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UPSC ESE 2025 Electronics and Telecommunications Paper

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Question : 146 of 150

Marks: +1, -0

The value of the integral

\oint\limits_{C} \frac{dz}{z (z^{2}+4)}

C

|z|=1

, is

0
$\frac{\pi i}{3}$
$2\pi i$
$\frac{\pi i}{2}$

Solution:

(d)

f(z)=\frac{1}{z (z^{2}+4)}

Poles of

f(z)

are

z=0

and

\pm 2i

Given contour is

c: |z|=1

(unit circuit)

Only

z=0

line in side

c

So

f(z)

is Non analytic in

c

Residue of

f(z)=\lim\limits_{z\to 0}(z-0)f(z)=\lim\limits_{z\to 0}\left(\frac{1}{z^{2}+4}\right)=\frac{1}{4}

By Cauchy's residue theorem,

I=\oint\limits_{c} f(z)\,dz=2\pi i(R)=2\pi i \left(\frac{1}{4}\right)=\frac{\pi}{2} i

Hence, the correct option is (d).

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