Homomorphism onto complex unit circle

Question

Let φ : $Z/n$ → $ C-\{0\} $ be a homomorphism. Why would Im(φ) ⊂ µn?

$Z/n$ is the group of addition modulo n, so $Z/n = \{[0] , [1 ] , ... , [n-1] \}$
We take $C-\{0\}$ under multiplication
So the set $\mu_n := \{\alpha \in C : \alpha^n = 1 \}$ Which is a set of points on the complex unit circle.

I do not see first of all how a such a homomorphism can only be mapped onto points of the complex unit circle. One example of a map would be $\phi(x) = e^{2\pi i}$ where points would always be mapped to 1 as the $x$ would always be an integer.

Any hints would be appreciated!

Answer 1

About your example, if $\phi(x)=2\pi i$ then certainly $Im(\phi)\subset \mu_n$, since $1^n=1$.

Hint: $$\phi([1])^n=\phi(n[1])=\phi([n])=\phi([0])=1$$

Answer 2

Since $φ([1])^n= φ(n[1])=φ([n])=1$. Thus, $Im(φ)=< φ([1])> ⊂ µn$.