Having a multiplicative homomorphism $\chi:(\mathbb{Z}/f\mathbb{Z})^\times\to\mathbb{C}^\times,$ why the series (I'm a bit vague here...) $$\sum_{n=1}^\infty\frac{\chi(n)}{n^s}$$ converges for $Re(s)>1$? I see that $\chi(n)/n^s\to0$ (since $(\chi(n))_{n\ge1}$ is bounded) so the necessary condition for convergence is satisfied, but that's all what I can say.
I saw somewhere this inequality $$\sum_{n=1}^\infty\left|\frac{\chi(n)}{n^s}\right|\le\sum_{n=1}^\infty\frac{1}{n^{Re(s)}},$$ but this would mean that $\chi$ maps $(\mathbb{Z}/f\mathbb{Z})^\times$ into $\{z\in\mathbb{C},|z|\le1\}$...
$G=(\Bbb Z/f\Bbb Z)^\times$ is a finite Abelian group. In a homomorphism $\chi:G\to\Bbb C^\times$, each element in the image has finite order, so is a root of unity. Therefore $|\chi(n)|=1$ for all $n$.