TITLE: Log of a negative number
QUESTION [17 upvotes]: Why is the log of $x$ equal to the log of the absolute value of $x$ plus $i$ times $\pi$?
$$\log(x)=\log(|x|)+i\pi\text{, for }x < 0$$
Where does the $\pi$ come from? Is it from a logarithmic identity?
I know it sounds silly, but I was not able to find an answer from existing solutions: here, here, and here.

REPLY [7 votes]: The standard explanation would go like this:

Any complex number $z$ can be written in polar form:
  $$z = |z|e^{i\theta}.$$
  So, if we want to define the complex logarithm, we do so as follows:
  $$\log (z) = \log(|z|e^{i\theta}) = \log(|z|) + \log(e^{i\theta}) = \log(|z|) + i\theta.$$
  In particular, the logarithm of a negative real number $x$ can then be calculated as
  $$\log (x) = \log(|x|e^{i\pi}) = \log(|x|) + \log(e^{i\pi}) = \log(|x|) + i\pi.$$

However, this explanation is not sufficient and the logarithm as presented is NOT a well-defined function. The angle $\theta$ is, well, an angle, and hence only defined up to multiples of $2\pi$: 
$$|z|e^{i\theta} = |z|e^{i(\theta+k2\pi)}$$
for all $k \in \mathbb{Z}$. Therefore, the complex logarithm is only defined up to multiples of $2\pi i$ !
For example:
$$\log (x) = \log(|x|e^{i\frac{3}{2}\pi}) = \log(|x|) + \log(e^{i\frac{3}{2}\pi}) = \log(|x|) + i\frac{3}{2}\pi,$$
or maybe 
$$\log (x) = \log(|x|e^{-i\frac{177}{2}\pi}) = \log(|x|) + \log(e^{-i\frac{177}{2}\pi}) = \log(|x|) - i\frac{177}{2}\pi.$$
But clearly $\pi \neq \frac{3}{2}\pi \neq -\frac{177}{2}\pi$. The point is: the complex logarithm is not a function, but what we call a multivalued function. To turn it into a proper function, we must restrict what $\theta$ is allowed to be, for example $\theta \in (-\pi,\pi]$. This is called the principal complex logarithm and is usually denoted by $\operatorname{Log}$ (capital L).
Technically, it doesn't matter to what range you restrict $\theta$, as long as the resulting logarithm is a proper function (not a multivalued function) and you are consistent in your restriction of $\theta$. For example, the following "proof" can be obtained if you're sloppy:
\begin{align}
e^{\pi i} = -1 & \implies (e^{\pi i})^2 = (-1)^2 & \text{ (square both sides)}\\
& \implies e^{2\pi i} = 1 & \text{ (calculate the squares)}\\
& \implies \log (e^{2\pi i}) = \log(1) & \text{ (take the logarithm)}\\
& \implies 2\pi i  = 0 & \text{ (calculate the logarithms)}
\end{align}
Clearly this is wrong!