Finding an isomorphism of groups 
Let $R$ be a commutative ring, $R_U$ be the group of units of R. Show that 
(i)
  $\mathbb{C}_U \cong \left(\left(\mathbb{R},+\right)/\mathbb{Z}\right)\times\left(\mathbb{R},+\right)$
(ii)
  $(\mathbb{Z}[x] / (x^2-x))_U \cong \mathbb{Z}_2 \times \mathbb{Z}_2$

where $\cong$ stands for isomorphic as groups.
(i) 
I thought about using the polar form of complex numbers. At first I had a problem with the fact that (for using with the angle) the real numbers can be negative. So I tried the Euler identity to the rescue: $e^{i\pi} = -1$. So if I map $\gamma : (r,\phi) \mapsto r e^{i\pi(1+\phi)}$, this seemed to work. But what about the zero? How can I get this to be injective?
(ii)
First as $\# \mathbb{Z}_2 \times \mathbb{Z}_2 = 4$, the target group will have 4 elements. But what are they?
I tried $\mathbb{Z}[x] / (x^2-x) = \{ax + b\ \big\vert\ a,b \in \mathbb{Z}\}$ (right?). 
$(ax+b)(cx+d) = 1$ didn't lead me to a useful solution.
Can you please help me to go on?

$\mathbb C_U=\mathbb C^{\times}\simeq(\mathbb R/\mathbb Z,+)\times(\mathbb R_{> 0},\cdot)$ by using the isomorphism you suggested. Moreover, $(\mathbb R_{>0},\cdot)\simeq(\mathbb R,+)$ by using logarithm.
$\mathbb{Z}[x] / (x^2-x)\simeq\mathbb Z[x]/(x-1)\times\mathbb Z[x]/(x)$ by CRT. Moreover, $\mathbb Z[x]/(x-1)\simeq\mathbb Z[x]/(x)\simeq\mathbb Z$, so $(\mathbb{Z}[x] / (x^2-x))_U\simeq\mathbb Z_U\times\mathbb Z_U$.
