# Logarithm of products in $\mathbb{C}$

For arbitrary $z_{1}, z_{2} \in \mathbb{C}$ $\log(z_{1}z_{2}) \neq \log(z_{1}) + \log(z_{2})$. What is the largest subset of $\mathbb{C}$ such that we do have equality? Is it just $\mathbb{R}$?

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For questions like this it is often helpful to write $z_1 = r_1e^{i\theta_1}$ and $z_2=r_2e^{i\theta_2}$. Then
$$z_1 z_2=r_1r_2e^{i(\theta_1+\theta_2)}$$
Now we bring in our requirement of equality \begin{align}\operatorname{Log}(z_1z_2)=&\operatorname{Log}(z_1)+\operatorname{Log}(z_2)\\ \log(r_1r_2)+\Theta i=&\log(r_1)+\theta_1i+\log(r_2)+\theta_2 i\\ \Theta=&\theta_1+\theta_2 \end{align} Where $\Theta=\theta_1+\theta_2 \pm 2\pi$ as necessary to keep it within $(-\pi,\pi].$ The answer should now be clear, the subset of $\mathbb C$ where this equality will hold is $\{re^{i\theta}:\theta\in(-\frac \pi 2,\frac \pi 2]\}\subset\mathbb C$, since within this subset we will not need to add or subtract $2\pi$. Any larger subset would lead to a counterexample.
Also note that whatever your set may be, for the logarithm to have an analytic determination, $z$ must be in a simply connected domain and have $f(z)$ non-vanishing there (though I imagine you could patch together a multiply connected domain with agreeing branches, but bounded by the above conditions.