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Latest Cauchy's Integral Theorem MCQ Objective Questions

Top Cauchy's Integral Theorem MCQ Objective Questions

Cauchy's Integral Theorem Question 1:

What is the value of ceiπz2z25z+2, where C is the circle |z| = 1 is with positive orientation

  1. π3
  2. 2π3
  3. π
  4. 2 πi

Answer (Detailed Solution Below)

Option 2 : 2π3

Cauchy's Integral Theorem Question 1 Detailed Solution

Concept:

For simple poles at z = a, b, c ⋯

Residue of f(z) = Res[f, a] =  limza(za)f(z)

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

Residue of f(z) = Res[f, a] = limza1(n1)![dn1dzn1(za)nf(z)]

Cauchy-Residue Theorem:

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

Cf(z)dz=2πiRi, where Ri's are the residues at the singular points within C where C is not analytic.

Solution:

Given,  ceiπz2z25z+2dz

C is the circle |z| = 1 is with positive orientation

Let f(z) =  eiπz2z25z+2

Putting 2z- 5z + 2 = 0

⇒ 2z- 4z - z + 2 = 0

⇒ 2z(z - 2) - 1(z - 2) = 0

⇒ (z - 2)(2z - 1) = 0

⇒ z = 2, 12

Now z = 12 lies inside C

∴ Res[f, 12] = limz12(z12)eiπz2z25z+2

limz12(2z1)eiπz2(z2)(2z1)

limz12eiπz2(12z)

limz12eiπ22(122)

i3

∴  ceiπz2z25z+2dz = 2πi∑Res(f(z)) 

= 2πi[i3]

2πi23

2π3 [∵ i2 = - 1]

∴ The value of the given integral is 2π3.

The correct answer is option 2.

Cauchy's Integral Theorem Question 2:

The value of cz24z(z2+9)is where C is a circle |z| = 1,

  1. 2πi9
  2. 4πi9
  3. 8πi9
  4. none of these

Answer (Detailed Solution Below)

Option 3 : 8πi9

Cauchy's Integral Theorem Question 2 Detailed Solution

Concept:

For simple poles at z = a, b, c ⋯

Residue of f(z) = Res[f, a] =  limza(za)f(z)

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

Residue of f(z) = Res[f, a] = limza1(n1)![dn1dzn1(za)nf(z)]

Cauchy-Residue Theorem:

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

Cf(z)dz=2πiRi, where Ri's are the residues at the singular points within C where C is not analytic.

Solution:

Given, cz24z(z2+9)dz

C is a circle, lzl=1

Let f(z) = z24z(z2+9)

Putting z(z+ 9) = 0

⇒ z = 0, 3i, - 3i

Now, z = 0 lies inside C.

⇒ Res[f, 0] = limz0(z - 0)f(z)

limz0z24(z2+9)

49

∴ cz4z(z2+9)dz = 2πi∑Res(f(z))

= 2πi [Res(f, 0)]

= 2πi[49]

8πi9

∴ The value of given integral is 8πi9.

The correct answer is option 3.

Cauchy's Integral Theorem Question 3:

Which theorem states that, "if an entire function is bounded for a values of z, then it is constant"?

  1. morera theorem
  2. Liouville's theorem
  3. Cauchy theorem
  4. Fundamental theorem

Answer (Detailed Solution Below)

Option 2 : Liouville's theorem

Cauchy's Integral Theorem Question 3 Detailed Solution

Option 1)-  Morera's Theorem- It states that If f (z) is continuous function in a 

simply connected domain D and if cf(z)dz=0 where, c is any 

rectifiable closed Jordan curve in D then f(z) is analytic in D. 

Option 2) - Liouville's Theorem - It states that entire bounded function is constant .

Option 3)-  If f(z) is analytic function in a simply connected region R then, 

cf(z)dz=0 for every closed contour C in R . 

Option 4)-  Let f(z) be single valued analytic function in a simple connected domain D, if a, b are in D, then

(∫f(z) dz) over the limits a and b = F(b)F(a)

Therefore, Correct Option is Option 2).

Cauchy's Integral Theorem Question 4:

The value of the integral

(6z2z43z3+7z23z+5)dz

evaluated over a counter-clockwise circular contour in the complex plane enclosing only the pole z = i, where 𝑖 is the imaginary unit, is

  1. (-1 + i) π
  2. (1 + i) π
  3. 2(1 - i) π
  4. (2 + i) π

Answer (Detailed Solution Below)

Option 1 : (-1 + i) π

Cauchy's Integral Theorem Question 4 Detailed Solution

Concept:

Residue theorem: if f(z) is an analytic function in a closed curve C except at a finite number of singular points within C then 

  • cf(z)dz=2πi × (sum of the residues at the singular point within curve C)

Residue for simple pole z = a:

  • Res f(a) = limza[(za)f(z)]

Calculation:

Given:

(6z2z43z3+7z23z+5)dz

pole z = i

Check for singularity at pole z = i

f(z) = 2z4 - 3z3 + 7z2 - 3z + 5

f(i) = 2(i)4 - 3(i)3 + 7(i)2 - 3i + 5 

f(i) = 2 ×1 - 3(-i) - 7 - 3i + 5 = 0

since, f(i) = 0 ⇒ z = i  is a singular point

From Residue theorem:

(6z2z43z3+7z23z+5)dz = 2πi (Residue at z = i )

Residue at z = i :

Res|z= i  limzi(zi)6z2z43z3+7z23z+5

at z = i , Res = 0/0 form, applying L'hospital rule 

Res|z= i  limzi12z  6i8z39z2+14z3 = 12i  6i8i39i2+14i3

Res|z= i  6i8i+9+14i3 = ii + 1

(6z2z43z3+7z23z+5)dz = 2πi × ii + 1

⇒  2πi2i + 1=2πi + 1×i  1i  1 = π(i  1)

(6z2z43z3+7z23z+5)dz = π(i  1) = π(-1 + i)

Cauchy's Integral Theorem Question 5:

C is a closed path in the z-plane given by |z| = 3. The value of the integral C(z2z+4jz+2j)dz is

  1. -4π (1 + j2)
  2. 4π (3 – j2)
  3. -4π (3 + j2)
  4. 4π (1 – j2)

Answer (Detailed Solution Below)

Option 3 : -4π (3 + j2)

Cauchy's Integral Theorem Question 5 Detailed Solution

Concept:

If f(z) is analytic within and on a closed curve, and if ‘a’ is any point within C then according to Cauchy Integral formula:

f(a)=12πief(z)(za)dz

Cf(z)(za)dz=2πif(a)

Application:

|z|=3

Cz2z+4jz+2j=?

Pole z = -2j, which lies Inside the given C i.e. |z| = 3

∴ Using the Cauchy Integral formula, we get:

Cz2z+4jz+2j=2πj[z2z+4j]atz=2j

Cz2z+4jz+2j=2πj[4+2j+4j]

=4π(3+j2)

Cauchy's Integral Theorem Question 6:

Value of integral cf(z)zdz, if f(z)=a+bz and the contour c is |z|=1

  1. 0
  2. 2πi(1 + a)
  3. 2πai
  4. 2πbi

Answer (Detailed Solution Below)

Option 3 : 2πai

Cauchy's Integral Theorem Question 6 Detailed Solution

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)=limza(za)f(z)

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

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

cf(z)dz=2πi×(SumofresidueatthepolesinsideoronC)

Calculation:

Given:

|z|=1 is unit circle

cf(z)zdz and f(z)=a+bz

cf(z)zdz=ca+bzzdz=caz+bz2dz

GATE-EM5 Complex Number Ques-5 A-1

cf(z)zdz=ca+bzzdz=caz+bz2dz

For pole, z2 = 0 i.e. z = 0, 0 (order 2)

cf(z)dz=2πi×(SumofresidueatthepolesinsideoronC)

Residue:

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

cf(z)dz=2πi{limz011!ddz[(z0)2az+bz2]}2πi[limz0(a)]=2πai

Cauchy's Integral Theorem Question 7:

Evaluate Ccosπz2z(z3)dz where C is the circle: |z| = 4

  1. 43πi
  2. 23πi
  3. 0
  4. 23πi

Answer (Detailed Solution Below)

Option 1 : 43πi

Cauchy's Integral Theorem Question 7 Detailed Solution

f(z)=cosπz2z(z3)

Cf(z)dz=2πi [Sum of residues at poles]

Poles: z = 0, 3

Both the poles are inside the |z| = 4

Residue at z = 0,

=ltz0zf(z)

=ltz0z.cosπz2z(z)3=13

Residue at z = 3,

=ltz3(z3)cosπz2z(z3)=cos9π3=13

f(z)dz=2πi[1313]=4πi3

Cauchy's Integral Theorem Question 8:

If C is a circle of radius r with center z0, in the complex z-plane and if n is a non-zero integer, then Cdz(zz0)n+1 equals

  1. 2πnj
  2. 0
  3. nj/2π
  4. 2πn

Answer (Detailed Solution Below)

Option 2 : 0

Cauchy's Integral Theorem Question 8 Detailed Solution

CONCEPT:

If f(z) is analytic within & on a closed curve C and “a” is a point inside the curve C then,

2πif(a)=Cf(z)dzza; we consider f(z)/(z-a) to be analytic at all points within C except at z=a.

If multiple singularities occur then we use this;

2πifn(a)=Cf(z)dz(za)n+1        ……(1)

CALCULATION:

We have to find this Cdz(zz0)n+1 

Comparing the above with equation ..(1), we conclude that f(z) =1.

If we take differentiation then f’(a) = 0.

This implies that

2πifn(a)=Cf(z)dz(za)n+1=0

Hence option 2 is correct 

Cauchy's Integral Theorem Question 9:

The value of the integral Cz+1z24dz in counter clockwise direction around a circle C of radius 1 with center at the point z = −2 is

  1. πi2
  2. 2πi
  3. πi2
  4. – 2πi

Answer (Detailed Solution Below)

Option 1 : πi2

Cauchy's Integral Theorem Question 9 Detailed Solution

Given that,

f(Z)=CZ+1Z24

C : |Z – (-2)| = 1

⇒ C : |Z + 2| = 1

GATE EE 2018 Techinical 54Q images Q13

poles of f(Z) are z2 – 4 = 0

⇒ Z = ±2

Z = 2 is lies outside the curve C.

f(Z) = 2πi [residue at Z = -2]

=2πiitZ2(Z+2)(Z+1)(Z+2)(Z2)

=2πi×(2+1)(22)=2πi(14)=πi2

Cauchy's Integral Theorem Question 10:

The value of the 1z2+16dz over the contour |z+j|=4

  1. π

  2. π/2

  3. π/4

  4. π

Answer (Detailed Solution Below)

Option 3 :

π/4

Cauchy's Integral Theorem Question 10 Detailed Solution

1z2+16=1(z+4j)(z4j)

The contour is shown:

F1 R.D D.K 14.09.2019 D1

∴Pole (0,4) lies inside |z+j|=4

∴ By Residue theorem

I=f(z)dz=2πi(sum of residue inside contour)

I=1z2+16dz

limzj4z+j4(z+j4)(zj4)= -1/j8

=2πj(4j4j)=π4

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