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Math Help - Complex Polar Forms, Sin and Cos Angles

  1. #1
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    Complex Polar Forms, Sin and Cos Angles

    Problem: Find all solutions of z^{3} = -8.

    I can do the entire problem except for one part. After putting it into the correct form,

    |z|^{3}(cos3\theta+isin3\theta),

    I do not know how to find the values of cos3\theta or sin3\theta. I know that |z|^3 is 8, but I can't figure out the values of cos and sin.

    Any help is appreciated.
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  2. #2
    MHF Contributor alexmahone's Avatar
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    Re: Complex Polar Forms, Sin and Cos Angles

    Quote Originally Posted by tangibleLime View Post
    Problem: Find all solutions of z^{3} = -8.

    I can do the entire problem except for one part. After putting it into the correct form,

    |z|^{3}(cos3\theta+isin3\theta),

    I do not know how to find the values of cos3\theta or sin3\theta. I know that |z|^3 is 8, but I can't figure out the values of cos and sin.

    Any help is appreciated.
    |z| = 2.
    So, cos\ 3\theta+isin\ 3\theta=-1.
    Equate real and imaginary terms to find \theta.
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  3. #3
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    Re: Complex Polar Forms, Sin and Cos Angles

    Thanks,

    So since |z|^3 = 8, I need cos3\theta+isin3\theta to equal -1 to satisfy the initial equation where z^3 = -8. The only way to get -1 from cos3\theta+isin3\theta is to have cos3\theta = -1 and sin3\theta = 0 to get rid of the imaginary number and return (-1 + i0).

    Correct?
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  4. #4
    MHF Contributor alexmahone's Avatar
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    Re: Complex Polar Forms, Sin and Cos Angles

    Quote Originally Posted by tangibleLime View Post
    Thanks,

    So since |z|^3 = 8, I need cos3\theta+isin3\theta to equal -1 to satisfy the initial equation where z^3 = -8. The only way to get -1 from cos3\theta+isin3\theta is to have cos3\theta = -1 and sin3\theta = 0 to get rid of the imaginary number and return (-1 + i0).

    Correct?
    Yes. You should get three values of \theta in the interval [0,\ 2\pi].
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  5. #5
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    Re: Complex Polar Forms, Sin and Cos Angles

    Quote Originally Posted by tangibleLime View Post
    Problem: Find all solutions of z^{3} = -8.
    Here is some notation: \exp(i\theta)=\cos{\theta)+i\sin(\theta)

    So we can write -8=8\exp(\pi i)
    The cube roots of that is then 2\exp \left( {\frac{{i\pi }}{3} + \frac{{2\pi ik}}{3}} \right),~~k=0,1,2
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