm{L}\mathrm{t}}[(z-a)f(z)]$

Calculation:

${\oint }_{\Gamma }^{}\frac{3z-5}{\left(z-1\right)\left(z-2\right)}dz=4\pi i$

${\oint }_1^{}\frac{3z-5}{\left(Z-1\right)\left(Z-2\right)}dz=4\pi i=2\pi i\left(2\right)$

∴ Sum of residue must be equal to 2

$\underset{z=1}{\mathrm{R}\mathrm{e}\mathrm{s}\mathrm{f}\left(\mathrm{z}\right)}=\underset{z\to 1}{lim}\left(z-1\right)\frac{\left(3z-5\right)}{\left(z-1\right)\left(z-2\right)}=\underset{z\to 1}{lim}\frac{\left(3z-5\right)}{\left(z-2\right)}=2$

$\underset{z=2}{\mathrm{R}\mathrm{e}\mathrm{s}\mathrm{f}\left(\mathrm{z}\right)}=\underset{z\to 2}{lim}\left(z-2\right)\frac{\left(3z-5\right)}{\left(z-1\right)\left(z-2\right)}=\underset{z\to 2}{lim}\frac{\left(3z-5\right)}{\left(z-1\right)}=1$

${\oint }_1^{}\frac{3z-5}{\left(Z-1\right)\left(Z-2\right)}dz=4\pi i=2\pi i\left(2\right)=2\pi i(Resf(1))$

∴ Z = 1 must lies inside C, Z = 2 lies outside C.

Concept:

Cauchy's Integral Formula:

For Simple Pole:

If f(z) is analytic within and on a closed curve c and if a (simple pole) is any point within c, then

${\oint }_c^{}\frac{f(z)}{z-a}dz=2\pi i.f(a)$

$\cosh z=\frac{e^z+e^{-z}}{2}$

Calculation:

Given:

${\oint }_C^{}\frac{\cosh 3z}{2z}dz$ where C represents unit circle i.e. radius is unity.

The above equation can be written in standard form i.e. ${\oint }_C^{}\frac{\frac{\cosh 3z}{2}}{z-0}dz$

Therefore $f(z)=\frac{\cosh 3z}{2}$ and a = 0.

The pole of the given function is at z = 0, and lie inside the circle.

$f(z)=\frac{\cosh 3z}{2}=\frac{e^{3z}+e^{-3z}}{2\times 2}=\frac{e^{3z}+e^{-3z}}{4}$

At z = 0,

$f(0)=\frac{e^0+e^{-0}}{4}=\frac{2}{4}=\frac{1}{2}$

Cauchy's Integral Formula:

${\oint }_c^{}\frac{f(z)}{z-a}dz=2\pi i.f(a)$

${\oint }_C^{}\frac{\frac{\cosh 3z}{2}}{z-0}dz=2\pi i\times f(0)$

${\oint }_C^{}\frac{\frac{\cosh 3z}{2}}{z-0}dz=2\pi i\times \frac{1}{2}=\pi i$

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, or if the simple closed curve does not contain any singular point of f(z) then,

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

Given,

The integral is $\underset{C}{\overset{}{\oint }}\frac{dz}{\left(z^2+6\right)}$ where C is |z -i| = 1

Now, for point singularity consider Z² + 6 = 0

Therefore point of singularity is a = ± √6i

For |z -i| = 1 we have,

radius r = 1,  centre c = (0,1),

a = ± √6i is out of circle,

By Cauchy's Integral Formula We have,

$\underset{C}{\overset{}{\oint }}\frac{dz}{\left(z^2+6\right)}$ = 0

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\frac{\sin x}{x^2+2x+2}dx$

Concept:

Cauchy Integral Theorem:

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

Where z = a be any point inside the close region.

Cauchy’s Residue Theorem:

$\frac{1}{2\pi i}\oint f\left(z\right)dz=$ Sum of Residue at Pole or singularity with in the region

Res at z = a

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

Calculation:

We know that $e^{ix}=\cos x+i\sin x$

Let x replace by z.

$\therefore I=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\frac{Imaginary\left[e^{iz}\right]}{z^2+2z+2}dz$

Now, z² + 2z + 2 = 0 then roots of z are:

$\therefore z=\frac{-2\pm \sqrt{{\left(2\right)}^2-8}}{2}=\frac{-2\pm \sqrt{4i^2}}{2}=-1\pm i$

z = -1 ± i is the only pole lying in f(z) > 0

Here n = 1

$Re{s}_{z=a}=\frac{1}{\left(n-1\right)!}\underset{z\to a}{lim}{\left(\frac{d}{dz}\right)}^{n-1}\left[{\left(z-a\right)}^{n}f\left(z\right)\right]$

$\therefore Re{s}_{atz=-1\pm i}=\frac{1}{\left(1-1\right)!}\underset{z\to -1\pm i}{lim}{\left(\frac{d}{dz}\right)}^{1-1}\left[\left(z-{\left(1\pm i\right)}^1\right).\frac{e^{iz}}{\left[z-\left(-1+i\right)\right]\left[z-\left(-1-i\right)\right]}\right]$

$\therefore Re{s}_{atz=-1+i}=\underset{z\to -1+i}{lim}\left[z-\left(-1+i\right)\right]\cdot \frac{e^{iz}}{(z-\left(-1+i\right)\left[z-\left(-1-i\right)\right]}$

$=\underset{z\to -1+i}{lim}\frac{e^{iz}}{z-\left(-1-i\right)}$

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

$\therefore {\oint }_c^{}\frac{Imaginary\left(e^{iz}\right)}{z^2+2z+2}dx=Imaginary\left[2\pi i\cdot \frac{e^{-i-1}}{2i}\right]$

$=Imaginary\left[\pi e^{-i}\cdot e^{-1}\right]$

$=Imaginary\left[\frac{\pi e^{-i}}{e}\right]\left[e^{ix}=\cos x+\sin x\right]$

$=\frac{\pi }{e}Imaginary\left[\cos 1-i\sin 1\right]$

Taking only imaginary part

Concept:

Cauchy integral formula:

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

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

Calculation:

Given

$f(z)=\frac{1}{z^2+6z+9}$

Contour is a unit radius circle with center is the origin

$f(z)=\frac{1}{(z+3{)}^2}$

Pole of f(z) is -3, -3 and both poles are outside of  the given unit circle (it is shown in below fig.)

Here all-poles of f(z) outside the circle so 

$\oint f(z)dz=0$

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:

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

Contour: |z|= 3

Simple pole, z = 2 and it is lies inside the contour.

Residue of f(z) at z = 2 is,

$\underset{z\to 2}{\mathrm{l}\mathrm{t}}\left(z-2\right)\frac{\left(z^3-2z+3\right)}{\left(z-2\right)}=2^3-2\left(2\right)+3=7$

f(z) = 2πi(7) = 14πi