The value of - Cdz1-z2 where C is the contour - z-12 -1 is

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The value of ...

Question

The value of $\oint_{C} \frac{d z}{(1 + z^{2})}$ where $C$ is the contour $∣ ∣ ∣ z - \frac{1}{2} ∣ ∣ ∣ = 1$ is

2 π

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π

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t a n^{- 1} z

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π i t a n^{- 1} z

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Solution

The correct option is B $π$
$I = \oint_{C} \frac{d z}{(1 + z^{2})} = \oint_{C} \frac{d z}{(z - i) (z + i)}$
Let $f (z) = \frac{1}{1 + z^{2}}$
Countour $C$ represents a circle, having radius $1$ unit and centre at $0, \frac{1}{2}$
The pole, $z = i$ lies inside the contour, while $z = - i$ outside

$∴$ The value of integral
$I = 2 π i [R e s f (z)$ at $z = i$ ]
$= 2 π i \times \frac{1}{i + i} = π$

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The value of ∮Cdz(1+z2) where C is the contour ∣∣∣z−12∣∣∣=1 is

The correct option is B πI=∮Cdz(1+z2)=∮Cdz(z−i)(z+i) Let f(z)=11+z2 Countour C represents a circle, having radius 1 unit and centre at 0,12 The pole, z=i lies inside the contour, while z=−i outside ∴ The value of integral I=2πi[Resf(z) at z=i] =2πi×1i+i=π

The value of $\oint_{C} \frac{d z}{(1 + z^{2})}$ where $C$ is the contour $∣ ∣ ∣ z - \frac{1}{2} ∣ ∣ ∣ = 1$ is