I was given this exercise: Show that $\mathbb{C}_0, \cdot$ (set of complex numbers without zero) is isomorphic to the direct product $W \times \mathbb{R}$, where we consider the groups $W = \{z \in \mathbb{C} \vert \|z\| = 1\}, \cdot$ and $\mathbb{R}, +$. This is the literal problem statement which I was given.
I have tried to define a map using exponential notation, but I am not able to find one. I also looked around on this site, but only find questions where the positive real numbers are considered under multiplication.
Any hints would be appreciated.
$\textbf{EDIT:}$ I think I have found a solution. Let us define a map $f: \mathbb{C} \to W \times \mathbb{R}: re^{i \theta} \mapsto (e^{i \theta}, \ln (r))$. This is possible since each element of $\mathbb{C}_0$ can be written uniquely as $re^{i \theta}$, where $r \in \mathbb{R}_{\geq 0}$. This map is a group morphism since we have the following equalities: \begin{align} f(r_1e^{i\theta_1} \cdot r_2e^{i \theta_2}) &= f(r_1r_2e^{i (\theta_1 + \theta_2)})\\ &= (e^{i (\theta_1 + \theta_2)}, \ln (r_1r_2))\\ &= (e^{i\theta_1} e^{i \theta_2}, \ln (r_1) + \ln (r_2))\\ &= (e^{i \theta_1}, \ln(r_1)) \ast ((e^{i \theta_2}, \ln(r_2)) \\ &= f(r_1e^{i\theta_1}) \ast f(r_2e^{i \theta_2}), \end{align} where $\ast$ is the binary operation on the direct product.
Moreover this map is injective: suppose $z \in \ker(f)$. We can write $z = re^{i \theta}$ and we have that $f(z) = (e^{i \theta}, \ln(r)) = (1,0)$. This implies that $e^{i\theta} = 1$ and $r = 1$, hence $z = 1$. Therefore the kernel is trivial, so $f$ is injective.
This map is also surjective. Given an element $(z,r) \in W \times \mathbb{R}$, we have that $z = e^{i \theta}$ and therefore we find that $f(e^re^{i \theta}) = (e^{i \theta}, \ln(e^r)) = (e^{i\theta}, r) = (z,r)$. This proves that $f$ is an isomorphism.