In a problem provided in my course (the assignment was due a while ago, this isn't a homework help request), we needed to show : $$\log(z_1z_2) = \log(z_1) + \log(z_2) + i2n\pi $$
I seem to understand the analytic branch of complex log, defined:
$$\log(z) = \ln|z| + i\arg(\theta)$$ where ln is the real natural log and $\arg(\theta) \in (-\pi, \pi)$.
In my professor's worked solution, he claims that this equality holds because : $$e^{\log(z_1z_2)} = z_1z_2 = e^{\log(z_1)}e^{\log(z_2)} =e^{\log(z_1)+\log(z_2)} $$ and because $e^{i2n\pi} = 1$ we get that the equality $\log(z_1z_2) = \log(z_1) + \log(z_2) + i2n\pi $ holds.
I proved the problem in a different way, but want to see how this method is valid. I understand you can always multiply by 1 on the rightmost equation by multiplying by $e^{i2n\pi}$, but why does that imply that the equality holds? To me, it only implies that $z_1z_2=z_1+z_2+i2n\pi$ by treating the analytic log as the inverse of the exponential function, but this is not necessarily correct. Any help in understanding this method would be much appreciated!
We have $e^{\log(z_1z_2)} = e^{\log(z_1)+\log(z_2)}$. That implies, $$ e^{\log(z_1z_2)-\log(z_1)-\log(z_2)} = 1$$ and the solution of $e^{\zeta} = 1$ is simply $z = 2n\pi i$. The result follows. There is no further need to treat the $\log$ as an inverse.