The value of the integral $\frac{1}{2 π j} \oint_{C} \frac{z^{2} + 1}{z^{2} - 1} d z$ where $z$ is a complex number and $C$ is a unit circle with center at $1 + 0 j$ in the complex plane is

1

Open in App

Solution

The correct option is A 1
Given:
$\frac{1}{2 π j} \oint \frac{z^{2} + 1}{z^{2} - 1} d z = \frac{1}{2 π j} \oint \frac{z^{2} + 1}{(z - 1) (z + 1)} d z$
Poles are $z = 1, - 1$ (Both simple)

Given $C$ is $(x - 1)^{2} + y^{2} = 1 \Rightarrow C; | z - 1 | = 1$
Clearly $1$ lies inside of $C$ and $- 1$ outside $C$ . Then Res $f (z)$ (at $z = 1$ )
${lim}_{Z \to 1} (z - 1) \frac{z^{2} + 1}{(z - 1) (z + 1)}$
${lim}_{Z \to 1} (\frac{z^{2} + 1}{z + 1}) = 1$
$∴$ By Cauchys Residue theorem
$\frac{1}{2 π j} \oint \frac{z^{2} + 1}{z^{2} - 1} d z = \frac{1}{2 π j} \times 2 π j \times 1 = 1$

Suggest Corrections

Similar questions

Q. The value of

\int_{c} \frac{z^{2}}{z^{4} - 1} d z

, using Cauchy's integral formula, around the circle

| z + 1 | = 1

where

z = x + i y

Q. The value of the integral is

(\frac{2 z + 1}{z^{2} + 1}) d z

where

c

| z | = \frac{1}{2}

Q. The value of complex intergral,

I = \oint c \frac{z + 1 d z}{z^{3} + z^{2} + (z + 1)} d z i s k π i w h e r e c i s a c i r c l e | z | = 2

Then the value of k is __________

The contour $C$ given below is on the complex plane $z = x + i y$ , where $j = \sqrt{- 1}$ .

The value of the integral $\frac{1}{π j} \oint_{C} \frac{d z}{z^{2} - 1}$ is

Q. If

z^{2} + z + 1 = 0

, where z is a complex number, then the value of

(z + \frac{1}{z})^{2} + (z^{2} + \frac{1}{z^{2}})^{2} + (z^{3} + \frac{1}{z^{3}})^{2} + . . . . + (z^{6} + \frac{1}{z^{6}})^{2}

Join BYJU'S Learning Program

The value of the integral 12πj∮Cz2+1z2−1dz where z is a complex number and C is a unit circle with center at 1+0j in the complex plane is 1

The value of the integral $\frac{1}{2 π j} \oint_{C} \frac{z^{2} + 1}{z^{2} - 1} d z$ where $z$ is a complex number and $C$ is a unit circle with center at $1 + 0 j$ in the complex plane is

1