Functional Equation and totally multiplicative functions

Question

Find all $f:\mathbb{C}\rightarrow\mathbb{C}$ s.t. $\forall a, b\in \mathbb{C}, f(ab) = f(a)f(b)$.

I could deduce a lot of things about what happens at the roots of unity, and 0, but I can't find out anything more.

Answer 1

In addition to what I said in the comments, let $z =e^{i\theta}$ and let $\{p_n\}$ be a sequence of rationals such that $p_n \to \theta/2\pi$. Then for every $p_n = \frac{a_n}{b_n}$, $$|f(e^{ip_n 2\pi}) |^{b_n} = |f(e^{ip_n 2\pi})^{b_n} | = |f(e^{ia_n 2\pi})| = |f(1)|=1$$ And thus $|f(e^{i p_n 2\pi})|=1$. If $f$ is continuous, $lim_{n\to\infty} |f(e^{i p_n 2\pi})| = |f(z)|=1$. That is: if $|z|=1$ then $|f(z)|=1$

If $f$ is bounded, then lets suppose $|f(z)| >1$ for some $z$. Thats absurd, because $lim\ |f(z^n)|=lim \ |f(z)|^n = \infty$. If $|f(z)|<1$ then $|f(z^{-1})|>1$ and thats also absurd.

So, if $f$ is bounded, $|f(z)| = 1$ for every $z$. And $f(z) = e^{i \theta(z)}$ with $\theta(x)+\theta(y) = \theta(xy)$