Contour Integral & Theorem MCQ Quiz - Objective Question with Answer for Contour Integral & Theorem - Download Free PDF

Concept:

Cauchy’s Integral Formula: If a complex function f(z) is analytic within and on a closed contour C inside a simply-connected domain, and if z₀ is any point inside C, then

 = 2πi f(z₀)

Explanation:

I = 

The only singular point inside C is z = 0

The function f(z) =  is analytic in C.

Then using Cauchy’s Integral Formula,

  = 2πi f(0) where f(z) = 

                   = 2πi ×  = πi

Option (2) is true.

Concept:

Argument theorem: Let f be meromorphic function and C be simple closed contour such that no zero or pole of f lies inside C. Let a₁, a₂,…,a_k be zeros of f of order n₁, n₂,…,n_k respectively and f(z) has no pole in C. Then
  = 

Explanation:

g(z) = z3 and and f(z) = z3 - z - 1.

Let a, b, c are the zeros of f(z) lies in C then

a + b + c = 0

a² + b² + c² = -1 and

abc = 1

Now, 

 = a³ + b³ + c³

                           = (a + b + c)(a² + b² + c² - ab - bc - ca) + 3abc

                           = 0(-1) + 3 = 3

Option (1) is true.

Concept:

Cauchy integral formula: If a complex function f(z) is analytic 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) = 

Explanation:

, C: |z + 1| = 2

singularities are given by

4 - z² = 0 ⇒ z = ± 2 out of which z = -2 lies inside C

SO, f(z) =  is analytic inside C.

Hence by Cauchy's integral formula, 

 = 2πi f(-2)  where f(z) = 

               = 2π i () =  2πi

(3) is true.

Explanation:

  where C is the triangular contour with vertices at 0 ,   , and  

The function    has a singularity at z = 1 , which is a simple pole

To apply the residue theorem, we check if z = 1 lies inside the triangular contour

The given contour has vertices at:    (which is 0.25 on the real axis) and    (which is 0.5 on the imaginary axis)

Since z = 1 is outside this triangular region, the contour does not enclose the singularity

Apply Cauchy's Residue Theorem : 

Since the singularity z = 1 is outside the contour, the value of integral will be zero:

Hence Option(3) is the correct answer.

Concept: 

Generalized Cauchy Integral Formula: 

For an analytic function f(z) inside and on a simple closed contour C , and for any point a inside C : 

Explanation:

Applying this Formula:

Here, we are given the integral:    

Comparing with the formula, we see that n = 2 , so we use:

Since 1! = 1 , we get: 

Rewriting the Integral in Terms of f'(z) :  

We now use Cauchy's Integral Formula for f'(a) :

Multiplying both sides by   , we get:

Now, comparing this with our given integral result:

 ,

we see that this exactly matches:

Hence Option(2) is the correct answer.

Concept:

Explanation:

C be the positively oriented circle in the complex plane of radius 3 centered at the origin.

Singularities of  is given by

z² = 0 ⇒ z = 0 and

e^z - e^-z = 0 ⇒ ez = e-z  ⇒ z = 0

Now,

  = 

          =  (Expansion of ez - e-z)

         = 

        =  (expansion of (1 + x)^-1)

So Residue of  = coefficient of 1/z =      

Hence  = 2πi(sum of residues) =  = −iπ/6

Option (4) is correct

Concept:

Cauchy Integral Theorem:

If a complex function f(z) is analytic 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) = 

Explanation:

Singular points are given by 

z² - 1 = 0 ⇒ z = 1, z = -1

The only singular point that lies inside γ is z = 1.

Let f(z) = 

Hence using Cauchy's Integral test

 = 2πi f(1) = 2πi × (e/2) = iπe

 Option (1) is correct. 

Concept:

Cauchy Integral Theorem:

If a complex function f(z) is analytic 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) = 

Explanation:

 = 

So poles are given

(z - 1)(z² + z +1) = 0 ⇒ z = 1, z =  = 

Poles inside γ is z  = 1

Hence f(z) =  is analytic inside γ

Therefore by Cauchy Integral theorem, 

 = f(1) =  = 1/3 

 Option (2) is correct 

Concept:

Modified Argument Principal

Let f be a holomorphic function with no zeros on the boundary of some simply connected region D and N + P denotes the number of zeros and poles of f inside D respectively, counted with multiplicity. Then, for an analytic fⁿ

 dz = ∑ g(a_i) n(D, ai) -∑ g(bi) n(D, b_i)

where, n(D, z) = winding number of D around z, z ∈ D. 

and a_i = all zeros of f with multiplicity

& b_i = Poles of f with orders.

Explanation:

Here, we have given, f has zero only at z₀, 0 0|

⇒ b_i = 0 ∵ f has no poles

(∵ f is holomorphic)

= g(z0).q (∵ z0 is only zero with, multiplicity)

 (∵ g(z) = z^p)

option (1) is true

Concept:

Explanation:

C be the positively oriented circle in the complex plane of radius 3 centered at the origin.

Singularities of  is given by

z² = 0 ⇒ z = 0 and

e^z - e^-z = 0 ⇒ ez = e-z  ⇒ z = 0

Now,

  = 

          =  (Expansion of ez - e-z)

         = 

        =  (expansion of (1 + x)^-1)

So Residue of  = coefficient of 1/z =      

Hence  = 2πi(sum of residues) =  = −iπ/6

Option (4) is correct

Concept:

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

f(a) = dz

Here, the integral should be taken in the positive sense around C.

Solution - Given , function

f(z) = 

and the function has singularity at z = i, z = - i

C : {z : |z - i|

So z = - i does not lie on the curve and z = i lies inside the curve

Hence

I = 2πi ×  = 2πi  = 2πi  = π (log i)³

Now, log (i) = log 1 + i tan^-1(1/0) = 0 + i = i

hence I = π = 

Therefore, Correct Option is Option 4).

Explanation:

Given: f is holomorphic fn on {z ∈ ℂ ∶ |z|

And, only zero of f in {z ∈ ℂ ∶ |z| ≤ 1} is a simple zero at the origin.

And, ν = {z ∈ ℂ ∶ |z| = 1}

Recall: Argument Theorem

dz = N − P; N = no. of zeros with multiplicity, P = no. of poles with order

Here N = 1, P = 0 (Because, we don't have poles ← only simple zeros)

∴ 

⇒  at z0 = 0

Hence option (3) is true.

Explanation:

Singularities are 0, i lies inside curve |z| = 6

Let f(z) = e^2iz, g(z) = z⁴

So using Cauchy integral formula

= 2πi( - ) 

= 2πi( - )

= 2πi(- +6)

= 

(3) correct

Concept:

Cauchy's integral formula: If a complex function f(z) is analytic within and on a close contour γ inside a simply connected domain and if z = a is any point inside γ then

Explanation:

Singularities are given by

z - 1 = 0 ⇒ z = 1 and

z - 2i = 0 ⇒ z = 2i

γ is positively oriented circle {z ∈ C: |z| = 3/2}

z = 1 is the pole inside the circle and z = 2i is outside the circle.

So f(z) is analytic at z = 2i.

Let 

By using the Cauchy Integral theorem,

dz = 2πi f(1) = 2πi × 

=  = 2πi ×  = 

= 

= 

= 2πi

So comparing with the given solution we get

C = 

So |C| =  =  = 1/5

∴ |C| is equal to 1/5.

Concept:

Argument theorem: Let f be meromorphic function and C be simple closed contour such that no zero or pole of f lies inside C. Let a₁, a₂,…,a_k be zeros of f of order n₁, n₂,…,n_k respectively and f(z) has no pole in C. Then
  = 

Explanation:

g(z) = z3 and and f(z) = z3 - z - 1.

Let a, b, c are the zeros of f(z) lies in C then

a + b + c = 0

a² + b² + c² = -1 and

abc = 1

Now, 

 = a³ + b³ + c³

                           = (a + b + c)(a² + b² + c² - ab - bc - ca) + 3abc

                           = 0(-1) + 3 = 3

Option (1) is true.