Is the homomorphism $f: n\mapsto e^{in\theta}$ injective?

Question

Would any of you mind just taking a look to see whether this is a valid proof?

I'm trying to prove whether the homomorphism $f: \mathbb{Z} \rightarrow U(1)$ defined by $f(n) = e^{i n \theta}$ is an injective homomorphism.

Here's my attempt:

Homomorphism Proof:

$f(n + m) = e^{i (n+m) \theta} = e^{i n \theta} e^{i m \theta}.$

Injectivity proof:

Let $ e^{i n \theta} = e^{i m \theta}$. Take the natural $\log$ of both sides, $i n \theta = i m \theta$, and therefore $n = m$.

Is this correct?

Answer 1

Homomorphism part is correct. However, you can't use the logarithm on complex numbers, as it is a multivalued function.

Injectivity Part

It is enough to check the kernel of the homomorphism is trivial. $e^{in\theta}=1$ implies $n\theta=2k\pi$ for some $k\in\mathbb{Z}$. So, the map is injective iff $\theta$ is not a rational multiple of $\pi$.