I have the following question, I really appreciate if someone can help me to clarify ideas and I apologize if is a stupid question: This is from Conway's complex analysis book:

Let $f: G \to \bf{C}$ and $g: G\to \bf{C}$ 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}$.

Maybe is a very stupid question but I have a bad time trying to understand. As long as I know $z^b = \exp(b \ell(z))$ where $\ell$ is a branch of the logarithm in some region $G$. Therefore if we fixed a branch $L(z)$ for the logarithm in the region we have $\ell(z)=L(z)+i2\pi k$ ($k\in \bf{Z}$ and fixed).

My idea goes as follows: Let $L(z)$ a branch of the logarithm in $G$. Then $f(z)= \exp(a(L(z)+i2\pi m))$ and $g(z)= \exp(b(L(z)+i2\pi n))$ ($m,n \in \bf{Z}$). Thus

$$f(z)\cdot g(z)= \exp(a (L(z)+i2\pi m))+b(L(z)+i2\pi n))$$ $$f(z)\cdot g(z)=\exp((a+b)L(z))\cdot \exp(i2\pi (am+bn))$$

Does this really defined a legitimate branch of $z^{a+b}$? I think the answer is no, because $a,b$ can be anything, not just integers, and for what I understand we need $\exp[(a+b)[L(z)+i2\pi k]]$ ($k\in \bf{Z}$) in order to be a branch for $z^{a+b}$.

For the other cases I have the same difficulties, at least that $f$ and $g$ are defined in the same branch of the logarithm, that is $m=n$, this seems to solve the difficulties but as written the statement does not mentioned that the branches are the same. I really need someone to help me to clarify ideas.

I have more question of the same sort but I think is not a good idea to put them all in one post.

Thanks in advance.

  • $\begingroup$ You are correct. $f$ and $g$ must use one definition of $\log$. $\endgroup$ – robjohn Mar 12 '15 at 20:19
  • $\begingroup$ @robjohn Thank you :) $\endgroup$ – Jose Antonio Mar 12 '15 at 20:51

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