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Math Help - Revision-complex number

  1. #1
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    Revision-complex number

    Find all roots of the equations
    (a) z^4 + 2z^2 + 4 = 0

    My step:

    Let u = z^2
    u^2 + 2u + 4 = 0

    u = -1 + (3)^1/2 and u = -1 - (3)^1/2

    i'm stuck at there...and tomorrow having complex number exam.
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  2. #2
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    Quote Originally Posted by wkn0524 View Post
    Find all roots of the equations
    (a) z^4 + 2z^2 + 4 = 0

    My step:

    Let u = z^2
    u^2 + 2u + 4 = 0

    u = -1 + (3)^1/2 and u = -1 - (3)^1/2

    i'm stuck at there...and tomorrow having complex number exam.
    Well, first your values for u are wrong. u= -1+ i 3^{1/2} and u= -1- i 3^{1/2}. You forgot the "i".

    Now use the fact that u= z^2.
    z^2= -1+ i3^{1/2} so z= \pm\sqrt{-1+ i 3^{1/2}}
    z^2= -1- i3^{1/2} so z= \pm\sqrt{-1- i3^{1/2}}.

    I presume you will want to use d'Moivre's formula to take those square roots. Since 1^2+ (3^{1/2})^2= 4 and \frac{3^{1/2}}{-1}= -\sqrt{3}= arctan(\pi/3}, in polar form, -1+ i3^{1/2}= 2(cos(\pi/3)+ i sin(\pi/3))= 2e^{i\pi/3} and -1- i3^{1/2}= 2(cos(\pi/3)- i sin(\pi/3))= 2e^{-i\pi/3}.
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  3. #3
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    Quote Originally Posted by HallsofIvy View Post
    Well, first your values for u are wrong. u= -1+ i 3^{1/2} and u= -1- i 3^{1/2}. You forgot the "i".

    Now use the fact that u= z^2.
    z^2= -1+ i3^{1/2} so z= \pm\sqrt{-1+ i 3^{1/2}}
    z^2= -1- i3^{1/2} so z= \pm\sqrt{-1- i3^{1/2}}.

    I presume you will want to use d'Moivre's formula to take those square roots. Since 1^2+ (3^{1/2})^2= 4 and \frac{3^{1/2}}{-1}= -\sqrt{3}= arctan(\pi/3}, in polar form, -1+ i3^{1/2}= 2(cos(\pi/3)+ i sin(\pi/3))= 2e^{i\pi/3} and -1- i3^{1/2}= 2(cos(\pi/3)- i sin(\pi/3))= 2e^{-i\pi/3}.

    I forgot put i, its mean -1. there are four solution.
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