Complex roots

  • Thread starter becka12
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  • #1
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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.
 

Answers and Replies

  • #2
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start by setting [tex] z^4=-4 [/tex] and solve for z taking the square root function twice (saving both roots)
 
  • #3
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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?
 
  • #4
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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
 
  • #5
HallsofIvy
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becka12 said:
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?
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.
 
  • #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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