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Thread: find complex numbers

  1. #1
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    find complex numbers

    find all complex numbers, z, which satisfy the equation $\displaystyle z^2 = 6 -8i$

    do i begin by writing $\displaystyle z^2 - 6 - 8i = 0$ and then solve?
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  2. #2
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    Quote Originally Posted by murielx View Post
    find all complex numbers, z, which satisfy the equation $\displaystyle z^2 = 6 -8i$

    do i begin by writing $\displaystyle z^2 - 6 - 8i = 0$ and then solve?
    Yes.

    $\displaystyle z^2 - 6 - 8i = 0$

    $\displaystyle z^2 - (6 + 8i) = 0$

    $\displaystyle z^2 - (\sqrt{6 + 8i})^2 = 0$

    $\displaystyle (z + \sqrt{6 + 8i})(z - \sqrt{6 + 8i}) = 0$


    $\displaystyle z = -\sqrt{6 + 8i}$ or $\displaystyle z = \sqrt{6 + 8i}$.


    Now the hard part is evaluating these square roots. I'd advise converting to Polars and using DeMoivre's Theorem.
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  3. #3
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    Another way to do this is to write z= x+ yi so that $\displaystyle z^2= x^2- y^2+ 2xyi= 6- 8i$

    Now solve the two equations $\displaystyle x^2- y^2= 6$ and $\displaystyle 2xy= -8$ for real x and y.
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  4. #4
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    I used the way suggested by HallsofIvy and I have become stuck at a certain point....

    $\displaystyle x^2 - y^2 = 6$

    $\displaystyle 2xy = -8$


    $\displaystyle y = - \frac{4}x$

    .:.
    $\displaystyle x^2 - (-\frac{4}x)^2 = 6$

    $\displaystyle 6 = x^2 + \frac{16}{x^2}$


    so $\displaystyle x^4 - 6x^2 + 16 = 0$


    here is where I am stuck and cant seem to find the two values of x

    I know what to do after finding these values.

    thanks
    Last edited by murielx; Mar 8th 2010 at 12:04 AM.
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