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Math Help - Find all complex roots

  1. #1
    Member Jones's Avatar
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    Find all complex roots

    Hi,

    I need to find all roots to the equation z^4=-16i

    This is what i tried:

    |-16|^{1/4}=2

    And the angle of -16i is 270 degrees or \frac{3\pi}{2}

    All the roots are on the form |z|^{1/n}\cdot cos\frac{\theta}{n}+isin\frac{\theta}{n}

    So the first root should be 2\bigg(cos(\frac{3\pi}{8})+isin(\frac{3\pi}{8})\bi  gg)

    But it's just not correct.
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  2. #2
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    Quote Originally Posted by Jones View Post
    Hi,

    I need to find all roots to the equation z^4=-16i

    This is what i tried:

    |-16|^{1/4}=2

    And the angle of -16i is 270 degrees or \frac{3\pi}{2}

    All the roots are on the form |z|^{1/n}\cdot cos\frac{\theta}{n}+isin\frac{\theta}{n}

    So the first root should be 2\bigg(cos(\frac{3\pi}{8})+isin(\frac{3\pi}{8})\bi  gg)

    But it's just not correct.
    Why do you say that? It is one of the roots.
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  3. #3
    Member Jones's Avatar
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    Are you sure?

    <br />
2\bigg(cos(\frac{3\pi}{8})+isin(\frac{3\pi}{8})\bi  gg)<br />

    gives  -2i, (-2i)^4 =16
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  4. #4
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    Quote Originally Posted by Jones View Post
    I need to find all roots to the equation z^4=-16i
    Here is the standard way to do all of these ‘root’ problems.
    -16i=16\exp\left(\frac{-\pi i}{2}\right).
    Let  \omega =2\exp\left(\frac{-\pi i}{8}\right)
    Let  \xi =\exp\left(\frac{\pi i}{2}\right)

    Now it is really simple to list the four roots: \omega \cdot \xi^k,~k=0,1,2,3
    Last edited by Plato; February 6th 2010 at 04:48 PM.
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  5. #5
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    Quote Originally Posted by Jones View Post
    Are you sure?

    <br />
2\bigg(cos(\frac{3\pi}{8})+isin(\frac{3\pi}{8})\bi  gg)<br />

    gives  -2i, \text{{\color{red} Mr F says: This is wrong. }} (-2i)^4 =16
    Of course I am sure.
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  6. #6
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    An alternative method is to remember that if you're taking the 4th root, there will be 4 roots, and they are all evenly spaced around a circle.
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  7. #7
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    Quote Originally Posted by Jones View Post
    Are you sure?

    <br />
2\bigg(cos(\frac{3\pi}{8})+isin(\frac{3\pi}{8})\bi  gg)<br />

    gives  -2i, (-2i)^4 =16
    No, it doesn't! cos(\frac{3\pi}{8})= 0.38268, not 0. cosine is 0 only at odd multiples of \frac{\pi}{2}.
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