Can non-zero translations be composed of two reflections in perpendicular lines?

[zcdfhn]

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

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

My guess at how to start this is to assume b = re^i$$\theta$$ where r = |b| and $$\theta$$=arg b, so then the direction of b is b/|b| = e^i$$\theta$$. Therefore the unit vector with the direction orthogonal to b would be c = e^{i($$\theta$$ + $$\pi$$/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 T_b but my work gets more and more complicated.

Thanks for your help.

 

[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.

 

[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 + e^i2$$\eta$$(zbar - z₀bar)

 

[HallsofIvy]

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 T_b(p) and draw lines L1 and L2 perpendicular to that line 1/3 and 3/4 of the way between p and T_b(p). If R1 is reflection in L1, R1(p) will be the point, p2, exactly half way between p and T_{b[/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).}