The value of the integral is 2z-1z2-1 dz where c is -z-12

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Standard XI

Mathematics

Distance formula using Complex Numbers

The value of ...

Question

The value of the integral is $(\frac{2 z + 1}{z^{2} + 1}) d z$ where $c$ is $| z | = \frac{1}{2}$

2 π i

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\frac{π}{2} i

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- 2 π i

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- \frac{π}{2} i

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Solution

The correct option is A $2 π i$
Poles are given by
$z^{2} + z = 0$
$z = 0, - 1$
$| z | = \frac{1}{2}$ is a circle with centre at origin and radius $\frac{1}{2}$ .
Therefore it encloses only one pole i.e. $z = 0$
$∴ \int_{0} \frac{2 z + 1}{z (z + 1)} d z = \int_{0} \frac{2 z + 1}{z + 1} d z = 2 π i$ (1)
$= 2 π i$

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The value of the integral is (2z+1z2+1)dz where c is |z|=12

The correct option is A 2πiPoles are given by z2+z=0 z=0, −1 |z|=12 is a circle with centre at origin and radius 12. Therefore it encloses only one pole i.e. z=0 ∴ ∫02z+1z(z+1)dz=∫02z+1z+1dz=2πi (1) =2πi

The value of the integral is $(\frac{2 z + 1}{z^{2} + 1}) d z$ where $c$ is $| z | = \frac{1}{2}$