Let $a$ be a non-zero complex number. Compute the contour integral $\int_C\ z^adz$ where $C$ is the semi-circle $|z| = 1$ going from 1 to −1 and $z^a$ is taken as the principal branch (or value) of the power.
Hi so I don't really know what my answer should be for this question. Question does not specify whether or not the curve is closed either. My professor has so far only briefly touched contour integrals in our class and has not gone over principle branchs/values of complex powers. I haven't been able to find many resources online to help me out, so here I am. So far, I've gotten up to the following but I'm not sure if I'm going in the right direction. Would greatly appreciate any help or guidance to point me in the right direction:
$$\int_C\ z^adz = \int_0^\pi i(e^{it})^ae^{it} dt = \begin{cases} \left| \frac{z^{a+1}}{a+1}\ \right|_{0}^{\pi},&a\ne-1 \\ \pi i, &a= -1\\\end{cases} $$
And for $ x\ne1 , \\ \left| \frac{z^{a+1}}{a+1}\ \right|_{0}^{1} = \frac{1}{a+1}\ [e^{it(a+1)} - e^{0}] = \frac{1}{a+1}\ (-e^{a\pi i}-1)=\frac{-1}{a+1}\ [cos(\pi a) + isin(\pi a)] $
Where $a = u+vi$, such that $u, v \in \mathbb{R}$
You have
$$i\int_0^\pi e^{it(a+1)}dt=\left.\frac i{a+1}e^{i(a+i)t}\right|_0^\pi=\frac1{a+1}\left(e^{i(a+1)\pi}-1\right)\;,\;\;\text{if}\;\;a\neq-1$$
For $\;a=-1\;$ you have
$$\int_c\frac{dz}z=i\int_0^\pi\frac{e^{it}dt}{e^{it}}=\pi i$$