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Rotating two different complex numbers

  1. Jul 7, 2010 #1


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    Let's say that I have two complex numbers, a and b, with different arguments. From a few "experiments" with a computer, I think that there always exists a positive integer n such that -pi/2 <= Arg(a^n) <= pi/2 and pi/2 <= Arg(b^n) <= 3pi/2.
    In other words, if Arg(a) = thetaA and Arg(b) = thetaB, then there exists an n such that
    -pi/2 <= n*thetaA <= pi/2 and pi/2 <= n*thetaB <= 3pi/2 up to a multiple of 2pi.
    Is this true? If so, how can I start to prove it?
  2. jcsd
  3. Jul 7, 2010 #2

    Filip Larsen

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    It's not true for a = -1 and b = 1. In general, if both a and b have rational arguments they will only have a finite number of integer power images so it seems likely that you may find a whole range of argument pairs for which your condition will never be satisfied.
  4. Jul 7, 2010 #3
    Do you just need them to be in different halves? In other words, is it sufficient that
    |\textrm{arg}(a^n)-\textrm{arg}(b^n)| = n\cdot |\textrm{arg}(a)-\textrm{arg}(b)| \geq \pi
    -modulo [itex] 2\pi[/itex] of course?

    As Flip Larsen commented, you can't necessarily pick which is in which half...the above is definitely true, though. The proof is pretty much immediate.
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