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Multiplying complex numbers

  1. Nov 23, 2012 #1
    I have solved the roots of a quadratic equation and want to "test" them by putting them back in for x. I am having a problem with the x^2 term. Of the two roots, I'm only trying so far the positive square root case. I am trying to avoid writing all my work out since that would be hell and I also think would not be easy to read.

    To evaluate x^2 where x = -2/3 + i√19, I first try the "from the ground up" method of just distributing, and eventually make a substitution of -1 for i^2. After making that substitution, I end up subtracting the term that had the i^2, from (9/4). After simplifying, I have (-5/2) + (-3i√19)/2

    The second way I try to evaluate x^2 term is by using the multiplication definition of complex numbers (a + bi)(c + di) = (ac - bd) + (bc + ad)i . I find that (ac - bd) results in (9/4) - (19i^2)/4 and when I change i^2 to -1, now the terms are being added! Then when I add the (bc + ad)i part, I finally end up with (14/2) + (-3i√19)/2 .

    Could someone point out the mistake here then? :S
     
  2. jcsd
  3. Nov 23, 2012 #2
    I am new to complex numbers so don't assume anything with me. I read somewhere on wiki that it is sometimes relevant for a +bi when b is negative to keep b itself negative as supposed to write as a - bi ???? I don't really remember about it much though, but just throwing it out there.
     
  4. Nov 23, 2012 #3

    Mark44

    Staff: Mentor

    If we can't see what you did, how can we provide help?
    This is incorrect. Since you didn't show your work, I have no way to point out where you went wrong.
    You have mistake here, as well, in both calculations.

    a = c = -2/3, and b = d = √19. Note that b and d are the coefficients of i. They do not include i.

    ac - bd = 4/9 - 19


    What do you get for bc + ad? That is the coefficient of i in the product.
     
  5. Nov 23, 2012 #4
    Ok i made some big mistakes in writing this firstly. Yikes. Need to start again...

    The roots of x^2 + 3x + 7 are x = -3/2 + (i√19)/2 and x = -3/2 - (i√19)/2

    x^2 when x = -3/2 + (i√19)/2
    = (-3 + i√19)/(2) x (-3 + i√19)/(2)
    = (9/4 - (3i√19)/4 - (3i√19)/4 - 19/4) switching i^2 with -1
    =(-10/4 - (6i√19)/4)
    =(-5/2 - (3i√19)/2)


    Now plugging in terms with the definition....
    ( (9/4 - (19i^2)/4) + ( (-3i√19)/4 - (3i√19)/4 ) )
    = ( (9 + 19)/4 + (-6i√19)/4 )
    = 28/4 - (6i√19)/4
    =14/2 - (3i√19)/4
     
  6. Nov 23, 2012 #5

    SteamKing

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    Staff Emeritus
    Science Advisor
    Homework Helper

    Try this: (a+b)^2 = a^2 + 2ab + b^2
    ((-3/2) + i*SQRT(19)/2)^2 = (-3/2)^2 + 2(3/2)(i*SQRT(19)/2) + (19/4)*i^2)
    = 9/4 + (-1)*(19/4) + (3/2)(i*SQRT(19)/2
    = (9-19)/4 + (3/4)*SQRT(19)*i
    = -5/2 + 3i*SQRT(19)/4

    You have to be careful handling - signs and i^2 in the same expression, or you might inadvertently make a mistake.
     
  7. Nov 23, 2012 #6
    OK i think I got it now.
    I had the right idea with my distributive method and it brings me to right answer. When it comes to just plugging in the needed values "using the definition", it boiled down to what mark mentioned; that being to exclude i from term b and d.

    Thanks
     
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