Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Reading, $\log(z)$ 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 $z=-1$.

share|cite|improve this question
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]( – 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
up vote 5 down vote accepted

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

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.

share|cite|improve this answer
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 $\pi i$.

share|cite|improve this answer
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
Depends on who you ask I suppose. – orlandpm Dec 11 '12 at 22:59

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.