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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.

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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 $f$.

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