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!
Best Answer
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.