If $\theta$ is a rational number, is $e^{i\pi\theta}$ algebraic?

Question

I want to know if $\theta$ is a rational number, is $e^{i\pi\theta}$ an algebraic number or not?

For the first step I tried to write it $(e^{i\pi})^\theta$, that equals $(-1)^\theta$, but I think powers are not commutative. Thanks for your help.

Answer 1

You are right to be concerned about which power laws you can rely on in the presence of complex numbers. I find it helps slightly to write the exponential function as $\exp(z)$ instead of $e^z$ as a reminder that all may not be as we're used to from the real case.

Now we hopefully already know, proved from first principles and whichever definition of the exponential function you're working with, that $$ \exp(z+w)=\exp(z)\exp(w) $$ and from this, by induction, we can prove for $z\in\mathbb C, n\in\mathbb N$: $$ \exp(z)^n = \exp(zn) $$ where the left-hand side is guaranteed to be meaningful because integer powers of arbitrary complex numbers are well defined.

Then, following Mathmo123's comment, we can say $$ \exp(i\pi\tfrac pq)^{2q} = \exp(i\pi\tfrac pq2q) = \exp(p\cdot 2\pi i) = 1 $$ where the last equality is from Euler's formula, which we also hopefully know already.

So $z=e^{i\pi\frac pq}$ is a root of the polynomial $z^{2q}-1$ and is therefore algebraic.