Finding branches of $z^{a+b}, z^{a-b}, z^{ab}$ This is a problem from Chapter III, Section 2, of John B. Conway's book Functions of One Complex Variable I.
Let $f:G\to \mathbb{C}$ and $g:G\to\mathbb{C}$ be branches of $z^a$ and $z^b$ respectively. Show that $fg$ is a branch of $z^{a+b}$ and $f/g$ is a branch  of $z^{a-b}$. Suppose that $f(G)\subset G$ and $g(G)\subset G$ and prove that both $f\circ g$ and $g\circ f$ are branches of $z^{ab}$.
I have two possible solutions. I would like if one of the two is correct. Or nothing ...
Solution 1. If $f$ branch of $z^a$ then $f(z)=\exp(a{f'}(z))$ with $f':G\to\mathbb{C}$ branch of logarithm on $G$ (This definition I'm not sure I understand it well
). Analogously $g(z)=\exp(b {g'}(z))$ with $g':G\to\mathbb{C}$ branch of logarithm on $G$.
Let $\log:G\to\mathbb{C}$ a branch of logarithm on $G$.
ByProposition 2.19 Conway, $f'(z)=\log(z)+2\pi k_1 i$ and $g'(z)=\log(z)+2\pi k_2 i$ for some $k_1,k_2\in\mathbb{Z}$.
Therefore
\begin{align*} f(z)g(z)&=\exp(a{f'}(z))\exp(b{g'}(z))\\
 &=\exp((af'(z)+bg'(z))\\
 &=\exp((a(\log(z)+2\pi k_1 i+b(\log(z)+2\pi k_2 i))\\
 &=\exp( (a+b)\log(z)+2\pi(ak_1+bk_2)i) \\
 &=\exp((a+b)\log(z))\exp(2\pi (ak_1+bk_2) i)\\
 &=z^{a+b}\exp(2\pi (ak_1+bk_2)i)\\
 &=z^{a+b}
 \end{align*}
Solution 2. If $f$ branch of  $z^a$ then  $f(z)=\exp(a{f'}(z))$ with $f':G\to\mathbb{C}$ a branch of logarithm on $G$. Analogously $g(z)=\exp(b {g'}(z))$ with  $g':G\to\mathbb{C}$ a branch of logarithm on $G$.
Let $\log:G\to\mathbb{C}$ a branch of logarithm on $G$.
By Proposition 2.19  Conway, $f'(z)=\log(z)+2\pi k_1 i$ and $g'(z)=\log(z)+2\pi k_2 i$ for some $k_1,k_2\in\mathbb{Z}$.
So
\begin{align*} f(z)g(z)&=\exp(a{f'}(z))\exp(b{g'}(z))\\
 &=\exp((af'(z)+bg'(z))\\
 &=\exp((a(\log(z)+2\pi k_1) i+b(\log(z)+2\pi k_2 i))\\
 &=\exp( (a+b)\log(z)+2\pi(ak_1+bk_2)i) \\
 &=\exp((a+b)\log(z))\exp(2\pi (ak_1+bk_2) i)\\
 &=z^{a+b}\exp(2\pi (ak_1+bk_2)i)\\
 &=z^{a+b}
 \end{align*}
Therefore $fg$ is a branch of $z^{a+b}$ on $G$.
(I use $\exp(2\pi (ak_1+bk_2)i)=1$ but this is true with a,b integers...
** Apparently, the statement should say that f and g have the same logarithm. With this additional assumption, we have **
Solution 3.  If $f$ is a branch of  $z^a$ then $f(z)=\exp(ah(z))$ con $h:G\to\mathbb{C}$ a branch of logarithm in $G$. Analogously $g(z)=\exp(b h(z))$\
So \begin{align*} f(z)g(z)&=\exp(ah(z))\exp(bh(z))\\
 &=\exp((a h(z)+b h(z))\\
 &=\exp( (a+b)h(z)) \\
  &=z^{a+b}
 \end{align*}
Therefore $fg$ is a branch of $z^{a+b}$ in $G$.
Pd1: Is true that $\exp( (a+b)h(z))=z^{a+b}$? I know that \exp((a+b)Log(z))=z with Log principal branch of logaritm but I do not know if with another branch of logarithm it continues to be fulfilled immediately.
pd2: If $h:G\to \mathbb{C}$ branch of logarithm in $G$ then $h(\exp(z))=z$? I know $\exp(h(z))=z$ but vice versa I don't know if it is true immediately.
 A: I believe Conway erred in the statement of this problem.
I'm sure someone will correct me if I'm wrong, but for $z^{a \pm b}$, he should have required $f$ and $g$ to be the same branch of $z^a$ and
$z^b$, resp.  For the composition, he should have stuck to just $g \circ f$ and required $g$ to be the principal branch of $z^b$
(or a modification thereof with a different cut).
Here is the argument:
Let $\log$ be the principal branch of the logarithm (or a modification thereof with a different cut).
$\log f = a(\log z + 2\pi i k_1). \log g = b(\log z + 2\pi i k_2). \log z^{a+b} = (a+b)(\log z + 2\pi i k_3)$.
Taking logs in the equality $fg = z^{a+b}$, we see that we must have $a k_1 + b k_2 = (a+b) k_3$.
If this is going to work for arbitrary $a$ and $b$, it is easy to see that this means that $k_1 = k_2 = k_3$.
In summary, it is necessary and sufficient to use the same branch for everything.
$\log z^{ab} = ab(\log z + 2\pi i k_4)$.
$\log(g \circ f) = b(\log f + 2\pi i k_2) = b(a(\log z + 2\pi i k_1)) + 2\pi i k_2)$.
Taking logs in the equality $g \circ f = z^{ab}$, we see that we must have $ab k_1 + b k_2 = ab k_4$.
If this is going to work for arbitrary $a$ and $b$, it is easy to see that this means that $k_2 = 0$ and $k_1 = k_4$.
In summary, it is necessary and sufficient to use a modified principal branch for $g$ and the same branch for $z^{ab}$ as for $f$.
