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 \(\displaystyle \frac{1}{2 \pi i} \int_{γ} \frac{d z}{z^{3}-1}\) 

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 

\(\displaystyle\int_{C} \frac{f(z)}{z-a}dz=2 π i .f(a).\eta(\gamma:a)\)

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 \(\displaystyle \frac{1}{2 \pi i} \int_{γ} \frac{d z}{z^{3}-1}\)

= \(\displaystyle \frac{1}{2 \pi i} \int_{γ} \frac{d z}{(z-1)(z^2+z+1)}\)

= \(\displaystyle \frac{1}{2 \pi i} \int_{γ} \frac{1/(z^2+z+1)}{(z-1)}dz\)

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 \(\int_C{f(z)\over z-a}dz\)

Explanation:

\(\oint_{|1-z|=1} \frac{e^z}{z^2-1} d z\)

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

\(\oint_{|1-z|=1} \frac{e^z}{z^2-1} d z\) = 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

\(\frac{1}{2 \pi i} \int_C{f(z)\over (z-a)}d z\) = 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

\(\frac{n!}{2 \pi i} \int_C{f(z)\over (z-a)^{n+1}}d z\) = fⁿ(a)

Explanation:

C denote the unit circle centered at the origin in ℂ

So C: |z| = 1 ⇒ \(z\bar z=1\) ⇒ \(\bar z=\frac1z\)

\(\frac{1}{2 \pi i} \int_C\left|1+z+z^2\right|^2 d z\)

= \(\frac{1}{2 \pi i} \int_C\left(1+z+z^2\right)(\overline {1+z+ z^2}) d z\)

= \(\frac{1}{2 \pi i} \int_C\left(1+z+z^2\right)( {1+\bar z+ \bar z^2}) d z\)

= \(\frac{1}{2 \pi i} \int_C\left(1+z+z^2+\bar z+z\bar z+z^2\bar z+\bar z^2+z\bar z^2+z^2\bar z^2\right)d z\)

=\(\frac{1}{2 \pi i} \int_C\left(1+z+z^2+\frac1 z+z\cdot \frac1z+z^2\cdot \frac1z+\frac1{z^2}+z\cdot\frac1{z^2}+z^2\cdot\frac1{z^2}\right)\)dz 

= \(\frac{1}{2 \pi i} \int_C\left(1+z+z^2+1+z+1+\frac1{z^2}+\frac1 z+\frac1{z}\right)\)dz

= \(\frac{1}{2 \pi i} \int_C\left(3+2z+z^2\right)dz\) + \(\frac1{2\pi i}\int_C\left(\frac1{z^2}+\frac2{z}\right)dz\)

= 0 + \(\frac1{2\pi i}\int_C\frac1{z^2}dz\) + \(\frac1{2\pi i}\int_C\frac2{z}dz\)(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 

\(\frac{1}{2 \pi i} \int_C \frac{ϕ^{\prime}(z)}{ϕ(z)} d z\) = 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 \(\frac{1}{2 \pi i} \int_C \frac{ϕ^{\prime}(z)}{ϕ(z)} d z\) = no + 1 - np 

⇒ \(\frac{1}{2 \pi i} \int_C \frac{(z f)^{\prime}}{z f} d z=n_0-n_p+1 .\) (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

\(\frac1{2π i}\int_C\frac{f'(z)}{f(z)}dz\) = 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:

\(\int_C\frac{z^{100}}{z^{101}+1}dz\)

= \(\frac{1}{101}\int_C\frac{101z^{100}}{z^{101}+1}dz\)

= \(\frac1{101}\int_C\frac{f'(z)}{f(z)}dz\) 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 × \(1\over 101\) × 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

\(\mathop \oint \limits_C f\left( z \right)dz = 0\)

Cauchy’s Integral Formula:

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

\(f\left( a \right) = \frac{1}{{2\pi i}}\mathop \oint \limits_C \frac{{f\left( z \right)}}{{z - a}}dz\)

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

Residue Theorem:

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

\(\mathop \smallint \limits_C f\left( z \right)dz = 2\pi i \times \left[ {{\rm{sum\;of\;residues\;at\;the\;singualr\;points\;within\;C}}} \right]\)

Formula to find residue:

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

\(Res\;f\left( a \right) = \mathop {\lim }\limits_{z \to a} \left[ {\left( {z - a} \right)f\left( z \right)} \right]\)

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

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

Application:

\(\mathop \smallint \nolimits_C \frac{{{z^2} + 1}}{{{z^2} - 2z}}dz\)

\( = \mathop \smallint \nolimits_C \frac{{{z^2} + 1}}{{z\left( {z - 2} \right)}}dz\)

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

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

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

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

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

Residue at x = 2i

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

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

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

 \(\mathop \oint \nolimits_C f\left( x \right)dx = \mathop \oint \nolimits_C \frac{{\sin \left( x \right)}}{{{x^2}\left( {{x^2} + 4} \right)}}\;dx = 2\pi i\) (sum of residues)

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

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

\(f\left( z \right) = \mathop \oint \limits_C \frac{1}{{{{\left( {z - 1} \right)}^3}\left( {z - 3} \right)}}\)

(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

\(\mathop \smallint \limits_c f\left( z \right)dz = 2\pi i\left[ {sum\;of\;the\;residues\;at\;the\;poles\;inside\;'C'} \right]\)

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

\(\frac{1}{{\left( {m - 1} \right)!}}\begin{array}{*{20}{c}} {lim}\\ {z \to a} \end{array}\frac{{{d^{m - 1}}}}{{d{z^{m - 1}}}}\left[ {{{\left( {z - a} \right)}^m}f\left( z \right)} \right]\)

Residue at z = 3 is,

\( = \begin{array}{*{20}{c}} {lim}\\ {z \to 3} \end{array}\left( {z - 3} \right)\frac{1}{{{{\left( {z - 1} \right)}^3}\left( {z - 3} \right)}}\)

\( = \begin{array}{*{20}{c}} {lim}\\ {z \to 3} \end{array}\frac{1}{{{{\left( {z - 1} \right)}^3}}} = \frac{1}{8}\)

Residue at z = 1 is,

\( = \frac{1}{{\left( {3 - 1} \right)!}}\begin{array}{*{20}{c}} {lim}\\ {z \to 1} \end{array}\frac{{{d^2}}}{{d{z^2}}}\left( {{{\left( {z - 1} \right)}^3}\frac{1}{{{{\left( {z - 1} \right)}^3}\left( {z - 3} \right)}}} \right)\)

\( = \frac{1}{{2!}}\begin{array}{*{20}{c}} {lim}\\ {z \to 1} \end{array}\frac{{{d^2}}}{{d{z^2}}}\left( {\frac{1}{{z - 3}}} \right)\)

\( = \frac{1}{2}\begin{array}{*{20}{c}} {lim}\\ {z \to 1} \end{array}2{\left( {z - 3} \right)^{ - 3}}\)

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

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

\(\oint_c \frac{{f\left( z \right)}}{{z - z_o}} = 2\pi i f(z_o)\)

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

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,

\(\mathop \smallint \nolimits_c^{} f\left( z \right)dz = 2\pi i \times ({\rm{sum}}\,{\rm{of}}\,{\rm{the}}\,{\rm{residues ~at}}\,{\rm{the}}\,{\rm{singular}}\,{\rm{points}}\,{\rm{within}}\,{\rm{C}})\)

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

\({\mathop{\rm Re}\nolimits} s{\rm{ }}f(a){\rm{ = }}\mathop {{\rm{Lt}}}\limits_{z \to a} [(z - a)f(z)]\)

Calculation:

\(\mathop \oint \nolimits_{\rm{{\rm Γ}}}^{} \frac{{3z - 5}}{{\left( {z - 1} \right)\left( {z - 2} \right)}}dz=4\pi i\)

\(\mathop \oint \nolimits_1^{} \frac{{3z - 5}}{{\left( {Z - 1} \right)\left( {Z - 2} \right)}}dz = 4\pi i = 2\pi i\left( 2 \right)\)

∴ Sum of residue must be equal to 2

\(\mathop {{\rm{Res\;f}}\left( {\rm{z}} \right)}\limits_{z = 1} = \mathop {\lim }\limits_{z \to 1} \left( {z - 1} \right)\frac{{\left( {3z - 5} \right)}}{{\left( {z - 1} \right)\left( {z - 2} \right)}} =\mathop {\lim }\limits_{z \to 1} \frac{{\left( {3z - 5} \right)}}{{\left( {z - 2} \right)}}= 2\;\)

\(\mathop {{\rm{Res\;f}}\left( {\rm{z}} \right)}\limits_{z = 2} = \mathop {\lim }\limits_{z \to 2} \left( {z - 2} \right)\frac{{\left( {3z - 5} \right)}}{{\left( {z - 1} \right)\left( {z - 2} \right)}} = \mathop {\lim }\limits_{z \to 2} \frac{{\left( {3z - 5} \right)}}{{\left( {z - 1} \right)}} = 1\)

\(\mathop \oint \nolimits_1^{} \frac{{3z - 5}}{{\left( {Z - 1} \right)\left( {Z - 2} \right)}}dz = 4\pi i = 2\pi i\left( 2 \right)=2\pi i\left( Res ~f(1) \right)\)

∴ 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:

\(\mathop \oint \nolimits_C^{} \frac{{\cosh 3z}}{{2z}}dz\) where C represents unit circle i.e. radius is unity.

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

Therefore \(f(z)=\frac{\cosh3z}{2}\) and a = 0.

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

\(f(z)=\frac{\cosh3z}{2}=\frac{e^{3z}+e^{-3z}}{2\times2}=\frac{e^{3z}+e^{-3z}}{4}\)

At z = 0,

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

Cauchy's Integral Formula:

\(\oint_{c}^{}\frac{f(z)}{z-a}dz=2\pi i.f(a)\)

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

\(\mathop \oint \nolimits_C^{} \frac{\frac{\cosh 3z}{2}}{{z-0}}dz=2\pi i\times \frac{1}{2}=\pi i\)

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,

\( \mathop \oint \limits_C f(z)~dz=0\)

Given,

The integral is \(\mathop \oint \limits_C^{} \frac{{dz}}{{\left( {{z^2}~ + ~6} \right)}}\) 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,

\(\mathop \oint \limits_C^{} \frac{{dz}}{{\left( {{z^2} ~+~ 6} \right)}}\) = 0

\(\mathop \smallint \limits_{ - \infty }^\infty \frac{{\sin x}}{{{x^2} + 2x + 2}}dx\)

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

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

Calculation:

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

Let x replace by z.

\(\therefore I = \mathop \smallint \limits_{ - \infty }^\infty \frac{{Imaginary\;\left[ {{e^{iz}}} \right]}}{{{z^2} + 2z + 2}}dz\)

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

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

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

Here n = 1

\(Re{s_{z = a}} = \frac{1}{{\left( {n - 1} \right)!}}\;\mathop {\lim }\limits_{z \to a} {\left( {\frac{d}{{dz}}} \right)^{n - 1}}\left[ {{{\left( {z - a} \right)}^n}f\left( z \right)} \right]\)

\(\therefore Re{s_{at\;z = - 1 \pm i}} = \frac{1}{{\left( {1 - 1} \right)!}}\mathop {\lim }\limits_{z \to - 1 \pm i} {\left( {\frac{d}{{dz}}} \right)^{1 - 1}}\left[ {\left( {z - {{\left( {1 \pm i} \right)}^1}} \right).\frac{{{e^{iz}}}}{{\left[ {z - \left( { - 1 + i} \right)} \right]\left[ {z - \left( { - 1 - i} \right)} \right]}}} \right]\)

\(\therefore Re{s_{at\;z = - 1 + i}} = \mathop {\lim }\limits_{z \to - 1 + i} \left[ {z - \left( { - 1 + i} \right)} \right] \cdot \frac{{{e^{iz}}}}{{(z - \left( { - 1 + i} \right)\left[ {z - \left( { - 1 - i} \right)} \right]}}\)

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

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

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

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

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

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

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 

\(\oint{f(z)d}z= 0\)

Concept:

For a given complex function with poles, the complex integral \(I = \mathop \oint \limits_C f\left( z \right)dz\) is given by

Residue theorem as;

\(I = \mathop \oint \limits_C f\left( z \right)dz = 2\pi \;i \times \left\{ {Sum\;of\;residue\;of\;poles\;inside\;or\;on\;C} \right\}\)

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,

\(\mathop {{\rm{lt}}}\limits_{z \to 2} \left( {z - 2} \right)\frac{{\left( {{z^3} - 2z + 3} \right)}}{{\left( {z - 2} \right)}} = {2^3} - 2\left( 2 \right) + 3 = 7\)

f(z) = 2πi(7) = 14πi