One of the parts of the question I'm working on goes something like this:
Express $z^i = \exp(i \log_I(z))$ in the form $u+iv$, where $u,v$ are real-valued functions, and the log is defined on the principal branch.
For some reason, I'm completely confused. I've seen other examples where we write $z=x+iy$, but if we do that here we get $\exp(i \log_I(x+iy))$. We know $\log_I(z)=\log(|z|)+i\arg_I(z)$, so we'd have $\log_I(x+iy)=\log(\sqrt{x^2+y^2})+i\arctan(y/x)$, as long as $\arctan(y/x)$ falls in $I=(-\pi,\pi]$.
Thus, we'd have $z^i=\exp(i (\log(\sqrt{x^2+y^2})+i\arctan(y/x)))$, but I'm not sure where to go from here/not sure if this is even on the right track.
I prefer hints over full answers. Thanks for your help.
Although $z^i$ is a multi-valued function, a branch of it is equal to $$ z^i=\exp (i\log z)=\exp\big(i\log\lvert z\rvert+i\log(z/\lvert z\rvert)\big) =\exp(-\vartheta)\big(\cos (\log \lvert z\rvert)+i\sin(\log \lvert z\rvert)\big), $$ where $$ i\vartheta=\log\left(\frac{z}{\lvert z\rvert}\right), $$ i.e., $$ \vartheta=\tan^{-1}(y/x), \,\, z=x+iy. $$