Show this is a branch for $z^{ab}$. Let $f: G \to \mathbb{C}$ and $g: G \to \mathbb{C}$ branches of $z^a$ resp. $z^b$. Suppose that $g(G) \subseteq G$ and $f(G) \subseteq G$. Show that 
$$f \circ g, g \circ f$$
are branches of $z^{ab}$.
Attempt:
Let $h: G \to \mathbb{C}$ be a branch of the logarithm such that $f(z) = \exp (a h(z))$ and $g(z) = \exp(b h(z))$ for all $z \in G$. Thus $$\forall z \in G: \exp (h(z)) = z$$
I then have to show that $\exp(ab h(z)) = f \circ g = g \circ f$.
$$f \circ g (z) = f (g(z)) = f(\exp(bh(z)) = \exp(ah(\exp(bh(z))))$$
and I'm not sure to proceed. Of course, one wants to use that
$$h(\exp(bh(z)) = bh(z)$$
but I'm not sure why this is true.
Any help wil be appreciated!
 A: Unless I am mistaken, the statement is not correct. Here is an example: Let $G = \Bbb C \setminus (-\infty, 0]$ be the complex plane without the negative real axis and $\log$ the principal branch of the logarithm on $G$. Then
$$
f(z) = e^{\frac 12 (\log z + 2\pi i)} \, , \, g(z) = e^{\frac 23 \log z}
$$
are holomorphic branches of $z^{1/2}$ and $z^{2/3}$, respectively. But
$$
 F(z) = f(g(z)) = e^{\frac 23 \log z + \pi i} = - e^{\frac 13 \log z}
$$
is not a holomorphic branch of $z^{1/3}$: 
$$
(F(z))^3 = -z^3 \ne z^3 \, .
$$
Therefore – unless the functions are somehow normalized, e.g. with $f(1) = g(1) = 1$ – we can conclude only that $f \circ g$ is a holomorphic branch of $z^{ab}$, multiplied with a constant  $c  = \exp(2 k \pi i a)$ for some integer $k$.
Note that we cannot assume that $f(z) = \exp (a h(z))$ and $g(z) = \exp(b h(z))$ with the same branch $h$ of the logarithm, only that
$$
f(z) = \exp (a h_1(z)) \, , \, g(z) = \exp(b h_2(z))
$$
where $h_1, h_2$ are both holomorphic branches of the logarithm in $G$. Then
$$ 
\exp(h_1(g(z)) = g(z) = \exp(b h_2(z))
$$
for all $z \in G$ and therefore
$$
h_1(g(z)) = b h_2(z) + 2k \pi i
$$
for some integer $k$. (Here I assume that $G$ is connected, otherwise the relation can hold with different $k$ in each component of $G$.)  Then
$$
 f(g(z)) = \exp(ab h_2(z) + 2k \pi i a) = c  \exp(ab h_2(z)) \, .
$$
