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Power of Complex numbers proof

  1. Sep 10, 2009 #1
    1. The problem statement, all variables and given/known data

    Prove that [tex]\left( {\frac{{1 + \cos x + i\sin x}}{{1 + \cos x - i\sin x}}} \right)^n} = \cos nx + i\sin nx[/tex]

    3. The attempt at a solution

    I thought it would be a good idea calling z = 1 + cos x + i*sen x, because then 1 + cos x - i*sen x would be [tex]\bar{z}[/tex], and then I would have something of the form [tex]{\left( {\frac{z}{{\bar z}}} \right)^n} = {\left( {\frac{{{z^2}}}{{{{\left| z \right|}^2}}}} \right)^n}[/tex]. If I use the Euler form for complex numbers, being [tex]\varphi[/tex] the argument of z, then z[tex]\left| z \right|{e^{i\phi }}[/tex].

    So [tex]{\left( {\frac{{{z^2}}}{{{{\left| z \right|}^2}}}} \right)^n} = {\left( {\frac{{{{(\left| z \right|{e^{i\phi }})}^2}}}{{{{\left| z \right|}^2}}}} \right)^n} = {\left( {\frac{{{{\left| z \right|}^2}{e^{i2\phi }}}}{{{{\left| z \right|}^2}}}} \right)^n} = {\left( {{e^{i2\phi }}} \right)^n} = {e^{i2n\phi }}\
    [/tex]. Which would all work just fine with the proof if I could somehow prove that [tex]\varphi[/tex] = x/2.

    My question is, is this OK? And if it is OK, how can I prove that [tex]\varphi[/tex] = x/2?

    Thanks
     
  2. jcsd
  3. Sep 10, 2009 #2

    Mark44

    Staff: Mentor

    Aren't the numerator and denominator equal to 1 + eix and 1 + e-ix respectively? That's the direction I would go. And the right side is ei*nx = (ei x)n = (cos x + i sin x)n,

    One comment about your work: You have the exponent wrong on the right side. Your equation should be as follows:
    [tex]{\left( {\frac{z}{{\bar z}}} \right)^n} = {\left( {\frac{{{z^2}}}{{{{\left| z \right|}^2}}}} \right)^{n/2}}[/tex]
     
  4. Sep 10, 2009 #3

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    That doesn't look right....
     
  5. Sep 10, 2009 #4
    Yes, that's basically what I did. Only that I recalled the whole numerator z so that I could deal with the division.

    No, why? [tex]\frac{z}{{\bar z}} = \frac{{{z^2}}}{{{{\left| z \right|}^2}}}[/tex]
     
  6. Sep 10, 2009 #5

    mheslep

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    Gold Member

    It is not right. Try, e.g., n=2 for z=(cos x + i sin x)n and we have

    (cos x + i sin x)2= [tex](cos^2x-sin^2x) + i(2cosxsinx)[/tex], but
    [tex]\left| z \right|^2 = \left| cos x + isinx \right|^2=1[/tex]
     
  7. Sep 11, 2009 #6

    Mark44

    Staff: Mentor

    [quote Originally posted by Mark44]
    One comment about your work: You have the exponent wrong on the right side. Your equation should be as follows:
    [tex]{\left( {\frac{z}{{\bar z}}} \right)^n} = {\left( {\frac{{{z^2}}}{{{{|z|}^2}}}} \right)^{n/2}}[/tex]
    [/quote]
    I agree. I didn't notice that the conjugate in the denominator on the left had changed to the square of the magnitude in the expression on the right.
     
  8. Sep 11, 2009 #7

    Mark44

    Staff: Mentor

    Back to the original problem...
    [tex]Prove~that~\left( {\frac{{1 + \cos x + i\sin x}}{{1 + \cos x - i\sin x}}} \right)^n} = \cos nx + i\sin nx[/tex]

    [tex]\left( {\frac{{1 + \cos x + i\sin x}}{{1 + \cos x - i\sin x}}} \right)^n} = \left(\frac{1 + e^{ix}}{1 + e^{-ix}}\right)^n[/tex]

    Inside the parentheses, multiply numerator and denominator by eix. Then factor eix out of the two terms in the numerator and cancel the factor that is common to numerator and denominator. It's then easy to show that what you have left equals cos(nx) + isin(nx) = einx = (eix)n.
     
  9. Sep 12, 2009 #8
    EDIT: I hadn't seen that comment from Mark44.

    Thanks, I wouldn't have thought that.
     
    Last edited: Sep 13, 2009
  10. Sep 13, 2009 #9
    Edit.
     
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