On complex powers of complex numbers

Question

For $z,s \in \mathbb{C}$ and $z\neq 0$, set $z^s = \exp(s\,\log z)$ and $-\pi < \arg z \leq \pi$. In this setting, I am worried about the cases where I have to be careful in assuming $(z_1z_2)^s = (z_1)^s(z_2)^s$. As long as $s \in \mathbb{Z}$, I am safe, but what about the rest? For example, if $z\in \mathbb{H}$, then \begin{equation} (-z)^s = e^{-\pi is}z^s, \end{equation} but \begin{equation} (-1)^sz^s = e^{\pi i s}z^s \end{equation} and both are not equal.

Answer 1

The rule $(zw)^s=z^sw^s$ depends on the equality $\log(zw)=\log z+\log w$. But you need the logarithm to be single valued, which requires limiting the argument to one interval, for instance $(-\pi,\pi]$ as you did; and this generates the following problem. If $t+u>\pi$, then $\log (e^{i(t+u)})=i(t+u-2\pi)$. Thus when $s=a+ib$, $z_j=r_je^{it_j}$, \begin{align} (z_1z_2)^s&=\exp((a+ib)(\log r_1r_2+ i(t_1+t_2-2k\pi))\\ \ \\ &=r_1^ar_2^a\,e^{ib\log r_1}\,e^{ib\log r_2}\,e^{iat_1}\,e^{iat_2}\,e^{-bt_1}\,e^{-bt_2}\,e^{2bk\pi}, \end{align} where $k=0$ if $t_1+t_2\leq\pi$, and $k=1$ when $t_1+t_2>\pi$. On the other hand you have $$ z_1^sz_2^s=r_1^ar_2^a\,e^{ib\log r_1}\,e^{ib\log r_2}\,e^{iat_1}\,e^{iat_2}\,e^{-bt_1}\,e^{-bt_2}. $$ The two expressions are equal when $t+u\leq\pi$, and differ by a factor of $e^{2b\pi}$ when $t+u>\pi$.

In summary, $(z_1z_2)^s=z_1^s+z_2^s$ (in your setup) when $\arg z_1+\arg z_2\leq\pi$. This is not terrible, but it prevents you from using the formula in general, when you don't know the arguments of $z_1$ and $z_2$.