Complex integration (zill exercise) Can any one help me with this path integral?
$$\int_C\dfrac{dz}{z^2-2i}$$ where $C$ is the right half of the circle $|z| = 6$ from $z = −6i$ to $z = 6i$.
Thanks.
 A: Note that $z^2-2i=0$ for $z_{\pm}=\pm\sqrt{2}e^{i\pi/4}$.
Then by Residue Theorem
$$\int_C\dfrac{dz}{z^2-2i}+\int_S\frac{dz}{z^2-2i}=2\pi i\cdot\mbox{Res}\left(\frac{1}{z^2-2i},\sqrt{2}e^{i\pi/4}\right)$$
where $S$ is the line segment from $6i$ to $-6i$.
In alternative, you may parametrize the half circle by $\gamma(t)=6 \cdot e^{i\pi t/2}$ for $t\in[-1,1]$ and calculate
$$\int_C\frac{dz}{z^2-2i}=3i\pi\int_{-1}^1\frac{e^{i\pi t/2}dt}{36e^{i\pi t}-2i}.$$
Note that the exercise asks for an upper bound for the absolute value of the given integral, which is an easier task:
$$\left|\int_C\frac{dz}{z^2-2i}\right|=3\pi\int_{-1}^1\frac{dt}{|36e^{i\pi t}-2i|}\leq 3\pi\int_{-1}^1\frac{dt}{36-2}= \frac{3\pi}{17}.$$
A: You may parametrize your curve $C$ by $\gamma(t) = 6 \cdot e^{it}$ for $t \in [-\frac{\pi}{2}, \frac{\pi}{2}]$. Then apply the definition of a complex path integral to transform it into a "common" Riemann integral.
By the way, this particular exercise actually does not ask you to evaluate this path integral, but to find an upper bound for the absolute value of it! ;-) This is somewhat simpler in my opinion, since the anti-derivative of the integrand you'd get is quite complicated...
