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Finding original algebraic equation

  1. Oct 7, 2009 #1
    1. The problem statement, all variables and given/known data

    if I have a root of a +/- bi

    i = imaginary number

    what should the original algebra eqution?


    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 7, 2009 #2

    Mark44

    Staff: Mentor

    The original polynomial should have factors of (x - (a + bi))(x - (a - bi)), or equivalently, a single factor of (x2 - 2ax + a2 + b2). If the degree of the polynomial you're working with is larger than 2, there will be other roots, as well.
     
  4. Oct 7, 2009 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    x- (a+ bi) and x- (a- bi) are factors. Multiply [itex](x- (a+bi))(x- (a- bi)= ((x-a)- bi)((x-a)+bi)= (x-a)^2+ b^2[/itex] and set it equal to 0.
     
  5. Oct 7, 2009 #4
    thanks Mark, shouldn't it be a^2 - b^2?
     
  6. Oct 8, 2009 #5

    Mark44

    Staff: Mentor

    Nope. The a2 + b2 comes from multiplying (-(a + bi))(-(a - bi)) = (a + bi)(a - bi) = a2 -b2i2 = a2 + b2.
     
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