# How to raise -1 to non-integer powers

How do you calculate $(-1)^x$ where $x$ is some real number. For example, what is $(-1)^{\sqrt{5}}$. This question came as I was trying to computer $e^{i\pi a}$ where $a$ is irrational.

• It’s not well-defined, but $e^{iπa}$ on the other hand is, see wiki on complex exponential. Commented Jan 13, 2015 at 19:49
• @Ahaan: Why did you create the tag (de-moivres-theorem) for this single question? Do you think we need this tag? Commented Jan 15, 2015 at 4:16
• If you go there, you may also rethink $1^x$ Commented Jan 15, 2015 at 4:49
• @EricStucky Yes, I did. And I think we do. There are several several questions on de Moivre's theorem. Commented Jan 15, 2015 at 14:55

One value is $$(-1)^x = \cos \pi x + i \sin \pi x$$
That was enough for my article, as this complex number can be used in Gelfond-Schneider. The conclusion is that if $x$ is real, irrational, but algebraic, then $\cos \pi x + i \sin \pi x$ is transcendental. I used one possible version of a contrapositive: I had both (real) $x$ and $\cos \pi x + i \sin \pi x$ algebraic, therefore $x$ was rational.
Note, however, that there are countably infinite logarithms of $(-1),$ so there are countably infinite different values of $(-1)^x.$ That's just life.
In general, for complex numbers (with $a \ne 0$) $a^b$ is defined as $e^{b \log(a)}$. However, $\log(a)$ is a multivalued function, and therefore so is $a^b$.
In general, $$e^{it}=\cos t+ i\sin t$$ This is known as Euler's formula. If we apply it to your example, we get $$(-1)^{\sqrt{5}}=\cos(\pi\sqrt 5)+i\sin(\pi \sqrt 5)$$ Since $-1$ has other representations as $e$ to some power (we can add any multiple of $2\pi i$ to its exponent), other numbers could also be called $(-1)^{\sqrt{5}}$, so it's better to work with $e^{i\pi\alpha}$.