Complex Numbers as External Direct Product

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

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

Let $\left({K, \times}\right)$ be the circle group.

Then:
 * $\left({\C_{\ne 0}, \times}\right) \cong \left({\R_{> 0}, \times}\right) \times \left({K, \times}\right)$

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

$\forall \left({a, b}\right) \in \R_{> 0} \times K:\exists z = a \times b \in \C$ such that
 * $\phi \left({z}\right) = \left({a, b}\right)$

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

So $\phi$ is surjective

To prove $\phi$ is injective, let $\phi\left({r_1e^{i\theta_1} }\right) = \phi\left({r_2e^{i\theta_2} }\right)$

So $\phi$ is injective, thus bijective.

Also:

So $\phi$ is a group homomorphism.

Since it is bijective, it is a group isomorphism.