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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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    Actually, in the note you referenced, they **do** define $\log$ on $\{0\}^C$?2012-12-11

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$\log{(-1)}$ 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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    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.2012-12-11
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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 $\pi i$.

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    Depends on who you ask I suppose.2012-12-11