$-1$ to the power of a irrational number

Question

According to Wolfram Alpha, $(-1)^\pi \approx -0.90 + 0.43i$.

But $\pi$ has proven to be irrational (we can't write $\pi$ in terms of a fraction $a/b$ with $a$ and $b$ integers) and, in the real numbers, a negative number raised to a fraction with even denominator is undefined*.

So why is $(-1)^\pi$ a complex number, if there is no even denominator?

*Of course there are exceptions, like $(-1)^{4/2} = (-1)^2 = 1$ and $(-1)^{6/2} = (-1)^3 = -1$, but consider that the numerator is not a multiple of the denominator.

Answer 1

This has to do with complex numbers. $$(-1)^{\pi}=\left(e^{i\pi}\right)^{\pi}=e^{i\pi^2}=\cos{\pi^2}+i\sin{\pi^2}$$ which is the value given by WolframAlpha.

However, since $-1=e^{(2n+1)\pi i}$ for every integer $n,$ there are actually infinitely many values of $(-1)^\pi.$ With complex numbers, exponentiation works differently from the way it does with real numbers.

Answer 2

For complex numbers, we define $a^b = \exp(b \log a)$. Of course $\log$ is "multivalued", so $a^b$ is also multivalued. Even $(-1)^{1/3}$ has three values, two of them are non-real. In case of $(-1)^\pi$, there are infinitely many values, all non-real.

Wolfram has a system of choosing a "principal value" in these cases, and it uses the logarithm with imaginary part in $(-\pi,\pi]$. So the principal value of $\log(-1)$ is $i\pi$. And the principal value of $(-1)^\pi$ is $\exp(\pi\log(-1)) = \exp(i\pi^2) = \cos(\pi^2)+i\sin(\pi^2)\approx -0.90 - i 0.43$.