I'm not sure what I'm missing from the following proof for that $\mathrm{Log}(z)$ is one-to-one for all $z\neq 0$. Here $\mathrm{Log}$ denotes the principal branch of the complex logarithm
Proof: Suppose that for $z_1, z_2 \in \mathbb{C}\setminus\{0\}$ it holds that $\mathrm{Log}(z_1) = \mathrm{Log}(z_2)$. Then $\ln|z_1| + i\mathrm{Arg}(z_1) = \ln|z_2| + i\mathrm{Arg}(z_2) \implies \begin{cases}\mathrm{Re}(z_1)^2 + \mathrm{Im}(z_1)^2 = \mathrm{Re}(z_2)^2 + \mathrm{Im}(z_2)^2\\\mathrm{Re}(z_1)/\mathrm{Im}(z_1) = \mathrm{Re}(z_2)/\mathrm{Im}(z_2)\end{cases}$, by bijectivity of the $\exp$ and $\tan$ functions. Thus $\mathrm{Re}(z_1) = (\mathrm{Re}(z_2)\mathrm{Im}(z_1))/\mathrm{Im}(z_2)$, so that $\mathrm{Im}(z_1) = \pm \mathrm{Im}(z_2)$, so that $\mathrm{Re}(z_1) = \pm \mathrm{Re}(z_2)$.
So here's the problem: How can we conclude that the $-$ case in $\pm$ is not possible?
Best Answer
If $|z_1| = |z_2|$ and $\operatorname{Arg}(z_1) = \operatorname{Arg}(z_2)$ then $$ z_1 = |z_1| e^{i \operatorname{Arg}(z_1)} = |z_2| e^{i \operatorname{Arg}(z_2)} = z_2 \, . $$
Or simply: All branches of the logarithm satisfy $\exp(\log(z)) = z$, and that implies injectivity: $$ \operatorname{Log}(z_1) = \operatorname{Log}(z_2) \implies z_1 = \exp(\operatorname{Log}(z_1)) = \exp(\operatorname{Log}(z_2)) = z_2 \, . $$