TITLE: Why isn't $\log(-1)=i\pi$?
QUESTION [7 upvotes]: Reading http://people.math.gatech.edu/~cain/winter99/ch3.pdf, $\log(z)$ is defined as $=\ln|z|+i\arg(z)$. Looking on the Wessel plane, isn't $\arg(-1)=\pi$ (more generally $\pi \pm 2 \pi n$)? And $e^0=1$, so surely $\ln|-1|=0$, making $\log(-1)=0+i(\pi \pm 2 \pi n)$?
My problem is that apparently $\log(z)$ is not defined for $z=x+i0, x<0$, and yet there seems no good reason why it shouldn't be, at least in the case of $z=-1$.

REPLY [3 votes]: Asking what $\log(-1)$ is is something like asking what $\arcsin(1/2)$ is.  To satisfy $sin(x)=1/2$, you can choose $x = \pi/6$, $5\pi/6$, $13\pi/6$, etc.  
Likewise, there are infinitely many answers $z$ in the complex plane that satisfy $e^z = -1$.  Namely, they are odd integer multiples of $\pi i$.