Power Function on Complex Numbers is Epimorphism

Theorem

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

Let $\struct {\C_{\ne 0}, \times}$ 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:

$\forall z \in \C_{\ne 0}: \map {f_n} z = z^n$

Then $f_n: \struct {\C_{\ne 0}, \times} \to \struct {\C_{\ne 0}, \times}$ 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 Group, $\struct {\C_{\ne 0}, \times}$ is a group.

Therefore $\struct {\C_{\ne 0}, \times}$ is closed by Group Axiom $\text G 0$: Closure.

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

 $\ds \map {f_n} {w \times z}$ $=$ $\ds \paren {w \times z}^n$ $\ds$ $=$ $\ds w^n \times z^n$ Power of Product of Commutative Elements in Group $\ds$ $=$ $\ds \map {f_n} w \times \map {f_n} z$

Thus $f_n$ is a group homomorphism.

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

Then $w = \map {f_n} z$ where:

$z = r^{1/n} \paren {\cos \dfrac \alpha n + i \sin \dfrac \alpha n}$

and so:

$\forall w: w \in \map {f_n} {\C_{\ne 0} }$

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:

$U_n = \set {e^{2 i k \pi / n}: k \in \N_n}$

which follows from Complex Roots of Unity in Exponential Form.

$\blacksquare$

Examples

Cube Function

Let $f_3: \C_{\ne 0} \to \C_{\ne 0}$ be the mapping from the set of complex numbers less zero to itself defined as:

$\forall z \in \C_{\ne 0}: \map {f_n} z = z^3$

Then $f_3: \struct {\C_{\ne 0}, \times} \to \struct {\C_{\ne 0}, \times}$ is a group epimorphism.

The kernel $U_3$ of $f_3$ is the set of complex $3$rd roots of unity:

$U_3 = \set {1, \omega, \omega^2}$

where:

 $\ds \omega$ $:=$ $\ds \cos \dfrac {2 \pi} 3 + i \sin \dfrac {2 \pi} 3$ $\ds \omega^2$ $:=$ $\ds \cos \dfrac {4 \pi} 3 + i \sin \dfrac {4 \pi} 3$

Hence for all $z \in \C_{\ne 0}$, the coset $z U_3$ is the set:

$z U_3 = \set {z, z \omega, z \omega^2}$

and multiplication on $\C_{\ne 0} / U_3$ of all such cosets satisfies:

$\set {z_1, z_1 \omega, z_1 \omega^2} \times \set {z_2, z_2 \omega, z_2 \omega^2} = \set {z_1 z_2, z_1 z_2 \omega, z_1 z_2 \omega^2}$

Hence the associated isomorphism $g: \C_{\ne 0} / U_3 \to \C_{\ne 0}$ takes the equivalence class:

$\set {z, z \omega, z \omega^2}$

into the cube:

$z^3 = \paren {z \omega}^3 = \paren {z \omega^2}^3$

of any one of its elements.