Identifying $\mathbb{C}^*/\mathbb{R}^{+}$ I define a  Surjective Homomorphism $\varphi:\mathbb{C}^*\to S^1$ by $z\mapsto {z\over |z|}$
Ker$\varphi=\{z\in\mathbb{C}^*:\varphi(z)=1\}\Rightarrow\{z: z=|z|\}=\mathbb{R}^{+}$
So, $S^1=\mathbb{C}^*/\mathbb{R}^{+}$
am I right?

Yes, you are right. You can visualize this result by identifying each element of $\mathbb{C}^*/\mathbb{R}^+$ with a ray out from $(0, 0)$ in $\mathbb{C}^*$. Then to multiply two rays, multiply representatives from them and take the ray containing the result. Clearly the product of two rays in this sense only depends on the angles, not the magnitudes, of the representatives so this operation is well-defined. So the chosen representatives might as well be the unique elements of $S^1$ on the two rays. Thus the group of rays is isomorphic to $S^1$.
