0
$\begingroup$

$\int_\Gamma \frac{\bar{z}}{z^3}\; dz $ , where $\Gamma$ is arc of the circumstance of radius 2 centered at the origin with initial point at $\sqrt{3}-i$ and terminal point at 2 that is traversed once in the counter clock wise direction.

My attempt: The circle of radius 2 centered at origin is parametrized by $z=2z^{i\theta}, 0\leq \theta \leq 2\pi$

Given that initial point is $\sqrt{3}-i$, $\theta =\frac{\pi}{3}$

terminal point is 2 then $\theta =\frac{\pi}{2}$

Is i am right can any one help this problem

$\endgroup$

1 Answer 1

1
$\begingroup$

The circle of radius 2 centered at origin is parametrized by $z=2{\color{red}{e}}^{i\theta}, 0\leq \theta \leq 2\pi.$

The initial point is $\theta_1=2\pi-\pi/6$ because $2e^{i\theta_1}=\sqrt{3}-i$. The terminal point is $\theta_2=2\pi$ because $2e^{i\theta_2}=2$.

Since $dz=2ie^{i\theta}d\theta$, $\bar{z}=2e^{-i\theta}$, $z^3=8e^{3i\theta}$, we have $$\int_\Gamma \frac{\bar{z}}{z^3}\; dz=\int_{\theta_1}^{\theta_2}\frac{2e^{-i\theta}2ie^{i\theta}d\theta}{8e^{3i\theta}} =\frac{i}{2}\int_{11\pi/6}^{2\pi}e^{-3i\theta}d\theta=\frac{1}{6}(-1+i)$$

$\endgroup$

You must log in to answer this question.