# Show that for any two circlines, $X, Y$, there is a Möbius transformation such that $M(X) = Y$

I understand how to show this if both $X, Y$ are lines or are circles, but I'm stuck on what to do if say $X$ is a circle and $Y$ is a line.

First, using the transformations that you would if $X, Y$ were circles, we set $M_1(x)X =$ the unit circle.

Then this next bit confuses me, it says

Take $\frac{1}{z - 1}$ : The image is $x = - \frac{1}{2}$. Then take $M_3$, where $x = - \frac{1}{2}$, = $Y$. Finally, do the composition $M_3 \circ \frac{1}{z - 1} \circ M_1$.

I don't understand why you start with $\frac{1}{z - 1}$ and then why the image is $x = - \frac{1}{2}$. The rest makes sense.

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Are you aware of the practice of accepting answers? If not, please check out the FAQ. –  joriki Oct 11 '12 at 13:30

## 1 Answer

All you really need for the "middle" transformation, between $M_1$ and $M_3$, is some Möbius transformation that takes a (known) circle to a (known) line. The particular transformation, $\frac1{z-1}$, was someone's idea of the simplest such transformation. Here's how you might think of it:

You want some circle to be sent to some line. To make life easy, try to do it for the unit circle (the simplest circle around). To send this to a line, you'll have to send one of the points on the unit circle to $\infty$; again, let's try to do it with the "simplest" point on the unit circle, namely 1. So the Möbius transformation we want should have $z-1$ in the denominator, so that it produces $\infty$ when $z=1$. What should go in the numerator? Almost any linear or constant function would do, as long as it doesn't vanish at $z=1$. ("Linear or constant" is to ensure that we get a Möbius transformation; non-vanishing at $z=1$ is to avoid canceling the work we did to get the denominator.) Again, let's take the simplest choice, 1. So we get $\frac1{z-1}$. And we know, from how we constructed it, that this will take the unit circle to some line. Which line? Well, a line is determined by any two points on it. So take two points (other than 1, which gets sent to $\infty$ by design) on the unit circle, see where the transformation maps them. The line you want is the line through these image points. I'm inclined to take the points $\pm i$; plugging these $z$ values in to $\frac1{z-1}$, I get $-\frac12\mp\frac i2$, so the line is $\text{Re}(z)=-\frac12$.

(After all my "simplest choices", why did I use $\pm i$ when $-1$ is simpler? Well, the computations for $\pm i$ are essentially just one computation, because everything in sight is invariant under complex conjugation. If I used $-1$, I'd still have to do a computation for a second point.)

The bottom line is that the proposed transformation $\frac1{z-1}$ seems to be the result of saying, several times, "let's take the simplest option". Other options would also work. You could start with a different circle, send a different point on it to $\infty$, and choose a different numerator for your Möbius transformation. But why work harder than necessary?

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