TITLE: Why is it that (negative number) ^ (irrational number) is nonreal?
QUESTION [5 upvotes]: We know that $(-1)^{\frac 12}$, $(-1)^{\frac 14}$, ... $(-1)^{\frac 1{even}}$ all result in complex answers, whereas $(-1)^{\frac 13}$, $(-1)^{\frac 15}$, ... $(-1)^{\frac 1{odd}}$ all result in real answers. I believe this comes from the definitions that $(-1)^{\frac 12} = \sqrt{-1} = i$ and $(-1)^{\frac 13}=\sqrt[3]{-1}=-1$. Moreover, of the set that provides real answers, if the numerator of the exponent is even, the result then becomes $+1$ instead of $-1$. This explains why the graph $y = (-1)^x$ results in an infinite amount of discontinuous points on $y=1$ and $y=-1$, only existing where the denominator of the exponent is odd:

But the sets I provided only explain $(-1)^{rational}$; in other words, this assumes that we are only raising $-1$ to a fraction. When I plug in $(-1)^e$, $(-1)^{\pi}$, $(-1)^{\sqrt{2}}$, or any other irrational number, the answer comes back undefined/imaginary.

This cannot come from the explanation that I provided earlier because, even though it is the fundamental explanation for where imaginary results come from, it assumes the exponent is a fraction. My intuition is that there would be some other rule that would sometimes result in an imaginary answer and sometimes result in a real number - similar to how rational exponents work - but it seems to always be imaginary. What, then, is the rule for this, and can we prove that this is the case for all irrational numbers?

REPLY [5 votes]: This is because the most general definition of $x^\alpha$ for complex $x,\alpha$ is $x^\alpha:=\exp(\alpha\log x)$, where $\exp$ is the exponential function and $\log$ is the (principal) natural logarithm. Pictorially, this means that taking $x$ to the power of $\alpha$ means multiplying the argument of $x$ by $\alpha$; that is, rotating $x$ with respect to the origin by an angle of $\alpha$ times the seperation between $x$ and the positive real axis. So if $x$ is a negative real number and $\alpha$ is an irrational real number, then we must rotate $x$ by some irrational multiple of $\pi$ (i.e. $180^\circ$). Obviously, irrational numbers are never integers, so the result after the rotation never ends up on the real axis; that is, $x^\alpha$ is not real.