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Isometries (Translations)

  1. Mar 31, 2009 #1
    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[tex]\in[/tex]C

    My guess at how to start this is to assume b = rei[tex]\theta[/tex] where r = |b| and [tex]\theta[/tex]=arg b, so then the direction of b is b/|b| = ei[tex]\theta[/tex]. Therefore the unit vector with the direction orthogonal to b would be c = ei([tex]\theta[/tex] + [tex]\pi[/tex]/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[tex]\circ[/tex]f and I should end up with Tb but my work gets more and more complicated.

    Thanks for your help.
    Last edited: Mar 31, 2009
  2. jcsd
  3. Mar 31, 2009 #2


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    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.
  4. Mar 31, 2009 #3
    I know M(z) = a(z-bar)+b, where |a| = 1 and a [tex]\neq[/tex] 1, a(b-bar) + b = 0

    and i also know M(z) = z0 + ei2[tex]\eta[/tex](zbar - z0bar)
  5. Apr 1, 2009 #4


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    You seem to be confusing "reflection" with "rotation". [itex]re^{i\theta}[/itex] will give a rotation by angle [itex]\theta[/itex] 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).
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