Moebius transform unit circle I have the following complex function
$$w=f(z)=\frac{2z+4\mathrm{i}}{2z-2\mathrm{i}}$$
The question is the mapping of the unit circle. I tried to set $z=\mathrm{e}^{\mathrm{i}t}$ and calculate the result but it was not helpful. My next step was to split the complex exponential to its real and imaginary part but I can not determine the map of a circle. The result should be $$(x+\frac{1}{2})^2+y^2=(\frac{3}{2})^2$$
Does anyone know how to calculate this?
 A: Starting with 
$$
w=f(z)=\frac{2z+4\mathrm{i}}{2z-2\mathrm{i}}
$$
we can solve for  $z = x + iy$ in terms of $w = u + iv$. I'm going to let you do that part. (Hint: matrix inversion is one way to get to the result), producing something that looks like
$$
z = x + iy = \frac{Aw+B}{Cw+D}
$$
for the inverse function. You then know that you want to find all $w$ for which this point $z$ is on the unit circle, i.e., for which $z\bar{z} = 1$. That amounts to saying that
$$
\left( \frac{Aw+B}{Cw+D} \right) \overline{\left( \frac{Aw+B}{Cw+D}\right)} = 1
$$
which you can write out in terms of $u$s and $v$s and do the multiplication, etc. to get an equation in $u$ and $v$ that $w$ must satisfy. 
I haven't done all the algebra for you, because ... well, it's your problem, and doing the algebra will give you time to reflect on what you're actually computing. 
Small warning: it's just possible that in the course of doing the algebra, you'll find yourself cancelling something like $\frac{3u + 4v}{3u + 4v}$ to get $1$. That's an OK thing to do...except that $3u + 4v$ might be zero. So the equation that you get for $u$ and $v$ might actually miss some $w$ that really does produce a point on the unit circle in $z$. If that happens, you should probably ask yourself "What happened in the algebra, and how does that match what happened in the geometric picture of things?" 
A: Note that$$\frac{2z+4i}{2z-2i}=\frac{z+2i}{z-i}=1+\frac{3i}{z-i}.$$So, see what's the image of the unit circle under $z\mapsto\dfrac{3i}{z-i}$ (problem: $i$ belongs to the unit circle) and add $1$ to that.
