Question

The value of the contour integral
$\frac{1}{2\pi j}\oint {(z+\frac{1}{z})}^{2}dz$
evaluated over the unit circle $|z|=1$ is

• A. 0

Answer 1

The correct option is A 0

The value of the contour integral
$\frac{1}{2\pi j}\oint {(z+\frac{1}{z})}^{2}dz,|z|=1$
$I=\frac{1}{2\pi j}\oint {(z+\frac{1}{z})}^{2}dz$
$I=\frac{1}{2\pi j}\oint {(z+\frac{z^2+1}{z})}^{2}dz$



There are two poles at $z=0$, (lying inside the contour)
so, by using $\oint \frac{f\left(z\right)}{(z-z_o{)}^n}dz=\frac{1}{(n-1)!}\frac{d^{n-1}}{d{z}^{n-1}}f\left(z\right)$
$I=\left(\frac{1}{2\pi j}\right)\left(\frac{1}{1!}\right)\frac{d}{dz}\left(\right(z^2+1{)}^2)$
$I=\frac{1}{2\pi J}\left(2\right(z^2+1\left)\right(2z\left)\right){|}_{z=0}$
$I=4z(z^2+1){|}_{z=0}$
$I=0$