Logarithm of $e^{z}$ For a complex number $z$, I came across a statement that $\ln(e^{z})$ is not always equal to $z$. Why is this true?
Thanks for the help.
 A: Every non-zero complex number has infinitely many logarithms, because if $e^z=w$, then $(\forall n\in\mathbb{Z}):e^{z+2\pi i}=w$. So, every number of the form $z+2n\pi i$, with $n\in\mathbb Z$, is a logarithm of $e^z$.
A: For the same reason that the square root of $x^2$ is not always $x$ for real values of $x$. You have two choices: define the square root to mean "all the square roots" (in which case $x$ is an element of the set of square roots) or pick one. 
Same deal with $\log$: because $e^{2\pi i} = 1$, you have $e^z = e^{z + 2n \pi i}$ for any integer $n$. So which should be the log? 
A: Hint: consider Euler's formula $e^{ix}=\cos x + i\sin x$ and what it means for the periodicity and thus invertibility of $e^z$.
A: This is so because $\ln (z)$ isn't a one-one function in argand plane. 
Since, every complex number, when rotated by $2\pi$ radians, results in the same complex number, that is each complex number $z$ can be expressed as $$z=re^{i(\theta +2k\pi)} \;, k \in \Bbb Z$$
This is the reason that $e^{2\pi i} =1$, but $\color{red}{2\pi i \neq 0}$.
