An integra I over a counterclockwise circle C is given by I-Cz2-1z2-1ezdzIf C is defined as -z-3- then the value of I is

Poles are z=± i As both z=± i lie inside the circle |z|=3 nbsp; Residue at nbsp; (z=i)=lim_z→ i(z i)(z^2 i)(z i)(z+i)e^z = 1 1zie^i=ie^i Residue at nbsp; ...

An integra I ...

Question

An integra $I$ over a counterclockwise circle $C$ is given by $I = \oint_{C} \frac{z^{2} - 1}{z^{2} + 1} e^{z} d z$
If $C$ is defined as $| z | = 3$ , then the value of $I$ is

- π i s i n (1)

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- 2 π i s i n (1)

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- 3 π i s i n (1)

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- 4 π i s i n (1)

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Solution

The correct option is D $- 4 π i s i n (1)$
Poles are $z = \pm i$
As both $z = \pm i$ lie inside the circle $| z | = 3$
Residue at
$(z = i) = {lim}_{z \to i} \frac{(z - i) (z^{2} - i)}{(z - i) (z + i)} e^{z}$
$= \frac{- 1 - 1}{z i} e^{i} = i e^{i}$
Residue at
$(z = - i) = {lim}_{z \to - i} \frac{(z - i) (z^{2} - i)}{(z - i) (z + i)} e^{z}$
$= - e^{- i}$
By Residue thorem
$I = 2 π i (i e^{i} - i e^{- i})$
$= - 2 π (c o s 1 + i s i n 1 - c o s 1 + i s i n 1)$
$= - 4 π i s i n (1)$

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An integra $I$ over a counterclockwise circle $C$ is given by $I = \oint_{C} \frac{z^{2} - 1}{z^{2} + 1} e^{z} d z$
If $C$ is defined as $| z | = 3$ , then the value of $I$ is