# Is it standard to say $-i \log(-1)$ is $\pi$?

I typed $\pi$ into Wolfram Alpha and in the short list of definitions there appeared

$$\pi = -i \log(-1)$$

which really bothered me. Multiplying on both sides by $2i$:

$$2\pi i = 2 \log(-1) = \log(-1)^2 = \log 1= 0$$

which is clearly false. I guess my error is $\log 1 = 0$ when $\log$ is complex-valued. I need to use $1 = e^{2 \pi i}$ instead.

So my question: is it correct for WA to say $\pi = -i \log(-1)$? Or should be they specifying "which" $-1$ they mean? Clearly $-1 = e^{i \pi}$ is the "correct" value of $-1$ here.

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It is not "clearly false", since the complex logarithm function is multivalued. It's not "which -1" that matters, but which branch of $\log$ they are using. They seem to be using the principal branch, with $\log(-1) = i\pi$ (the imaginary part of the answer should lie in $(-\pi,\pi]$ to be in the principal branch), which seems reasonable. – Arturo Magidin Mar 7 '11 at 1:57
My "clearly false" refers to $2\pi i = 0$. – Fixee Mar 7 '11 at 2:30

It is not which $-1$ but which $\log$ which is the issue. Wolfram Alpha has used some kind of principal value of the complex logarithm. But using this, $a \log (b) = \log (b^a)$ is no longer always true.

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The (principal value) of the complex logarithm is defined as $\log z = \ln |z| + i Arg(z)$.

Therefore,

$$\log(-1) = \ln|-1| + i Arg(-1) = 0 + i \pi.$$

and then, one simply gets

$$-i \log(-1) = -i (i \pi) = \pi.$$

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