Let $f: G \to \bf{C}$ and $g: G\to \bf{C}$ be branches of $z^a$ and $z^b$ respectively ($a,b\in \Bbb C$). Suppose that $f(G) \subset G$ and $g(G) \subset G$. Prove that both $f\circ g$ and $g \circ f$ are branches of $ z^{ab}$.
My attempt is that $f(z)=e^{af_1(z)}$ and $g(z)=e^{bg_1(z)}$, where $f_1$ and $g_1$ are branches of the logarithm on $G$, i.e. $e^{f_1(z)}=z=e^{g_1(z)}$ on $G$. Now when I start computing $f(g(z))$, I am confused.
Now $$f(g(z))=f(e^{bg_1(z)})=e^{af_1(e^{bg_1(z)})}$$ and I know that $$e^{f_1(e^{g_1(z)})}=z=e^{g_1(e^{f_1(z)})},$$ but completely lost. I have searched in MSE but didn't find any appropriate answer to this. Any help/hint is appreciated.
Since $\exp(f_1(\exp(z))) = \exp(z)$, you have $f_1(\exp(z)) = z + 2ik_f(z)\pi$ where $k_f : G \to \Bbb Z$. If $G$ is connected then $k_f(z)$ is a constant $k$.
Then,
$\exp(af_1(\exp(bg_1(z)))) = \exp(a(bg_1(z)+2ik\pi)) = \exp(abg_1(z) + 2aik\pi)$
If $f\circ g$ was a branch cut of $z^{ab}$ you would have $f(g(z)) = \exp(ab h(z))$ where $ \exp(h(z)) = z$.
So you would have $abg_1(z) + 2aik\pi = abh(z) + 2ik'\pi$ with some other integer constant $k'$, and so $h(z) = g(z) + 2i\pi(k/b-k'/ab)$.
Hence after taking $\exp$ of both sides, $z = z\exp(2i\pi(k/b-k'/ab))$ and so $(k/b-k'/ab) = k''$ for some integer $k''$
Since there may not exist integers $k',k''$ such that this relation holds, $f\circ g$ may not be a branch cut of $z^{ab}$.
Which is why you should never write things like $z^a$ or $z^{ab}$