Find a möbius transformation from the line to the unit circle?

Question

Let us take the line $\gamma=\{1+\lambda i:\lambda\in \Bbb{R}\}$ I want to find a möbius transformation to the unit circle.

My idea was to map $1\mapsto 1$, $\frac{1}{2}\mapsto 0$ and $\frac{3}{2}\mapsto \infty$. Solving a small system of linear equalities I would get that $$f(z)=\frac{-4z+2}{4z-6}=\frac{-2z+1}{2z-3}$$

Is this correct so?

Answer 1

The most intuitive way to do this is to use inversion!

If a line $L$ does not pass through the center of the circle of inversion, then inversion maps it to a circle that passes through center. Note that inversion is a self inverse transformation, so it occurs the opposite way too (circle through center-> line not through center). (pg-127)

To get the line to a circle centered at origin, all we have to do is invert it by the unit circle centered at origin and then reflect along x axis

$$f(z) = \frac{1}{z}$$

We can now check that this transformation sends the line to a circle at center $.5$ with radius $.5$, by a sequence of a translation, by shifting and dilating, we find the final transformation as:

$$ T(z) = 2( \frac{1}{z} - .5)$$

Note: There is no unique moebius transformation here, your equation is correct too! One can compose the final eqtn I got with a further maps which preserve the unit circle to get other moebius transformations doing the same.