If $f$ and $g$ are branches of $z^a$ and $z^b$ respectively show that $fg$ is a branch of $z^{a+b}$

Question

Suppose $f: G\rightarrow \mathbb{C}$ is a branch of $z^a$. Then $f(z) = e^{ah_1(z)}$ for all $z\in G$. Likewise if $g:G\rightarrow \mathbb{C}$ is a branch of $z^b$ then $g(z) = e^{bh_2(z)}$ for all $z\in G$.

Here $h_1$ and $h_2$ are branches of the logarithm on $G$, and hence they differ by $2\pi k i$ with $k\in \mathbb{Z}$.

However then $fg(z) = f(z)g(z) = e^{ah_1}\cdot e^{b(h_1+2\pi k i)}= e^{(a+b)h_1+2\pi k b i}$, which is not a branch of $z^{a+b}$.

Question: Should I be using the same branch of $\log$ when defining the branches of $z^a$ and $z^b$. That is, should I write $g(z) = e^{bh_1(z)}$, beacause then everything works out. Also, why should I be choosing the same $\log$. Thanks.

Answer 1

The functions $z^a, z^b$ are multi-valued (if $a$ or $b$ are not integers), and in general $z^{a+b}$ is a subset of $z^a \cdot z^b$. Working with a single branch of the logarithm (say $\mathrm{Log}(r e^{i\theta}) = \log r + i \theta$), and defining the branch of $z^a$ to be $\exp(a \cdot \mathrm{Log} (z))$ for all complex $a \in \mathbb{C}$ allows you to preserve some of the familiar properties of the power function (which is ultimately the goal). For example, you still have $z^a z^b = z^{a+b}$, but you no longer have $(z_1 z_2)^a = z_1^a \cdot z_2^a$.