Integrating $z^i$ over the unit circle

Question

Why is there a difference between integrating it over a unit circle parametrized over $t \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ and $t \in \left[-\frac{\pi}{2}, \frac{3\pi}{2}\right]$?

Answer 1

The difference comes from $e^{i\theta}$ giving different signs of $\pm i$ for the different angles.

The first approach I would always take on these problems is by looking for branch cuts in the complex logarithm by doing the sneaky move of replacing as,

$$ z^i = e^{i\ln{z}} = e^{i\ln{re^{i\theta}}} $$ But $r=1$ on the unit circle,

$$ e^{i\ln{e^{i\theta}}} $$

Now at this point I'll link to an explanation on branch cuts (by someone related to me!) with relevance to the complex phase $e^{i\phi}$, which should give you an awareness of some of the concepts as you have a complex phase with a phase angle of $\phi=\ln{e^{i\theta}}$

Also note that $\ln(\pm i)=\pm \frac{i\pi}{2}$

As it turned out your phase wasn't actually even complex as all the values you provide give a nice $\phi = \ln{(\pm i)}$ so you will obtain something like $e^{\mp \pi /2}$ where the difference is due to my opening assertion