Homomorphism $ z \rightarrow\frac{z}{|z|}$ For my algebra assignment, I have to analyze the homomorphism $f: \mathbb C^* \rightarrow \mathbb C^*$ given by $$z \rightarrow\frac{z}{|z|}$$
I have to give the kernel, image, cosets and a plot of the homomorphism. 
I know the kernel is the set of $z$ for which $f(z)=e$ (identity), so is that just all complex numbers with length $1$? Or the complex numbers $z = 1$?
I also know the image is given by $\{f(z): z \in \mathbb C^*\}$, so is this the set $\{\frac{z}{|z|}: z \in \mathbb C^*\}$?
It would be nice if someone could tell me if im on the right path :)
Thanks in advance.
 A: The kernel is given by
$
\{z\in\mathbb C^*~:~f(z)=1\}.
$
Next, we check $$f(z)=1\Leftrightarrow \frac{z}{|z|}=1\Leftrightarrow z=|z|.$$
Since the RHS is real we conclude that $z$ has to be real too. And real numbers which are equal to their absolute value are ...
You are right that the image of $f$ is given by
$$
img(f)=\left\{\frac{z}{|z|}~:~z\in\mathbb C^*\right\}.
$$
What happens geometrically if you divide a vector by its length? (Hint: What is the length of the elements in $img(f)$?) 
Using the hint, you might find a better description of the set, which has a name, because it is very special!
A: Well, for the kernel, it's all complex numbers with $f(z) = e$, that is, $f(z)=1$.
Now, $f(z)=\frac z{|z|}$, so this means $\frac z{|z|}=1$, that is, $z=|z|$. 
Which complex numbers have $z=|z|$? 
A: The identity element of $(\mathbb{C}\setminus\{0\},\times)$ is $1$. Therefore,$$\ker f=\left\{z\in\mathbb{C}\,\middle|\,f(z)=1\right\}.$$
And, yes, the image of $f$ is $\left\{\frac z{|z|}\,\middle|\,z\in\mathbb{C}\setminus\{0\}\right\}$, but that's not the best description of that set.
