$\log(z_1z_2)=\log(z_1)+\log(z_2)$ where $z_1,z_2\in \mathbb{C}$\{0}

Question

I need to prove the set identity of the complex logarithm $\log(z_1z_2)=\log(z_1)+\log(z_2)$ where $z_1,z_2\in \mathbb{C}$.

$\log(z_1)+\log(z_2)=${$\log|z_1|+\log|z_2|+i(\text{Arg}(z_1)+\text{Arg}(z_2)+4k\pi i|k\in \mathbb{Z}$}$\implies$

{$\log|z_1z_2|+i((\text{Arg}(z_1)+2k\pi i)+(\text{Arg}(z_2)+2k\pi i)|k\in \mathbb{Z}$}$\implies$

{$\log|z_1z_2|+i(\text{arg}(z_1)+\text{arg}(z_2))|k\in \mathbb{Z}$}$\implies$

{$\log|z_1z_2|+i(\text{Arg}(z_1z_2)+2k\pi i)|k\in \mathbb{Z}$}$\implies$$\log(z_1z_2)$

This is wclearly wrong I guess but this is how I tried to approach it. Would aprreciate any answers.Thanks a lot

Answer 1

Right away your first line is not correct, because $$\log z_1 + \log z_2 = \log |z_1| + \log |z_2| + i(\operatorname{Arg} z_1 + \operatorname{Arg} z_2) + 2\pi i k_1 + 2 \pi i k_2,$$ where $k_1, k_2 \in \mathbb Z$; that is, the branch of the two logs do not necessarily come from the same value of $k$.