I am looking to find a counterexample for the equation $e^{z_{1}z_{2}}=(e^{z_{1}})^{z_{2}}$. Obviously this is wrong due to the choice of branch of multivalued complex function can be different. However, I am troubled in finding one counterexample for it. Choosing $z_{1}=z_{2}$ a same complex number seems easier, but which number can be chosen?
By the way I have tried $z_{1}=z_{2}=\text{Log}(i)$.
EDITTED after Joe's comment:
Defining $z^w$ where $z$ and $w$ are both complex numbers.
\begin{equation} z^w=e^{w\text{Log}(z)}=e^{w(\text{Log}\vert z\vert+i\text{Arg}(z))} \end{equation} Here, we choose the principal branch, and the principal argument is $(-\pi,\pi]$.
For all $z_1,z_2\in\mathbb C$, $\DeclareMathOperator{\Log}{Log}$ \begin{align} \exp(z_1z_2)=\exp(z_1)^{z_2}&\iff \exp(z_1z_2)=\exp(z_2\Log(\exp(z_1)) \\[5pt] &\iff \exists n\in\mathbb Z: z_1z_2=z_2\Log(\exp(z_1))+2\pi in \, . \end{align} Hence, if $\Log(\exp(z_1))=z_1$, then it is true that $\exp(z_1z_2)=\exp(z_1)^{z_2}$. So to find a counter-example, we need to find a $z_1$ for which this does not hold. For instance, if we let $z_1=2\pi i$, then \begin{align} \exp(z_1z_2)=\exp(z_1)^{z_2}&\iff \exists n\in \mathbb Z:(2\pi i)z_2=2\pi i n \\[5pt] &\iff z_2\in\mathbb Z \, . \end{align} Now let $z_2$ be a non-integer. Then, $\exp(z_1z_2)\neq\exp(z_1)^{z_2}$.