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Linear algebra - find all solutions with complex numbers

  1. Sep 30, 2007 #1
    (a)Find all t [tex]\epsilon C[/tex] such that [tex]t^{2}[/tex] + 3t + (3-i) = 0. Express your solution(s) in teh form x+iy where x,y [tex]\epsilon R.[/tex]

    (b) Prove that | 1+iz | = | 1-iz | if and only if z is real.

    Okay so I tried to use the quadratic formula to find the roots to find the solutions, but I am stuck because I have a complex number within the square roots.

    t = -b +/- sqrt(b^2 - 4ac) / 2a
    t = -3 +/- sqrt[(-3)^2 - 4(1)(3-i)] / 2(1)
    t = -3 +/- sqrt(-3 + 4i) / 2

    what do i do?

    also for question (b), where do I even start?
  2. jcsd
  3. Sep 30, 2007 #2

    matt grime

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    Homework Helper

    1. Well, you should know how to take square roots of imaginary numbers, if you're supposed to use that method. Since you do not, it seems, why not retry the question without using the quadratic formula which you probably weren't supposed to use anyway. For example, I hope you wouldn't use the quadratic formula on x^2 - 5x +6 to find the roots of 2 and 3.

    2. Iz z=x+iy, you want to show y=0. Well, what does the condition |1-iz|=|1+iz| imply?
  4. Sep 30, 2007 #3
    I have tried finding the roots like a normal quadratic, but the last term (3-i) is throwing me off.
  5. Sep 30, 2007 #4
    is there anyway to solve this instead of finding the roots like a normal quadratic.. is there an actual format to do that.. i seem to be just guessing..
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