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(Algebra) Isometries on the complex plane

  1. Feb 19, 2012 #1
    So this is the problem as written and I'm totally lost. Any help or explanation would be greatly appreciated.

    "Viewing ℂ=ℝ2 , we can identify the complex numbers z = a+bi and w=c+di with the vectors (a,b) and (c,d) in R2 , respectively. Then we can form their dot product, (a,b)[itex]\bullet[/itex](c,d)=ac+bd.
    Prove that ζ[itex]\bar{c}[/itex] + c = 0 iff c is orthogonal to [itex]\sqrt{ζ}[/itex]."

    I feel like there are too many things undefined - or maybe I just don't get what things are supposed to be. [itex]\bar{c}[/itex] is supposed to be the conjugate of c, I know that much. And in another problem ζ was defined as cosθ + i sinθ, but I'm not sure how to use this information (or if it even applies to this problem). I know that c and ζ being orthogonal means they're both vectors and their dot product is zero. It's in the chapter on isometries of ℝ and ℂ.

    I've been trying to figure this problem out for hours and am frustrated to the point of tears. Please help.
     
  2. jcsd
  3. Feb 19, 2012 #2

    micromass

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    c orthogonal to [itex]\sqrt{\xi}[/itex]? That doesn't even make sense. Those two things are numbers, what does it even mean that they're orthogonal??
     
  4. Feb 19, 2012 #3
    You can see my frustration then. I checked to make sure I copied the problem word for word. I'm so confused by it - it's making me crazy.

    The only thing I can think is ζ is a rotation (as it had been used before). So it makes sense to me that ζ[itex]\bar{c}[/itex] = -c, where ζ rotates the conjugate -90°. but the c orthogonal to √ζ is what's killing me.

    I'm totally lost.
     
  5. Feb 19, 2012 #4

    morphism

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    Assuming [itex]\zeta = \cos\theta + i \sin\theta[/itex] and [itex]\sqrt \zeta = \cos\frac \theta 2 + i \sin\frac \theta 2[/itex], and writing [itex]c = a+ib[/itex], we see c is orthogonal to [itex]\zeta[/itex] iff [itex]a \cos\frac \theta 2 + b\sin\frac \theta 2 = 0[/itex] iff [itex]\tan\frac\theta 2 = -\frac ab[/itex]. Now use the usual formulas expressing [itex]\sin\theta[/itex] and [itex]\cos\theta[/itex] in terms of [itex]\tan\frac\theta 2[/itex] to see that [itex]\zeta \bar c + c = 0[/itex] iff [itex]\tan\frac\theta 2 = -\frac ab[/itex].
     
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