Conditions required for $({z}_{1}{z}_{2}{)}^{\omega}={z}_{1}^{\omega}{z}_{2}^{\omega}$, 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:

$({z}_{1}{z}_{2}{)}^{\omega}\equiv {z}_{1}^{\omega}{z}_{2}^{\omega}\phantom{\rule{2em}{0ex}}\mathrm{\forall}\omega \in \mathbb{C}$

My first step was to rewrite the equation as:

${e}^{\omega \mathrm{Log}({z}_{1}{z}_{2})}={e}^{\omega \mathrm{Log}({z}_{1})}{e}^{\omega \mathrm{Log}({z}_{2})}$

We have that:

$\mathrm{Log}({z}_{1}{z}_{2})\equiv \mathrm{Log}({z}_{1})+\mathrm{Log}({z}_{2})+2ni\pi ,\phantom{\rule{2em}{0ex}}n\in \mathbb{Z}$

And therefore, equating both sides of the equation, we get:

$\mathrm{Log}({z}_{1})+\mathrm{Log}({z}_{2})+2ni\pi =\mathrm{Log}({z}_{1})+\mathrm{Log}({z}_{2})$

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 $(\mathrm{\Re}(z)\notin {\mathbb{R}}_{-}\wedge \mathrm{\Im}(z)=0)\vee \mathrm{\Im}(z)\ne 0$ because this is the point at which $\mathrm{Log}(z)$ is discontinuous, however, I am unable to prove that this is the case.

I am having trouble finding the conditions on ${z}_{1}$ and ${z}_{2}$ in order for:

$({z}_{1}{z}_{2}{)}^{\omega}\equiv {z}_{1}^{\omega}{z}_{2}^{\omega}\phantom{\rule{2em}{0ex}}\mathrm{\forall}\omega \in \mathbb{C}$

My first step was to rewrite the equation as:

${e}^{\omega \mathrm{Log}({z}_{1}{z}_{2})}={e}^{\omega \mathrm{Log}({z}_{1})}{e}^{\omega \mathrm{Log}({z}_{2})}$

We have that:

$\mathrm{Log}({z}_{1}{z}_{2})\equiv \mathrm{Log}({z}_{1})+\mathrm{Log}({z}_{2})+2ni\pi ,\phantom{\rule{2em}{0ex}}n\in \mathbb{Z}$

And therefore, equating both sides of the equation, we get:

$\mathrm{Log}({z}_{1})+\mathrm{Log}({z}_{2})+2ni\pi =\mathrm{Log}({z}_{1})+\mathrm{Log}({z}_{2})$

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 $(\mathrm{\Re}(z)\notin {\mathbb{R}}_{-}\wedge \mathrm{\Im}(z)=0)\vee \mathrm{\Im}(z)\ne 0$ because this is the point at which $\mathrm{Log}(z)$ is discontinuous, however, I am unable to prove that this is the case.