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:
 * $\phi \left({r e^{i \theta} }\right) = \left({r, e^{i \theta} }\right)$

Obviously $\phi$ is a bijection.

Also:

So $\phi$ is a group homomorphism.

Since it is bijective, it is a group isomorphism.