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Thread: 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 $\displaystyle u=z^2$

    Then: $\displaystyle u^2+2iu+1=0$ or:

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

    then:

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

    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 $\displaystyle z^2$, with solutions $\displaystyle z^2 = -i\bigl(1\pm\sqrt2\bigr)$. The square roots of $\displaystyle -i$ are $\displaystyle \pm\tfrac1{\sqrt2}(1-i)$, so the solutions for z are $\displaystyle z = \pm\tfrac1{\sqrt2}(1-i)\sqrt{1+\sqrt2}$ and $\displaystyle z = \pm\tfrac1{\sqrt2}(1-i)i\sqrt{\sqrt2-1}$. 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 $\displaystyle z^2$ are $\displaystyle z^2 = -i\bigl(1+\sqrt2\bigr)$ and $\displaystyle z^2 = i\bigl(\sqrt2-1\bigr)$. These have polar form $\displaystyle z^2 = re^{i\theta}$, where for the first root $\displaystyle r=1+\sqrt2$ and $\displaystyle \theta = -\pi/2$. For the second root $\displaystyle r = \sqrt2-1$ and $\displaystyle \theta = \pi/2$. 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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