I'm working on some past exam papers and I wanted to see if I'm thinking of the following in the correct way;
Given a contour integral $\int_{\gamma}\frac{1}{z}dz$ I want to find all of it's possible values given that $\gamma$ is the semi-circle starting at $1$ and going to $-1$
My thinking is that as the radius of the semicircle is $1$ then $|z|=1$ and so $z=e^{i\theta}$.
We can also parametrise $\gamma$ as $e^{i\theta}, \theta \in [-\pi,\pi]$
This means we can express our integral as;
$\int_{\gamma}\frac{1}{z}dz=\int_{\pi}^{-\pi}e^{-i\theta}ie^{i\theta}d\theta=\int_{\pi}^{-\pi}id\theta=0$
So what does the question mean by find all the values of $\int_{\gamma}\frac{1}{z}$ ?
It seems to me there is only one ?
Given a Contour integral $\int_{\gamma}\frac{1}{z}dz$ where $\gamma$ is the semicircle going from $1$ to $-1$, then there two arcs which we can integrate over.
There is the lower semi-circle, where we go from $1$ to $-1$ in a clockwise direction, and also the upper semi-circle , where we go from $1$ to $-1$ in an anti-clockwise direction .
With that in mind there are two semi-circles we can choose from;
1) moving in a clockwise direction in which case $\gamma$ can be parametrised as $e^{-i\theta}, \theta \in[0,\pi]$
So here we have $\int_{\gamma}\frac{1}{z}dz=\int_{0}^{\pi}\frac{-ie^{-i\theta }d\theta}{e^{-i\theta}}$, as z=$e^{-i\theta}$ when moving counter-clockwise
$\int_{0}^{\pi}\frac{-ie^{-i\theta }d\theta}{e^{-i\theta}}=-i\int_{0}^{\pi}d\theta=-i[\theta]^{\pi}_{0}=-i\pi$
2) moving in a counter clockwise direction , in which case we can parametrise $\gamma$ as $e^{i\theta}, \theta \in [0,\pi]$
So for this we have $\int_{\gamma}\frac{1}{z}dz=\int_{0}^{\pi}\frac{ie^{i\theta }d\theta}{e^{i\theta}}$, as z=$e^{i\theta}$
$\int_{0}^{\pi}\frac{ie^{i\theta }d\theta}{e^{i\theta}}=i\int_{0}^{\pi}d\theta=i[\theta]^{\pi}_0=\pi i$