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.

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