Describe image of unit circle $|z| = 1$ under $ f(z) = \frac{z+1}{z-\alpha}$ I have the following Möbius transformation:
$$
f(z) = \frac{z+1}{z-\alpha} \quad \alpha \neq -1
$$
And I have to describe the image of the unit circle ($|z| = 1$).
I can use symmetries in Riemann sphere and I don't have to do it analytically. So I found a solution for $ |\alpha| = 1 $ (just a line).
But I can't find the center and the radius of the circle for $ |\alpha| \neq 1 $. How to use symmetry for circles?
Thank you!
 A: One good method for such questions is to perform the transformation in easy steps. Viz:
Subtract $\alpha$, invert, multiply by $\alpha +1$, add $1$.
The only 'difficult' step is the second of these. The circle before that stage is $(z+\alpha)(z^*+\alpha^*)=1$. Let $w=\frac{1}{z}$, then $(\frac{1}{w}+\alpha)(\frac{1}{w^*}+\alpha^*)=1$ and so $(\alpha w+1)(\alpha^*w^*+1)=ww^*$.
Put this in the form $(\alpha \alpha^*-1)ww^* +\alpha^*w^*+\alpha w+1=0$. This gives us the centre and radius of the circle. For example, the centre is at $-\frac{\alpha}{\alpha \alpha^*-1}$.
The rest of the transformation is now straightforward.
A: Let $w = f(z)=\frac{z+1}{z-\alpha} $. Then $z(w-1) = \alpha w +1$. Given $|z|=1$, we have
$$|w-1|^2 = |\alpha w +1|^2\implies (w-1)(\bar{w}-1)=(\alpha w +1)(\bar{\alpha}\bar{ w} +1)$$
Rearrange,
$$(1-|\alpha^2|)|w|^2 -2Re[(1+\alpha)w] = 0$$
and express it in the form,
$$|w|^2 - 2Re\left(\frac{1+\alpha}{1-|\alpha^2|}w\right)
+\bigg|\frac{1+\alpha}{1-|\alpha^2|}\bigg|^2=\bigg|\frac{1+\alpha}{1-|\alpha^2|}\bigg|^2$$
or, explicitly, in the form of a circle,
$$\bigg|w-\frac{1+\alpha}{1-|\alpha^2|}\bigg|^2=\bigg|\frac{1+\alpha}{1-|\alpha^2|}\bigg|^2$$
Thus, the center of the circle is $\frac{1+\alpha}{1-|\alpha^2|}$ and its radius is $\frac{|1+\alpha|}{1-|\alpha^2|}$.
A: Let me try to expand upon what Lubin said in a comment to give a hint. 
Since we have a  Möbius transformation, we know that the unit circle must be mapped to either a line or a circle. Let's map three points from the unit circle. 
First notice that:
$$
f(z) = \dfrac{z+1}{z-\alpha} = z + \dfrac{1}{z} - \alpha
$$
So we have:
$$
f(1) = 2-\alpha\\
f(-1) = -\alpha-2\\
f(i) = -\alpha
$$
Let's write $\alpha = x+iy$, with $x, y \in \mathbb{R}$. 
So we could also write our three points as:
$$
f(1) = (-x+2)-iy\\
f(-1) = (-x-2)-iy\\
f(i) = -x-iy
$$ 
Now you just need to write the equation of the line/circle defined by these three points.
