Prove $A$, the set of all complex numbers with modulus $1$, is a subgroup of $\Bbb C^*$. Prove also that $\Bbb C^*\cong\Bbb R^+\times A$.

Let $\Bbb C^*$ be the set of all nonzero complexes. Prove that the set $A$ of all complexes with modulus $1$ is a multiplicative subgroup of the group $\Bbb C^*$. Prove that the multiplicative group $\Bbb C^*$ is isomorphic to $\mathbb{R}^{+}\times A$.

I was able to show that it is an abelian group, but I have not been able to find an application that makes it an isomorphism.

Define the map $ \varphi:\mathbb{C^{\times}\to\mathbb{R}}^{+}\times A $

By: $ \varphi\left(c\right)=\left(|c|,e^{i\text{Arg}\left(c\right)}\right) $

Where we decide that $Arg(c) \in [-\pi,\pi) $

This map is homomorphism (It's easy to verify that it preserves multipication). Also it is injective since $ \varphi(c_1) = \varphi(c_2) $ implies $|c_1|=|c_2| $ and $Arg(c_1)=Arg(c_2)$, so that $c_1=|c_1|e^{iArg(c_1)} = |c_2|e^{iArg(c_2)}=c_2$

And it is also surjective since given $(r,e^{ik})$ take $ c=r\cdot e^{ij}$ Where $j=k-2\pi m $ for $m\in \mathbb{Z} $ such that $j\in [-\pi,\pi) $.

All in all this is an isomorphism

Use the one-step subgroup test.

Since $|1|=1$, we have $A\neq \varnothing$.

By definition, $A\subseteq \Bbb C^*$.

Let $a,b\in A.$ Then $a=e^{i\theta}, b=e^{i\varphi}$ for $\theta,\varphi\in\Bbb R$ by Euler's formula. Now

$$\begin{align} ab^{-1}&=e^{i\theta}(e^{i\varphi})^{-1}\\ &=e^{i\theta}e^{-i\varphi}\\ &=e^{i\theta-i\varphi}\\ &=e^{i(\theta-\varphi)}, \end{align}$$

but then $|ab^{-1}|=1$, so $ab^{-1}\in A$.

Hence $A\le \Bbb C^*$.

For the isomorphism, as in @FreeZe's answer, consider

$$\begin{align} \psi: \Bbb C^*&\to \Bbb R^+\times A\\ re^{i\varphi}&\mapsto (r, e^{i\varphi}). \end{align}$$