Complex exponentiation identity $a^{z}b^{z} = (ab)^z$ for $a, b > 0, z\in\mathbb{C}$

Question

This is likely a trivial question, but I can only recall a list of general failures for the complex exponentiation rules from by CA class and not any "here's a list of the 'regular' rules that work in these and these settings". In any case, in which situation does $a^zb^z = (ab)^z$ hold for general $z\in \mathbb{C}$ and $a, b > 0$ two positive real numbers? If I am not mistaken, the definition $z^w = e^{w\mathrm{log}(z)}$ for $\mathrm{log}(z) = \mathrm{ln}|z| + i(\mathrm{Arg}(z) + 2\pi k), k\in\mathbb{Z}$ gives:

$$a^zb^z = \exp(z\mathrm{log}(a))\exp(z\mathrm{log}(b))\Leftrightarrow$$

$$a^zb^z = \exp(z(\ln|a| + 2\pi i k_1)\exp(z(\ln|b| + 2\pi i k_2)\Leftrightarrow$$

$$a^zb^z = \exp(z\ln(ab) + 2\pi i (k_1+k_2)z)\Leftrightarrow$$

$$a^zb^z = (ab)^ze^{z2\pi i (k_1 + k_2)}$$

for $k_1, k_2\in\mathbb{Z}$. And I am not really sure how you can get rid of the last $e^{z2\pi i (k_1 + k_2)}$ term.

Answer 1

It is a matter of (common) convention. We can read for instance in Complex Analysis by L. Ahlfors:

(L. Ahlfors): By convention the logarithm of a positive number shall always mean the real logarithm, unless the contrary is stated. The symbol $a^b$, where $a$ and $b$ are arbitrary complex numbers except for the the condition $a\ne 0$, is always interpreted as an equivalent of $\exp(b\log a)$. If $a$ is restricted to positive numbers, $\log a$ shall be real, and $a^b$ has a single value. Otherwise $\log a$ is the complex logarithm, and $a^b$ has in general infinitely many values which differ by factors $e^{2\pi i nb}$.