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Complex Numbers : Urgent

  1. Sep 14, 2007 #1
    1. The problem statement, all variables and given/known data
    I have an exam coming up on Monday, and I cant seem to solve this question. Please point me in the right direction.

    [tex]x=e^{i\alpha}[/tex], [tex]y=e^{i\beta}[/tex] [tex]z=e^{i\gamma}[/tex].

    If [tex]x+y+z=xyz[/tex], prove that,

    [tex]cos(\beta -\gamma) + cos(\gamma -\alpha) + cos(\alpha -\beta) + 1=0[/tex]
    2. Relevant equations

    [tex]e^{ix}=cosx +i sinx[/tex]

    3. The attempt at a solution

    I tried to go from x+y+z=xyz by using the eular form but couldnt get anywhere.

    Then I multiplied the result by two to get [tex]2cos(\beta -\gamma) + 2cos(\gamma -\alpha) + 2cos(\alpha -\beta) + 2=0[/tex].

    Splitting the cosine terms into the eular form and then rearranging the equation, I got:


    Now, if I can prove
    [tex]e^{2i\alpha}+e^{2i\beta}+e^{2i\gamma}-(e^{3i\alpha}+e^{3i\beta}+e^{3i\gamma}=-2[/tex], then by implication, x+y+z=xyz as

    [tex]e^{i\alpha}+e^{i\beta}+e^{i\gamma}=x+y+z[/tex] and [tex]e^{i(\alpha+\beta+\gamma)}=xyz[/tex].

    How do I go about this?
  2. jcsd
  3. Sep 14, 2007 #2


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    Homework Helper

    The way I'd approach it... I'd try to simplify (y/z) + (z/x) + (x/y)... and see what comes out... the real part of this sum is:

    [tex]cos(\beta -\gamma) + cos(\gamma -\alpha) + cos(\alpha -\beta)[/tex]...
  4. Sep 14, 2007 #3
    I just tried that, here's where Im getting stuck with it:

    [tex]\frac{xy^2+yz^2+zx^2+xyz}{xyz}[/tex]. Where do I go from here?
  5. Sep 14, 2007 #4


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    Homework Helper

    Yeah, I wasn't able to get anywhere with that either...

    using x + y + z = xyz

    you know that... cos(alpha) + cos(beta) + cos(gamma) = cos(alpha + beta + gamma)

    you also know that sin(alpha) + sin(beta) + sin(gamma) = sin(alpha + beta + gamma)

    square both equations and add... with the cos(A-B) identity... this will give you the result.
  6. Sep 15, 2007 #5
    Thank you. Thats excellent! You're a lifesaver!
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