Describe image of unit circle $|z| = 1$ under $ f(z) = \frac{z+1}{z-\alpha}$

Question

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!

Answer 1

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.

Answer 2

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.

Answer 3

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 $\frac{|1+\alpha|}{1-|\alpha^2|}$.