Residue Theorem MCQ Quiz in বাংলা - Objective Question with Answer for Residue Theorem - বিনামূল্যে ডাউনলোড করুন [PDF]

Last updated on Apr 11, 2025

পাওয়া Residue Theorem उत्तरे आणि तपशीलवार उपायांसह एकाधिक निवड प्रश्न (MCQ क्विझ). এই বিনামূল্যে ডাউনলোড করুন Residue Theorem MCQ কুইজ পিডিএফ এবং আপনার আসন্ন পরীক্ষার জন্য প্রস্তুত করুন যেমন ব্যাঙ্কিং, এসএসসি, রেলওয়ে, ইউপিএসসি, রাজ্য পিএসসি।

Latest Residue Theorem MCQ Objective Questions

Top Residue Theorem MCQ Objective Questions

Residue Theorem Question 1:

The residues of a complex function X(z)=12zz(z1)(z2)at its poles are

  1. 12,12and1
  2. 12,12and1
  3. 12,1and32
  4. 12,1and32

Answer (Detailed Solution Below)

Option 3 : 12,1and32

Residue Theorem Question 1 Detailed Solution

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=2πi×[sumofresiduesatthesingualrpointswithinC]

Formula to find residue:

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

Resf(a)=limza[(za)f(z)]

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

Resf(a)=1(n1)!{dn1dzn1[(za)nf(z)]}z=a

Calculation:

Given:

X(z)=12zz(z1)(z2)

Poles are simple and located at z = 0, z = 1, and z = 2

Resf(a)=limza[(za)f(z)]

At z = 0, the residue is:

Resf(0)=limz0(z0){12zz(z1)(z2)}

Resf(0)=12×0(01)(02)=12

At z = 1, the residue is:

Resf(1)=limz1(z1){12zz(z1)(z2)}

Resf(1)=12×11(12)=1

At z = 2, the residue is:

Resf(2)=limz2(z2){12zz(z1)(z2)}

Resf(2)=12×22(21)=32

 

Residue Theorem Question 2:

The residue of f(z)=z3(z1)4(z2)(z3) at z = 3 is

  1. -8
  2. 10116
  3. 0
  4. 2716

Answer (Detailed Solution Below)

Option 4 : 2716

Residue Theorem Question 2 Detailed Solution

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=2πi×[sumofresiduesatthesingualrpointswithinC]

Formula to find residue:

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

Resf(a)=limza[(za)f(z)]

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

Resf(a)=1(n1)!{dn1dzn1[(za)nf(z)]}z=a

Calculation:

z = 3 is a simple pole of f(z)

The residue at z = 3 is,

=limz3(z3)z3(z1)4(z2)(z3)

=33(31)4(32)=2716

Residue Theorem Question 3:

Let c[1(z2)4(a2)2z+4]dz=4π, where the close curve C is the triangle having vertices at i, (1i2) and (1i2), the integral being taken in anti-clockwise direction. Then one value of a is

  1. 1 + i
  2. 2 + i
  3. 3 + i
  4. 4 + i

Answer (Detailed Solution Below)

Option 3 : 3 + i

Residue Theorem Question 3 Detailed Solution

c[1(z2)4(a2)2z+4]dz=4π 

The given curve c is the triangle having vertices at i, (1i2),(1i2)

The poles of given function are,

z = 0, z = 2

z = 2 is outside the curve C.

Now, the integral will be

C((a2)2z+4)dz=4π 

Residue of f(z) at z = 0 is,

ltz0zf(z)=ltz04z(a2)2=(a2)2

Now the value of integral will be

2πi [-(a - 2)2] = 4π

⇒ (a - 2)2 = 2i

From the options, a = 3 + i satisfies the above condition.

Residue Theorem Question 4:

The value of the integral ∫C (z – 2)3 dz, where C is a circle with |z – 2| = 4, is:

  1. –64
  2. 64i
  3. 0
  4. –128

Answer (Detailed Solution Below)

Option 3 : 0

Residue Theorem Question 4 Detailed Solution

Cauchy integral formula:

If f(z) is an analytic function within a closed curve C and let 'a' is a point inside C then,

Cf(z)zadz=2πif(a)

where f(a) is the residue of the f(z)

The residue is given by:

Case 1: For a simple pole (z = a):

f(a)=zaLt  [(za)f(z)]

Case 2: For the repeated pole (z = a...........n times):

f(a)=zaLt  1(n1)![dn1dzn1(za)nf(z)]

Calculation:

Given, Closed curve (C) = |z – 2| = 4

| x + i y - 2 | = 4

(x2)2+y2=16

This represents a circle with a center (2,0) and radius = 4.

F(z) = ∫C (z – 2)3 dz

Given function does not have any poles lying in the circle.

Hence, Cf(z)zadz=2πi(0)

F(z) = ∫C (z – 2)3 dz = 0

Residue Theorem Question 5:

For f(z)=z2z2+a2, which of the following conclusions is/are correct?

  1. z = ia is a simple pole of f(z)
  2. z = ia is not a simple pole of f(z)
  3. 12ia is a residue at z = ia of f(z)
  4. 13ia is a residue at z = ia of f(z)

Answer (Detailed Solution Below)

Option :

Residue Theorem Question 5 Detailed Solution

f(z)=z2z2+a2 

This can be written as:

f(z)=z2(z+ia)(zia) 

z = ia is a simple pole of f(z)

Now, the residue at z = ia will be:

=limzia(zia)f(z) 

=limzia(zia)z2(zia)(z+ia) 

=limziaz2z+ia=a22ia=12ia 

Residue Theorem Question 6:

The value of the integral 12πiCz2+1z21dz where z is a complex number and C is a unit circle with centre at 1 + 0j in the complex plane is __________ .

Answer (Detailed Solution Below) 1

Residue Theorem Question 6 Detailed Solution

12πiCz2+1z21dz

Simple poles, z = ±1

According to Cauchy’s residue theorem,

Cf(z)dz=2πi[sumofresidues]

f(z)=z2+1z21

At, z = +1, residue is

limz1(z1)z2+1(z+1)(z1)=1

At, z = -1 residue is zero as z = -1 lies outside the curve C.

12πiCz2+1z21dz=12πi×2πi[1]=1

Residue Theorem Question 7:

The value of the contour integral 12πj(z+1z)2dz evaluated over the unit circle |z| = 1 is _______.

Answer (Detailed Solution Below) 0

Residue Theorem Question 7 Detailed Solution

12πj(z+1z)2dz

12πj(z2+1z2+2)dz

12πj[z2dz+1z2dz+2dz]

The first and third term are analytic. Hence integral = 0

For integral

1z2dz

limzz01(n1)!dn1dzn1(zz0)nf(z)

limzz01(21)!ddz(z0)21(z)2

limzz011!ddz(1)

= 0

Hence, 2πi(z+1z)2dz=0

|z| = 1

Residue Theorem Question 8:

The residue at the singular point z = -2 of f(z)=1+z+z2(z1)2(z+2)

  1. 1/2
  2. 1/3
  3. 4/3
  4. 3/2

Answer (Detailed Solution Below)

Option 2 : 1/3

Residue Theorem Question 8 Detailed Solution

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=2πi×[sumofresiduesatthesingualrpointswithinC]

Formula to find residue:

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

Resf(a)=limza[(za)f(z)]

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

Resf(a)=1(n1)!{dn1dzn1[(za)nf(z)]}z=a

Calculation:

Res(z=2)=limz2(z+2)1+z+z2(z1)2(Z+2)

12+432= 1/3

Residue Theorem Question 9:

The Residue of 1(z2+a2)2 for z = ia is

  1. 14a3
  2. i4a3
  3. 14a3
  4. i4a3

Answer (Detailed Solution Below)

Option 2 : i4a3

Residue Theorem Question 9 Detailed Solution

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] = limzz0[(zz0)f(z)]

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

Res [f(z) : z = z0=1(m1)!{limzz0[dm1dzm1(zz0)mf(z)]}

Calculation:

Given,

f(z)=1(z2+a2)2=1(z+ia)2(zia)2

Pole of f(z) has order "2"

Res [f(z) : z = ia] =1(21)!{limzia[d(21)dz(21)(zia)21(z+ia)2(zia)2]}

=limzia{ddz1(z+ia)2}

=limzia2(z+ia)3

=i4a3

The residue of f(z) is i4a3

Residue Theorem Question 10:

Find f(z)=zsin(1z)dz

C is unit circle

Answer (Detailed Solution Below) 0

Residue Theorem Question 10 Detailed Solution

sin(1z)=(1z13!(1z)3+15!(1z)5+)

zsin(1z)=113!z2+15!1z4+

From definition

Residue = coefficient of 1/z in Laurent series expansion

Here coefficient of 1z=0

Residue = 0

zsin(1z)dz=2πi(0) = 0
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