TITLE: What is the value of $\ln \left(e^{2 \pi i}\right)$
QUESTION [6 upvotes]: I know that $$e^{2 \pi  i} = 1$$
so by taking the natural logarithm on both sides
$$\ln \left(e^{2 \pi  i}\right)=\ln (1)=0$$
however, why isn't this $2 \pi i$ as expected? Is it beacuse logarithms can only provide real values?

REPLY [8 votes]: The problem is that in the complex numbers, the exponential function is no longer one-to-one, since $\exp(z + 2\pi i) = \exp z$. Thus it's no longer possible to define a proper inverse for it: if all I tell you about a number is its exponential, you don't know how many multiples of $2\pi i$ I've added.
There are a few possible fixes to this. One is to consider the logarithm as a multi-valued function, or more correctly a relation, so that $\log 1 "=" 2\pi i k$ for every integer $k$. This is sometimes convenient e.g. when solving equations.
Another is to define not the logarithm but a logarithm, that is to pick an arbitrary choice of imaginary part to get a well-defined function you can use. You may, for example, declare you're interested only in value with imaginary part between $-\pi$ and $\pi$ (let's say exclusive on the lower bound and inclusive on the upper) or $0$ and $2\pi$, or anything else you like. Sadly, this forces your logarithm function to be discontinuous, but if you slice a line out of the complex plane from $0$ down the negative reals, you can at least make the logarithm continuous everywhere other than on that line – you'll have something like imaginary part $-\pi$ on one side of the line and imaginary part $\pi$ on the other side. This is called a "branch cut" and you are choosing a "branch" of the logarithm.