Natural log operation on $e^{i \tau}$

Question

I have a layman's question:

$e^{i\pi} = -1$ ... euler's identity

and $e^{i\tau}$ = 1 where $\tau = 2\pi$

so if I assume that $e^x=y$ implies $ln(y)=x$, which it should, then

$ln(1) = i\tau$

but

$ln(1) = 0 \ne i\tau$

What am I doing wrong?

Answer 1

For complex values, $\log$ is a multi-branched function. In order to get a well-defined function, you have to first identify which branch of $\log$ you are working with. For example, one might choose to define $\log$ as

$$\log (r e^{i \theta}) = \ln r + i \theta$$

where $- \pi < \theta < \pi$.

Answer 2

It is incorrect to say $$e^x=y \implies \ln y = x$$ The correct statement would be $$e^x=y \implies \ln y = x + 2k\pi \space \textrm{ for some } k\in\mathbb Z$$ (here the choice of $k$ would depend partly on the choice of the branch of the logarithm).