Isomorphism of $\mathbb{C}/\mathbb{Z}$ and $\mathbb{R}$ Using the fundamental homomorphism theorem show the isomorphism:$$(\mathbb{C},+)/\mathbb{Z}\cong (\mathbb R_{\ne 0},*)$$
I tried to use the exp function somehow, but I can't get the whole $\mathbb R_{\ne 0}$ as the image. 
 A: The additive group structure of $\Bbb C$ is that of $\Bbb R\oplus\Bbb R$ via $z=x+iy\sim (x,y)$. Quotienting by $\Bbb Z\oplus\{0\}$ which is the canonical copy of $\Bbb Z\subseteq\Bbb C$ we see that
$$\Bbb C/\Bbb Z\cong \left(\Bbb R\oplus\Bbb R\right)/\left(\Bbb Z\oplus\{0\}\right)\cong \left(\Bbb R/\Bbb Z\right)\oplus\left(\Bbb R/0\right)$$
$$\cong S^1\oplus\Bbb R$$
Using the isomorphism $(\Bbb R,+)\cong (\Bbb R^+,\cdot)$ via $x\longleftrightarrow e^x$ we get
$$\Bbb C/\Bbb Z\cong (S^1,\cdot)\oplus (\Bbb R^+,\cdot)$$
and from the polar decomposition of $\Bbb C^\times$
$$re^{i\theta}\longleftrightarrow (e^{i\theta}, r)$$
with $r>0$ and $e^{i\theta}\in S^1$, we see that
$$\Bbb C/\Bbb Z\cong S^1\oplus \Bbb R^+\cong\Bbb C^\times.$$

Addendum:  Your original formulation cannot be right.
Reason:  $(\Bbb R_{\ne 0},\cdot)\cong (\{\pm 1\},\cdot)\oplus (\Bbb R^+,\cdot)\cong (\{\pm 1\},\cdot)\oplus(\Bbb R,+)$ via the map
$$x\longleftrightarrow (\text{sgn }(x),|x|)\longleftrightarrow (\text{sgn }(x),\log|x|)$$
However, this group has only $2$-torsion, but if you look at the point $\left({1\over n}, 0\right)={1\over n}+0i\in\Bbb C$ in the quotient this is $n$-torsion for any $n\in\Bbb N$, because $n\left({1\over n}+0i\right)=1\in\Bbb Z$.
