# Conditions required for (z_(1)z_(2))^(omega)=z_(1)^(omega)z_{2}^(omega), where z_(1),z_(2),omega in bbb(C) I am having trouble finding the conditions on z_1 and z_2 in order for:

Conditions required for $\left({z}_{1}{z}_{2}{\right)}^{\omega }={z}_{1}^{\omega }{z}_{2}^{\omega }$, where ${z}_{1},{z}_{2},\omega \in \mathbb{C}$
I am having trouble finding the conditions on ${z}_{1}$ and ${z}_{2}$ in order for:
$\left({z}_{1}{z}_{2}{\right)}^{\omega }\equiv {z}_{1}^{\omega }{z}_{2}^{\omega }\phantom{\rule{2em}{0ex}}\mathrm{\forall }\omega \in \mathbb{C}$
My first step was to rewrite the equation as:
${e}^{\omega \mathrm{Log}\left({z}_{1}{z}_{2}\right)}={e}^{\omega \mathrm{Log}\left({z}_{1}\right)}{e}^{\omega \mathrm{Log}\left({z}_{2}\right)}$
We have that:
$\mathrm{Log}\left({z}_{1}{z}_{2}\right)\equiv \mathrm{Log}\left({z}_{1}\right)+\mathrm{Log}\left({z}_{2}\right)+2ni\pi ,\phantom{\rule{2em}{0ex}}n\in \mathbb{Z}$
And therefore, equating both sides of the equation, we get:
$\mathrm{Log}\left({z}_{1}\right)+\mathrm{Log}\left({z}_{2}\right)+2ni\pi =\mathrm{Log}\left({z}_{1}\right)+\mathrm{Log}\left({z}_{2}\right)$
However, this appears to me to be true $\mathrm{\forall }{z}_{1},{z}_{2}\in \mathbb{C}$, yet if we set ${z}_{1}={z}_{2}=-1$ and $\omega =-i$, then the equality does not hold, so I'm not sure where I'm making my error.
I believe that the condition required is when $\left(\mathrm{\Re }\left(z\right)\notin {\mathbb{R}}_{-}\wedge \mathrm{\Im }\left(z\right)=0\right)\vee \mathrm{\Im }\left(z\right)\ne 0$ because this is the point at which $\mathrm{Log}\left(z\right)$ is discontinuous, however, I am unable to prove that this is the case.
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devilvunga
"I am having trouble finding the conditions on ${z}_{1}$ and ${z}_{2}$ in order for:
$\left({z}_{1}{z}_{2}{\right)}^{\omega }\equiv {z}_{1}^{\omega }{z}_{2}^{\omega }\phantom{\rule{2em}{0ex}}\mathrm{\forall }\omega \in \mathbb{C}$"
This question makes no sense as formulated since $\omega ↦{z}^{\omega }$ does not define a function on $\mathbb{C}$ except for some very specific values of $z$. Fortunately, the question is soon reformulated as:
"Find some conditions on ${z}_{1}$ and ${z}_{2}$ to ensure that, for every $\omega$ in $\mathbb{C}$,
${e}^{\omega \mathrm{Log}\left({z}_{1}{z}_{2}\right)}={e}^{\omega \mathrm{Log}\left({z}_{1}\right)}{e}^{\omega \mathrm{Log}\left({z}_{2}\right)}$"
and this new version does make sense. Thus, one asks that, for every $\omega$ in $\mathbb{C}$
$\omega \mathrm{Log}\left({z}_{1}{z}_{2}\right)-\omega \mathrm{Log}\left({z}_{1}\right)-\omega \mathrm{Log}\left({z}_{2}\right)$
is an integer multiple of $2\mathrm{i}\pi$. Define a function $n$ on $\mathbb{C}$ by
$n\left(\omega \right)=\frac{\omega \mathrm{Log}\left({z}_{1}{z}_{2}\right)-\omega \mathrm{Log}\left({z}_{1}\right)-\omega \mathrm{Log}\left({z}_{2}\right)}{2\mathrm{i}\pi }.$
Then the function $n$ is continuous, integer valued, and $n\left(0\right)=0$. This shows that $n\left(\omega \right)=0$ for every $\omega$, for example, $n\left(1\right)=0$, that is,
$\mathrm{Log}\left({z}_{1}{z}_{2}\right)=\mathrm{Log}\left({z}_{1}\right)+\mathrm{Log}\left({z}_{2}\right).$
This necessary condition is obviously sufficient. It holds if and only if
$-\pi <\mathrm{Arg}\left({z}_{1}\right)+\mathrm{Arg}\left({z}_{2}\right)⩽\pi .$