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Math Help - Trignotmatric Problem

  1. #1
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    Trignotmatric Problem

    find all solution of Z^3/2=(1/2-i).
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  2. #2
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    Quote Originally Posted by winsome View Post
    find all solution of Z^3/2=(1/2-i).
    This is unreadable.

    Is the left hand side z^{\frac{3}{2}} or \frac{z^3}{2}?


    Is the right hand side \frac{1}{2} - i or \frac{1}{2-i}?
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  3. #3
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    Quote Originally Posted by Prove It View Post
    This is unreadable.

    Is the left hand side z^{\frac{3}{2}} or \frac{z^3}{2}?


    Is the right hand side \frac{1}{2} - i or \frac{1}{2-i}?

    I re-write this problem as below
    Find all solution of Z^(3/2)=[(1/2) - i]
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  4. #4
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    z^{\frac{3}{2}} = \dfrac{1}{2}-i \Leftrightarrow z = \left(\dfrac{1}{2}-i\right)^{\frac{2}{3}}.
    Last edited by TheCoffeeMachine; June 9th 2010 at 05:44 AM.
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  5. #5
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    Quote Originally Posted by TheCoffeeMachine View Post
    z^{\frac{3}{2}} = \dfrac{1}{2}-i \Leftrightarrow z = \left(\dfrac{1}{2}-i\right)^{\frac{3}{2}}.
    Actually it's

    \left(\frac{1}{2} - i\right)^{\frac{2}{3}}.

    And this is only one solution. There will be two others.

    Convert to polars, use DeMoivre's Theorem to simplify, then recall that the three roots are evenly spaced around a circle to find the other two.
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  6. #6
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    Quote Originally Posted by Prove It View Post
    Actually it's

    \left(\frac{1}{2} - i\right)^{\frac{2}{3}}.
    Thanks. That's what I had thought I have written.
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