# Why isn't $\log(-1)=i\pi$?

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 $z=-1$.

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That's typically where we take the branch cut of the logarithm. You can certainly have other branches of the logarithm where $\log(-1) = i\pi$. –  EuYu Dec 11 '12 at 18:53
Why? Is there some more advanced definition of the logarithm from which it is natural to cut out the real negatives, as from my naive stance it seems silly? –  Alyosha Dec 11 '12 at 18:55
$\log$ can be defined on $\mathbb{C}\setminus \{0\}$ such that it will satisfy $e^{\log x} = x$. However, it must be discontinuous somewhere, with the usual $\log$ if you approach $-1$ from 'below' the imaginary part will approach $-\pi$, if you approach $-1$ from 'above' the imaginary part will approach $+\pi$. There are more satisfying answers involving analytic continuation, but it is not as simple as just defining $\log$ on $x \leq 0$. –  copper.hat Dec 11 '12 at 19:03
As you observed, there are multiple possibilities for $\log -1$, so it can't be a function. However, there is something called principal branch, for example see [here](en.wikipedia.org/wiki/Logarithm#Complex_logarithm). –  dtldarek Dec 11 '12 at 19:16
@Alyosha: Any '$\log$'-like function will have similar discontinuity. $\log$ has some useful properties when restricted to simply connected domains that do not contain $0$. Nothing is failing here. The choice of the negative real line is arbitrary to some degree, but we like to have it match the $\mathbb{R}$ $\log$ on the positive real axis, and a human preference for symmetry dictates the choice of the negative line. I do not understand what your second question means, nor what ailments are at issue. To reiterate, you can define $\log$ on $\{0\}^C$, it just will be discontinuous. –  copper.hat Dec 11 '12 at 19:49

$\log{(-1)}$ does equal $i\pi$, for the reasons you described.

http://www.wolframalpha.com/input/?i=log%28-1%29

But it mainly depends on the universe in which you are taking the logarithm. If you decide to only work in the reals, then $\log{(-1)}$ wouldn't be defined. But it's perfectly okay to work in the complexes, too.

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So the log function is continuous (when using complex numbers) everywhere except $z=0$? Thanks. –  Alyosha Dec 11 '12 at 19:00
Nope, see my comment above. –  copper.hat Dec 11 '12 at 19:07
The log function is continuous when using complex numbers, but you have to choose a branch of $arg(z).$ This amounts to saying that log is well defined and continuous on $\mathbb{C} \setminus L$ where L is any line starting at the origin and going out to $\infty.$ If you don't cut out a line like this, when you follow the values of log around a circle they increase by $2\pi,$ so log can't be defined on $\mathbb{C} \setminus 0.$ This is what it really means when we say $arg(-1) = \pi +/- 2n\pi.$ Here choosing $n$ amounts to choosing a branch of log. –  mck Dec 11 '12 at 19:09
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 $\pi i$.
Although isn't a similarly arbitrary constraint of $\sin(x)$ being defined from $-\pi$ to $\pi$ in place? –  Alyosha Dec 11 '12 at 20:07
I was taught the convention that $\sin^{-1}$ refers to the principal branch and that $\arcsin$ doesn't necessarily. I have no idea if this is generally accepted, and it doesn't really matter. I think the only important thing is to realize that the meaning of $\log$ is trickier over complex numbers. –  orlandpm Dec 11 '12 at 20:38
Wait, $\sin^{-1}$ and $\arcsin$ aren't the same thing? –  Alyosha Dec 11 '12 at 22:55