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I am having trouble finding the conditions on $z_{1}$ and $z_{2}$ in order for: $$(z_{1}z_{2})^{\omega}\equiv z_{1}^{\omega}z_{2}^{\omega}\qquad \forall\omega\in\mathbb{C}$$

My first step was to rewrite the equation as:

$$e^{\omega \operatorname{Log}(z_{1}z_{2})}=e^{\omega \operatorname{Log}(z_{1})}e^{\omega \operatorname{Log}(z_{2})}$$

We have that:

$$\operatorname{Log}(z_{1}z_{2})\equiv \operatorname{Log}(z_{1})+\operatorname{Log}(z_{2})+2ni\pi, \qquad n\in\mathbb{Z}$$

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

$$\operatorname{Log}(z_{1})+\operatorname{Log}(z_{2})+2ni\pi=\operatorname{Log}(z_{1})+\operatorname{Log}(z_{2})$$

However, this appears to me to be true $\forall z_{1},z_{2}\in\mathbb{C}$, 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 $(\Re(z)\not\in\mathbb{R}_{-}\land \Im(z)=0) \lor \Im(z)\neq 0$ because this is the point at which $\operatorname{Log}(z)$ is discontinuous, however, I am unable to prove that this is the case.

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I am having trouble finding the conditions on $z_{1}$ and $z_{2}$ in order for: $$(z_{1}z_{2})^{\omega}\equiv z_{1}^{\omega}z_{2}^{\omega}\qquad \forall\omega\in\mathbb{C}$$

This question makes no sense as formulated since $\omega\mapsto z^\omega$ 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 $\mathbb C$, $$e^{\omega \operatorname{Log}(z_{1}z_{2})}=e^{\omega \operatorname{Log}(z_{1})}e^{\omega \operatorname{Log}(z_{2})}$$

and this new version does make sense. Thus, one asks that, for every $\omega$ in $\mathbb C$, $$ \omega \operatorname{Log}(z_{1}z_{2})-\omega \operatorname{Log}(z_{1})-\omega \operatorname{Log}(z_{2}) $$ is an integer multiple of $2\mathrm i\pi$. Define a function $n$ on $\mathbb C$ by $$ n(\omega)=\frac{\omega \operatorname{Log}(z_{1}z_{2})-\omega \operatorname{Log}(z_{1})-\omega \operatorname{Log}(z_{2})}{2\mathrm i\pi}. $$ Then the function $n$ is continuous, integer valued, and $n(0)=0$. This shows that $n(\omega)=0$ for every $\omega$, for example, $n(1)=0$, that is, $$ \operatorname{Log}(z_{1}z_{2})=\operatorname{Log}(z_{1})+ \operatorname{Log}(z_{2}). $$ This necessary condition is obviously sufficient. It holds if and only if $$ -\pi\lt\operatorname{Arg}(z_{1})+ \operatorname{Arg}(z_{2})\leqslant\pi. $$

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  • $\begingroup$ Thank you, this makes sense; it didn't occur to me at first that $$-\pi<\operatorname{Arg}(z_1)+\operatorname{Arg}(z_2)\leq \pi$$ Was the way of expressing the requirement for $\operatorname{Log}(z_1 z_2) = \operatorname{Log}(z_1)+\operatorname{Log}(z_2)$. $\endgroup$ – Thomas Russell Apr 21 '14 at 16:18
  • $\begingroup$ Also, is it the case that $$\operatorname{Log}\left(\prod_{i=1}^{n}z_{i}\right)=\sum_{i=1}^{n} \operatorname{Log}(z_i)$$ Holds iff: $$-\pi < \sum_{i=1}^{n}\operatorname{Arg}(z_{i})\leq \pi?$$ $\endgroup$ – Thomas Russell Apr 21 '14 at 16:26
  • $\begingroup$ Yes. Same proof. $\endgroup$ – Did Apr 21 '14 at 16:48

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