The group homomorphism $z \mapsto (z/|z|)^2$ with an application of the homomorphism theorem. Question 
Let $G = (\mathbb{C}\setminus\{0\}, \cdot)$. 
1) Show that $N = (\mathbb{R}\setminus  \{0\},\cdot)$ is a normal subgroup. 
2) Show that $f:G \rightarrow G$ with $z \mapsto (z/|z|)^2$ is a group homomorphism. 
3) Determine $H$ s.t. $G/N \cong H$ (using $f$ from 2.)  
My solution 
I think 1) is obvious bc. $N$ is a subgroup of $G$ and $G$ is commutative. Therefore $N$ is a normal subgroup. For 2) i take $w,z \in G$ and calculate $f(wz)$ and $f(w)f(z)$ and look if they are identical. 
$f(wz)=(wz/|wz|)^2$ 
$f(w)f(z)=(w/|w|)^2 \cdot (z/|z|)^2 = (wz/|wz|)^2$, 
so $f$ is a homomorphism. 
in 3) i would say that $H = \{z : |z| = 1\}$, so i would say $H$ is the unit circle bc i only get complex numbers with absolute value $1$ and i get every point on the unit circle. 
Is my solution correct?
 A: The first isomorphism theorem tells you that if you have a homomorphism from $G$ to $H$ with exactly $N$ as kernel, then $G/N$ is isomorphic to the image of this homomorphism.
Therefore, your task should be to locate a homomorphism which has kernel exactly $N$, and then find the image of this homomorphism.
Now, what is the kernel of $f$? Note that $|z|^2 = z \bar z$, so $\frac{z^2}{|z|^2} = \frac z{\bar z}$ ,which equals $1$ if and only if $z = \bar z$, if and only if $z \in \mathbb R$. In other words, the kernel of $f$ is indeed $N$.
As for the image, as you have pointed out, since $|z|= |\bar z|$, we see that the image is certainly contained within $H = \{|x|= 1\}$. However, the converse needs to be shown i.e. that $H$ is contained in the image. This is not very difficult : let $y \in H$, and let $z$ be a square root of $y$, which always exists by the fundamental theorem of algebra (or using the polar form). Then $|z| = 1$, and $f(z) = y$.
Therefore, your statements are correct, but here is the formal verification anyway. 
