# 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.

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).
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 $$f$$.