# Homework Help: Symmetry with respect to a circle

1. Oct 25, 2008

### mrbohn1

1. The problem statement, all variables and given/known data

Could someone please just explain a step in a proof for me?

Suppose x and y are symmetric with respect to a circle, and a,b and c are three points on the circle. This means that:

(x,a,b,c) = ($$\bar{y}$$,$$\bar{a}$$,$$\bar{b}$$,$$\bar{c}$$)

The writer then says:

"symmetry only depends on the circle, and not on the points, so if x and y are symmetric then:

$$\frac{x-b}{a-b}$$ = $$\frac{\bar{y}-\bar{b}}{\bar{a}-\bar{b}}$$"

I understand why this would be if the points were on a straight line, as then you could choose c to be the point at infinity, and the cross ratio would reduce to this. But why is it also true for a circle?

2. Oct 25, 2008

### Staff: Mentor

You need to give us some more information. Are we talking about a circle in the real plane? What's the significance of $$\bar{y}$$? That suggests conjugates, so maybe we're not talking about circles in the real plane. And what does (x, a, b, c) denote?

3. Oct 26, 2008

### mrbohn1

Fair enough!

This is a complex analysis problem - the circle is in the complex plane.

If x, a, b, c are complex numbers, then (x,a,b,c) denotes the cross ratio:

$$\frac{(x-b)(a-c)}{(x-c)(a-b)}$$

Two points x and y being symmetric with respect to a circle means that if a, b, c are any three points on that circle, then:

(x,a,b,c) = $$\overline{(y,a,b,c)}$$

(ie. the cross ratio of x, a, b, c is equal to the complex conjugate of the cross ratio of y, a, b, c ).

So: $$\frac{(x-b)(a-c)}{(x-c)(a-b)}$$ = $$\frac{\overline{(y-b)}\overline{(a-c)}}{\overline{(y-c)}\overline{(a-b)}}$$

But the text says that we can choose the points a, b, c such that some of these terms cancel and we get:

$$\frac{(x-b)}{(a-b)}$$ = $$\frac{\overline{(y-b)}}{\overline{(a-b)}}$$

I don't understand this - as far as I can see this would only be possible by choosing c to be infinity, but the circle does not contain the poiint at infinity.

Last edited: Oct 26, 2008
4. Oct 27, 2008

### mrbohn1

Is it considered bad form to post a reply to your own question to get it back to the top of the board?

5. Oct 28, 2008

### Staff: Mentor

I had one class on complex analysis in grad school, but I have to confess we didn't cover any of the ideas you've presented, such as symmetry wrt a circle or the cross-ratio business.

So looking at what you have and where you want to go, it seems that you want the a -c and x -c factors to cancel in the ratio on the left, and the a -c and y -c (conjugate) factors to cancel in the ratio on the right. For the one on the right, if you choose a = y those factors cancel, but on the left what seems to work is choosing a = x. You said that one could choose a, b, and c, but I don't think you can choose a to be simultaneously equal to both x and y, so that doesn't seem to help.

What bothers me is that the word "symmetry" is used in this problem, but I don't have a clear picture in mind of where x and y are located on the circle. I would think that x and y being symmetric with respect to the circle means one of the following:
1. Re(x) = Re(y) (the points are directly across the real axis from one another so that x = conj(y))
2. Im(x) = Im(y) (the points are directly across the imaginary axis from one another)
3. Arg(x) = Arg(y) + k*pi, where k is an integer (symmetry about the origin)
Is symmetry with respect to a circle what I have described here?

Sorry I'm not able to provide more help.