I'm looking at example 7.54 in the book by Ponnusamy, available publicly here, on page 240 (249 in the pdf viewer).
Example 7.54. Consider $$\int_{|z|=1} f(z)\ dz,\quad f(z)=1/z^{1/2}.$$ Then $z=0$ is a branch point of $f(z)$. If we choose principal branch, then $$z^{1/2}=e^{(1/2)\operatorname{Log}z}=e^{(1/2)(\ln|z|+i\operatorname{Arg} z)}$$ so that for $z=e^{i\theta}$, we have $z^{1/2}=e^{i(1/2)\operatorname{Arg}z}=e^{(1/2)i\theta}$ and $$\int_{-\pi}^\pi\frac{ie^{i\theta}}{e^{i\theta/2}}\ d\theta=i\int_{-\pi}^{\pi}e^{-i\theta/2}\ d\theta=4i.$$
In the above example, I do not understand two things:
1) Does $\frac{e^{i\theta}}{e^{i\theta/2}}$ reduce to $e^{-i\theta/2}$, generally?
2) Why are the integration limits $\pi$ and $-\pi$ and not $0$ and $2 \pi$? (as you'll notice, the limits of integration affect the result if the answer to 1) is "No".)
1) $ \frac{e^{2a}}{e^{a}}=e^{2a}e^{-a}=e^{2a-a}=e^a.$
2) A possible parametrization of $\{z: |z|=1\}$ is $z( \theta)=e^{i \theta}$ for $\theta \in [- \pi, \pi].$