Show that two groups are isomorphic $$\varphi : \mathbb{C}^{*}/\mathbb{R}_{>0} \to \ S =  \left \{ z\in {C}^{*}: |z|=1 \right \}$$
So far, I've done this: $\phi : \mathbb{C}^{*} \to S$ where $\phi(z)=|z|$. $\phi$ is a homomorphism and is surjective, since for every z in S there is a z in $\mathbb{C}^{*}$. Therefore, $S\simeq \mathbb{C}^{*}/\mathbb{R}_{>0}$. 
Is this proof sufficient? I also have a feeling I defined $\phi$ wrong.
I would appreciate any help! Thank you. 
 A: You have the right idea, but the wrong $\phi$. The map you've defined doesn't even land in the right place! $2\in\Bbb C^\times$, but $\left|2\right| = 2\not\in S^1$.
The map I think you intended is the map projecting each infinite ray from the origin in $\Bbb C^\times$ to the unique element on the ray with absolute value one: this map is given by
\begin{align*}
\phi : \Bbb C^\times&\to S^1\\
z&\mapsto\frac{z}{\left|z\right|}.
\end{align*}
Now this is a surjective group homomorphism (exercise), and we have
\begin{align*}
\ker\phi &= \{z\in\Bbb C^\times\mid\phi(z) = 1\}\\
&= \{z\in\Bbb C^\times\mid z/\left|z\right| = 1\}\\
&= \{z\in\Bbb C^\times\mid z = \left|z\right|\}\\
&= \Bbb R^+.
\end{align*}
So, by the first isomorphism theorem, $\Bbb C^\times/\Bbb R^+\cong S^1$.
A: There's already a good answer. Here's an overview that may explain the confusion in the statement of the problem.
Writing nonzero complex numbers in polar form as $re^{i\theta}$ essentially establishes an isomorphism between $\mathbb{C}^*$ and the direct product $\mathbb{R}^+ \times S^1$. The projections onto each factor exhibit each factor as the quotient of $\mathbb{C}^*$ modulo the other factor.
