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Math Help - Complex roots

  1. #1
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    Complex roots

    Having some difficulty with finding the four complex roots to the equation

    Z^4 + 2i*Z^2 + 1


    thanks
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  2. #2
    Junior Member SirOJ's Avatar
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    From what i understand you Factorize first, then use the difference of two square rule.

    If in doubt just use the -b forumula though
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  3. #3
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    Not that simple, but thanks.

    I have broken it into two quadratics, but as its complex I need to represent the alternative roots in terms of complex exponentials.

    I am having some trouble on this stage as I think i should square root something twice, whereas my answers tells me I don't need to


    thanks again for trying
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  4. #4
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    Quote Originally Posted by Bryn View Post
    Having some difficulty with finding the four complex roots to the equation

    Z^4 + 2i*Z^2 + 1


    thanks
    That's just a quadratic right?

    Let u=z^2

    Then: u^2+2iu+1=0 or:

    u=\frac{-2i\pm i 2 \sqrt{2}}{2}

    then:

    z=\sqrt{-i(1\pm \sqrt{2})}

    So you got two cases in the square root, and each case has two square root.
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  5. #5
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    The equation is a quadratic in z^2, with solutions z^2 = -i\bigl(1\pm\sqrt2\bigr). The square roots of -i are \pm\tfrac1{\sqrt2}(1-i), so the solutions for z are z = \pm\tfrac1{\sqrt2}(1-i)\sqrt{1+\sqrt2} and z = \pm\tfrac1{\sqrt2}(1-i)i\sqrt{\sqrt2-1}. I don't think there is any way of avoiding those repeated square roots.
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  6. #6
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    Yes,

    Thanks, I didn't explain exactly my problem I guess.

    I need to write the solutions in terms of complex exponentials which is what I was finding tricky - anychance you could try that

    thanks
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  7. #7
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    Quote Originally Posted by Bryn View Post
    I need to write the solutions in terms of complex exponentials which is what I was finding tricky - any chance you could try that
    The solutions for z^2 are z^2 = -i\bigl(1+\sqrt2\bigr) and z^2 = i\bigl(\sqrt2-1\bigr). These have polar form z^2 = re^{i\theta}, where for the first root r=1+\sqrt2 and \theta = -\pi/2. For the second root r = \sqrt2-1 and \theta = \pi/2. To find the values of z, you need the square roots of these. That means taking the square root of r and halving the value of theta.
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