Question

Using Cauchy's integral formula, the value of the integral (integration beign taken in counter clockwise direction) $\oint \frac{z^3-6}{3z-i}dz$ is within the unit circle

• A. $\frac{2\pi }{81}-4\pi i$
• B. $\frac{\pi }{8}-6\pi i$
• C. $\frac{4\pi }{81}-6\pi i$
• D. $1$

Answer 1

The correct option is A $\frac{2\pi }{81}-4\pi i$

$\oint \frac{z^3-6}{3z-i}dz={\oint }_{C}f\left(Z\right)dz$
$C$ is $|Z|=1$
$f\left(z\right)$ has a pole
$z=i/3$ inside $c$.
residue of $f\left(z\right)$ at (z=i/3\)
$={lim}_{z\to i/3}\left\{(z-\frac{i}{3})f\right(z\left)\right\}$
$={lim}_{z\to i/3}\left\{\frac{(z^3-6)}{3}\right\}$
$=\frac{1}{3}[\frac{-i}{27}-6]=-\frac{i}{81}-2$
By cauchy's residue theorem
$\oint f\left(z\right)dz=2\pi i(\frac{-i}{81}-2)=\frac{2\pi }{81}-4\pi i$