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Math Help - Complex Number Problem

  1. #1
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    Complex Number Problem

    Ok so the problem is as follows:

    Find all real number solutions y of:

    e^(iy) + e^(2iy) + e^(3iy)=0

    Firstly I tried letting e^(iy) equal some variable X. Then:

    X + X^2 + X^3=0 ==> X(1 + X + X^2)=0.

    I used the quadratic formula on (1 + X + X^2) to get X=-1/2 (sqrt(3)/2)*i

    So e^(iy)=
    -1/2 (sqrt(3)/2)*i

    Where do I go from here? Am I even on the right track? Any help much appreciated!
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  2. #2
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    Re: Complex Number Problem

    e^{i y}+e^{2i y}+e^{3i y}=e^{2i y}\left(1+e^{i y}+e^{-i y}\right)=e^{2i y}(1+2\cos  y)
    Last edited by Idea; September 27th 2013 at 10:31 AM.
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  3. #3
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    Re: Complex Number Problem

    Quote Originally Posted by CrispyPlanet View Post
    Ok so the problem is as follows:

    Find all real number solutions y of:

    e^{iy} + e^{2iy} + e^{3iy}=0
    That is: \cos(y)+\cos(2y)+\cos(3y)+i(\sin(y)+\sin(2y)+\sin(  3y))=0

    So you need \cos(y)+\cos(2y)+\cos(3y)= 0~\&~\sin(y)+\sin(2y)+\sin(3y)= 0
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  4. #4
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    Re: Complex Number Problem

    Idea and Plato have used different methods but what you have done is perfectly valid.

    Of course, iy= ln(-1/2+ i\sqrt{3}/2, in the first case, and iy= ln(-1/2- i\sqrt{3}/2. To find that logarithm, write the numbers in "polar" or exponential form: \sqrt{(-1/2)^2+ (\sqrt{3}/2)^2}= \sqrt{1/4+ 3/4}= 1 and arctan(\frac{\sqrt{3}/2}{1/2})= arctan(\sqrt{3})= \frac{\pi}{3} so that we have ln(e^{i\pi/3})= i\pi/3 so that iy= i\pi/3 and then y= \pi/3.

    The logarithm, like most functions extended to the complex numbers, is not "single valued". e^{i(x+ 2k\pi)}= e^{ix}, for any integer k, so that ln(re^{ix})= ln(re^{i(x+ 2k\pi)}= ln(r)+ (x+  2\pi) i. that is, we can add any multiple of 2\pi to the solution above.
    Last edited by HallsofIvy; September 27th 2013 at 11:04 AM.
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  5. #5
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    Re: Complex Number Problem

    Hi HallsofIvy,

    When I plug in pi/3 into the polar formula (cosy+cos2y+cos3y+i(siny+sin2y+sin3y)=0), the result is not zero. What am I missing? I've tried working with the cosines and sines separately, making them equal to zero, but I'm not getting anywhere with that either.

    CP
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  6. #6
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    Re: Complex Number Problem

    Quote Originally Posted by Idea View Post
    e^{i y}+e^{2i y}+e^{3i y}=e^{2i y}\left(1+e^{i y}+e^{-i y}\right)=e^{2i y}(1+2\cos  y)
    Hi Idea, many thanks for your response. I worked out your method and got y=2pi/3, which makes the equation equal to zero. So the solution is y=(2pi/3 +2k*pi).
    What I'm not certain about is why my method yielded y=pi/3.

    CP
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