Question

The countour $C$ in the adjoining figure is described by $x^2+y^2=16$. Then the value of ${\int }_c\frac{z^2+8}{\left(0.5\right)z-\left(1.5\right)j}dz$

• A. $-2\pi j$
• B. $-2\pi j$
• C. $4\pi j$
• D. $-4\pi j$

Answer 1

The correct option is D $-4\pi j$

$C=x^2+y^2=16\Rightarrow |z|=4$
Now ${\int }_c\frac{z^2+8}{\left(0.5\right)z-\left(1.5\right)j}dz$
Pole of f(z) is $z=3j$ (simple pole) which lies with in the given contour $={lim}_{Z\to 3j}(z-3j)f\left(z\right)$
$={lim}_{Z\to 3j}\left[\frac{z^2+8}{1/2}\right]=-2$
So by Cauchy residue theorem
$I={\int }_{C}f\left(x\right)dz=2\pi j$ [sum of residues]
$=-4\pi j$