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Complex roots

  1. Sep 19, 2006 #1
    the book asks to fnd the four roots of z to the fourth power + 4 = 0 and then use to demonstrate that z to the fourth power + 4 can be factored into two quadratics with real coefficinets. I am clueless on where to start. Please help.
     
  2. jcsd
  3. Sep 19, 2006 #2
    start by setting [tex] z^4=-4 [/tex] and solve for z taking the square root function twice (saving both roots)
     
  4. Sep 19, 2006 #3
    Ok so I did what you said and then did the quadratic formula so i have 4 roots but how do I form quadratic formulas?
     
  5. Sep 22, 2006 #4
    so let a,b,c,d be your roots. the general form of your equation can be written as (z-a)(z-b)(z-c)(z-d)=0. From that you can pick any two, multiply out and you'll have a quadratic ie: [tex] (z^2-bz-az+ab)(z-c)(z-d)=0 [/tex]. Since your roots are complex and you need real coefficients you might have to try different combinations
     
  6. Sep 22, 2006 #5

    HallsofIvy

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    Let y= z2 so that you have a quadratic equation
    y2+ 4= 0. What are the two solutions to that? Then, for each of the two values of y, set z2= y and take square roots.
     
  7. Sep 22, 2006 #6
    Ok, I noticed something while playing with your equation, you can factor out z^4 + 4 = 0 into
    (z^2 + 2i)(z^2 - 2i)
    then
    (z^2 + 2i) = (z + 1 - i)(z - 1 + i)
    (z^2 - 2i) = (z + 1 + i)(z - 1 - i)

    multiply
    (z + 1 - i)(z + 1 + i) = z^2 + 2z + 2
    (z - 1 + i)(z - 1 - i) = z^2 - 2z + 2

    and there you have your two real number quadratics
     
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