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Math Help - de Moivre's Theorem

  1. #1
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    de Moivre's Theorem

    Hello.

    when you see the\theta which is unknown

    i need to solve the following by Moivres theorem

    r^3(cos 3\theta + i sin 3\theta )=1..how can i find the \theta ?
    i thought that the Real part equals 1 as you can see...and the imaginary part is 0 since is not there...so
    r^3 cos3\theta =1
    r^3 sin3\theta =0

    so to find the angle \theta i tried r^3 sin3\theta devided by r^3 cos3\theta equals 0 devided by 1...then the next step would be sin3\theta devided cos3\theta equals 0...therefore tan3\theta =0...from now on i am stuck...but i am not sure it was the right step to follow...any advices how to get ahead?
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  2. #2
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    Quote Originally Posted by domenfrandolic View Post
    when you see the\theta which is unknown
    i need to solve the following by Moivres theorem
    r^3(cos 3\theta + i sin 3\theta )=1..how can i find the \theta ?
    It seems as if you have just thrown us into the middle of a problem.
    There is really no way to help you if you do not give us the original question in its whole form. Please state the entire question being solved.
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  3. #3
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    z^3-1=0...but it has to be solved by using the moivre's theorem
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    Quote Originally Posted by domenfrandolic View Post
    z^3-1=0...but it has to be solved by using the moivre's theorem
    Had you done that to begin with, would have gotten help much sooner.

    1=\cos(0)+\mathbf{i}\sin(0). That is one cube root of 1.
    The three cube roots are
    \cos\left(\frac{2k\pi}{3}\right)+\mathbf{i}\sin \left(\frac{2k\pi}{3}\right),~k=0,1,2
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  5. #5
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    Quote Originally Posted by domenfrandolic View Post
    Hello.

    when you see the\theta which is unknown

    i need to solve the following by Moivres theorem

    r^3(cos 3\theta + i sin 3\theta )=1..how can i find the \theta ?
    i thought that the Real part equals 1 as you can see...and the imaginary part is 0 since is not there...so
    r^3 cos3\theta =1
    r^3 sin3\theta =0

    so to find the angle \theta i tried r^3 sin3\theta devided by r^3 cos3\theta equals 0 devided by 1...then the next step would be sin3\theta devided cos3\theta equals 0...therefore tan3\theta =0...from now on i am stuck...but i am not sure it was the right step to follow...any advices how to get ahead?
    Since r^3 cos(3theta)= 1, r cannot be 0 and so from r^3 sin(theta)= 0, you have sin(theta)= 0. Either theta= 0 or theta= pi. If theta pi, then r^3 cos(theta)= -r^3= 1 which is impossible because r is never negative: theta= 0 so that r^3 cos(theta)= r^3= 1 and r= 1. That, is one solution to z^3= 1 is z= 1!!! If you found that difficult then finding the other two solutions is going to be really difficult.
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  6. #6
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    Quote Originally Posted by Plato View Post
    Had you done that to begin with, would have gotten help much sooner.

    1=\cos(0)+\mathbf{i}\sin(0). That is one cube root of 1.
    The three cube roots are
    \cos\left(\frac{2k\pi}{3}\right)+\mathbf{i}\sin \left(\frac{2k\pi}{3}\right),~k=0,1,2
    how did you come out for those cube roots?it looks very good though...so with it then i find 3 solutions right?
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  7. #7
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    Quote Originally Posted by HallsofIvy View Post
    Since r^3 cos(3theta)= 1, r cannot be 0 and so from r^3 sin(theta)= 0, you have sin(theta)= 0. Either theta= 0 or theta= pi. If theta pi, then r^3 cos(theta)= -r^3= 1 which is impossible because r is never negative: theta= 0 so that r^3 cos(theta)= r^3= 1 and r= 1. That, is one solution to z^3= 1 is z= 1!!! If you found that difficult then finding the other two solutions is going to be really difficult.
    so basically yeah...so why you put that theta is either 0 or pi?for the rest is clear
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    Quote Originally Posted by domenfrandolic View Post
    how did you come out for those cube roots?it looks very good though...so with it then i find 3 solutions right?

    Writing z^3 = r(\cos(\theta+2k\pi)+i\sin(\theta+2k\pi)=1, you get by de Moivre's Formula

    that z = r^{1/3}\left(\cos\left(\frac{\theta+2k\pi}{3}\right)+i\  sin\left(\frac{\theta+2k\pi}{3}\right)\right)=1 , k=0,1,2

    (why can we take only those three integer values of k?) , and from here that the answer you've been given.

    Tonio
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