I am studying about the function $\log(z)$
where $\log(z)=\ln r+i\theta$ and as far as I've learnt, defining argument of $\theta$ is base on the given branch. But I am confused how for example for
for $3\pi/4 <\theta<11\pi/4$ :
$\log(-1)$ will have $\theta=\pi$
$\log(i^2)$ will have $\theta=5\pi/2$
honestly I dont understand how different periods affect $\theta$ and I will appreciate any help.
$\log z = \ln |z| + i\arg z$, where $\arg z$ is a suitably chosen argument for $z$. You apparently want the branch where $\frac{3\pi}{4} < \arg z < \frac{11\pi}{4}$.
For example, with this choice $\log(-1) = i(\pi + 2\pi k)$ where $k$ is an integer chosen so that $\frac{3\pi}{4} < \pi + 2\pi k < \frac{11\pi}{4}$, i.e. $k=0$ so that $\log(-1) = i\pi$.
Your second equality doesn't make sense: $\log(i^2) = \log(-1) = i\pi$ as well.