Complex Numbers as External Direct Product

Theorem
Let $\struct {\C_{\ne 0}, \times}$ be the group of non-zero complex numbers under multiplication.

Let $\struct {\R_{> 0}, \times}$ be the group of positive real numbers under multiplication.

Let $\struct {K, \times}$ be the circle group.

Then:
 * $\struct {\C_{\ne 0}, \times} \cong \struct {\R_{> 0}, \times} \times \struct {K, \times}$

Proof
Let $\phi: \C_{\ne 0} \to \R_{> 0} \times K$ be the mapping:
 * $\map \phi {r e^{i \theta} } = \paren {r, e^{i \theta} }$

$\forall \tuple {a, b} \in \R_{> 0} \times K:\exists z = a \times b \in \C$ such that:
 * $\map \phi z = \tuple {a, b}$

by Complex Multiplication is Closed and $\R \subset \C$.

So $\phi$ is surjective.

To prove $\phi$ is injective, let $\map \phi {r_1 e^{i \theta_1} } = \map \phi {r_2 e^{i \theta_2} }$.

So $\phi$ is injective, thus bijective.

Also:

So $\phi$ is a group homomorphism.

Since it is bijective, it is a group isomorphism.