3
$\begingroup$

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.

  • 1
    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.2011-03-07
  • 0
    My "clearly false" refers to $2\pi i = 0$.2011-03-07

2 Answers 2

8

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.

7

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

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