$\log(z_1z_2)=\log(z_1)+\log(z_2)$ where $z_1,z_2\in \mathbb{C}$\{0} 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$$\log(z_1z_2)$
This is wclearly wrong I guess but this is how I tried to approach it. Would aprreciate any answers.Thanks a lot
 A: Right away your first line is not correct, because $$\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,$$ where $k_1, k_2 \in \mathbb Z$; that is, the branch of the two logs do not necessarily come from the same value of $k$.
A: It is true that 
$$\log(z_1z_2)=\log(z_1)+\log(z_2) \tag1$$
when $(1)$ is interpreted as a set equivalence.  
This means that any value of $\log(z_1z_2)$ can be expressed as the sum of some value of $\log(z_1)$ and some value of $\log(z_2)$.  And conversely, it means that the sum of any value of $\log(z_1)$ and any value of $\log(z_2)$ can be expressed as some value of $\log(z_1z_2)$.
Note that $(1)$ is true since $\log(|z_1z_2|)=\log(|z_1|)+\log(|z_2|)$ and $\arg(z_1z_2)=\arg(z_1)+\arg(z_2)$. 
To see that $\arg(z_1z_2)=\arg(z_1)+\arg(z_2)$, we write $z_1 = |z_1| e^{i\phi_1}$ and $z_2=|z_2|e^{i\phi_2}$.  Clearly, $z_1z_2=|z_1||z_2|e^{i(\phi_1+\phi_2)}$.  So any argument of $z_1$ plus any argument of $z_2$ is an argument of $z_1z_2$.  On the other hand consider any argument of $z_1z_2$, which is of the form $\phi_1+\phi_2+2n\pi$ for some integer $n$.  We could simply take $\arg(z_1)=\phi_1$ and $\arg(z_2)=\phi_2+2n\pi$ and have the stated equality hold.


NOTE: It is important to understand that $(1)$ does not hold in general if $\log(z)$ is taken on the Principal branch of the complex logarithm (or any other designated branch since then we lose a degree of freedom).


As an example, take $z_1=i$ and $z_2=-i$.  Then, if we choose $\log(i)=i\pi/2$ and $\log(-i)=-i\pi/2$, then we must choose the branch of $\log(z)$ for which $\log(1)=0$.  If on the other hand we choose $\log(i)=-i3\pi/2$ and $\log(-i)=-i\pi/2$, then we must choose the branch of $\log(z)$ for which $\log(1)=-i2\pi$.
Conversely, if we choose the branch of $\log(z)$ for which $\log(1)=0$, then for $\log(i)=-i3\pi/2$, we must have $\log(-i)=-i\pi/2$.
