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Complex number problem

  1. May 2, 2005 #1
    Hi,
    I'm looking at a question from my Pure 6 textbook (united kingdom), it's not actually for homework but I'd like to figure it out.
    First part of the question goes like this:

    If

    2 cos θ = z + z^-1

    prove that (if n is a positive integer)

    2 cos n θ = z^n + z^-n.


    I can get a solution as follows:

    Let z = cos θ + i sin θ
    then z^-1 = cos (-θ) + i sin (-θ) by de Moivre's theorem
    so z^-1 = cos θ - i sin θ
    so z + z^-1 = 2 cos θ

    similarly z^n = cos nθ + i sin nθ
    z^-n = cos -nθ + i sin -nθ
    z^-n = cos nθ - i sin nθ
    so z^n + z^-n = 2 cos n θ


    However I'm not sure if this is valid because of the first line where I let z=cos θ + i sin θ. The question does not state what z is, so can it be assumed to be any complex number? If so then I don't think z=cos θ + i sin θ is valid because that only works when the modulus of z is 1, right? Or can cos θ + i sin θ represent complex numbers of modulus other than 1, maybe if θ itself is complex? But then would de Moivre's theorem still be appliciable?

    Or does the fact that the question states 2 cos θ = z + z^-1 imply that z must be representable by cos θ + i sin θ ?

    Thanks very much for any help with this.
     
  2. jcsd
  3. May 2, 2005 #2

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    Given theta, you have a second order equation for z. This means there will be two solutions. It's not hard to see they will be reciprocals of eachother. So once you find all the zs and prove what you need to with all of them, you have the general proof. And yes, a complex theta will give complex numbers of modulus other than 1, and your proof extends to cover them, but I doubt that was intended by the question.

    By the way, if you want to be a little more rigorous, you can let [itex]z = r e^{i \theta} [/itex] and show that r must equal one if the imaginary part of z+1/z is to vanish, assuming theta is real.
     
    Last edited: May 2, 2005
  4. May 3, 2005 #3
    There are many ways!

    1. By POMI: proove it for n = 2, and assume for n=k and show it holds for n=k+1 also.

    :smile:

    2. Use z=r*exp[i{theta}].
    So,
    2cos{theta} = r*exp[i{theta}] + r*exp[-i{theta}]
    So, we can see that
    Re[r*exp[i{theta}] + r*exp[-i{theta}]] = 2cos{theta}
    So, that means...
    rcos{theta}+(1/r)cos{theta} = 2cos{theta}
    => r + 1/r = 2
    Or
    => cos{theta} = {pi}/2 !

    Now I guess you are OK?
     
  5. May 3, 2005 #4
    Thanks for the assistance!
     
  6. May 3, 2005 #5
    Yeah, the way the problem was stated it totally looked like it was setting you up for a proof by induction.

    But you wouldn't need to prove it true for n=2, the problem allows you to assume it's true for n=1 (no proof necessary) so proving it true for n=k+1 when assuming true for n=k would be sufficient. That part I'm not sure how to do.
     
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