Parametrize the contours of integration and write the integrals in terms of the parametrizations. Do not calculate them.
$$\int\frac{\bar(z)}{z^3}dz$$ where $$\Gamma$$ is the arc of the circle of radius $$\sqrt{2}$$ centered at the origin with initial point at 1+i and terminal point at 1-i that lies in the right half-plane and is transversed once.
I asked a question similar to this last night and felt that everyone was very helpful. I'm hoping to see the solution to this problem as I feel completely lost.
The circle of radius $2$ centered at the origin is parametrized by $z = \sqrt{2}e^{i\theta}$, $0 \le \theta \le 2\pi$. However, we want just the arc $\Gamma$ of the circle. So the range of $\theta$ will be restricted. In polar form, $1 + i = \sqrt{2}e^{i\pi/4}$ and $1 - i = \sqrt{2}e^{-i\pi/4}$. So $\theta$ ranges from $-\pi/4$ to $\pi/4$. Further, since $\Gamma$ is traversed from $1 + i$ to $1 - i$, it follows that $\Gamma$ is oriented clockwise. Thus
$$\int_{\Gamma} \frac{\bar{z}}{z^3}\, dz = -\int_{-\pi/4}^{\pi/4} \frac{\sqrt{2}e^{-i\theta}}{2^{3/2}e^{3i\theta}} i\sqrt{2}e^{i\theta}\, d\theta = -\frac{i}{\sqrt{2}}\int_{-\pi/4}^{\pi/4} e^{-3i\theta}\, d\theta.$$