# Isometries (Translations)

1. Mar 31, 2009

### zcdfhn

Prove that every non-zero translation Tb is a composition of two reflections in lines which are perpendicular to the direction of vector b.

Note: Tb(z) = z+b where z,b$$\in$$C

My guess at how to start this is to assume b = rei$$\theta$$ where r = |b| and $$\theta$$=arg b, so then the direction of b is b/|b| = ei$$\theta$$. Therefore the unit vector with the direction orthogonal to b would be c = ei($$\theta$$ + $$\pi$$/2). From there, I am shaky about what to do. I attempted to create two reflection f,g that reflection over lines with the same direction as c and I attempted to do g$$\circ$$f and I should end up with Tb but my work gets more and more complicated.

Last edited: Mar 31, 2009
2. Mar 31, 2009

### Dick

What kind of formula do you know for a reflection? The reflections you are talking about not just reflections through the origin. You'll never get a translation out of that.

3. Mar 31, 2009

### zcdfhn

I know M(z) = a(z-bar)+b, where |a| = 1 and a $$\neq$$ 1, a(b-bar) + b = 0

and i also know M(z) = z0 + ei2$$\eta$$(zbar - z0bar)

4. Apr 1, 2009

### HallsofIvy

Staff Emeritus
You seem to be confusing "reflection" with "rotation". $re^{i\theta}$ will give a rotation by angle $\theta$ together with an expansion (or contraction) by r, not a reflection. Given any point p, draw the straight line between p and Tb(p) and draw lines L1 and L2 perpendicular to that line 1/3 and 3/4 of the way between p and Tb(p). If R1 is reflection in L1, R1(p) will be the point, p2, exactly half way between p and Tb[/sup](p), and reflection of p2 in L2 will be Tb will be Tb(p) itself. Show that successive reflections of any point x in L1 and then L2 will give Tb[/sup](x).