Power Function on Complex Numbers is Epimorphism

Theorem
Let $n \in \Z_{>0}$ be a strictly positive integer.

Let $\left({\C_{\ne 0}, \times}\right)$ be the multiplicative group of complex numbers.

Let $f_n: \C_{\ne 0} \to \C_{\ne 0}$ be the mapping from the set of complex numbers less zero to itself defined as:
 * $\forall z \in \C_{\ne 0}: f_n \left({z}\right) = z^n$

Then $f_n: \left({\C_{\ne 0}, \times}\right) \to \left({\C_{\ne 0}, \times}\right)$ is a group epimorphism.

The kernel of $f_n$ is the set of complex $n$th roots of unity.

Proof
From Non-Zero Complex Numbers under Multiplication form Abelian Group, $\left({\C_{\ne 0}, \times}\right)$ is a group.

Therefore $\left({\C_{\ne 0}, \times}\right)$ is closed by group axiom $G0$.

Let $w, z \in \C_{\ne 0}$.

Thus $f_n$ is a group homomorphism.

Now suppose $w = r \left({\cos \alpha + i \sin \alpha}\right)$, expressing $w$ in polar form.

Then $w = f_n \left({z}\right)$ where:
 * $z = r^{1/n}\left({\cos \dfrac \alpha n + i \sin \dfrac \alpha n}\right)$

and so:
 * $\forall w: w \in f_n \left({\C_{\ne 0}}\right)$

That is, $f_n$ is a surjection.

Being a group homomorphism which is also a surjection, by definition $f_n$ is then a group epimorphism.

The kernel of $f_n$ is the set:
 * $U_n = \left\{{e^{2 i k \pi / n}: k \in \N_n}\right\}$

which follows from Complex Roots of Unity in Exponential Form.