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Complex division:

  1. Mar 6, 2007 #1
    I'm working on the proof that two complex numbers can be divided from Alhford and I'm completely s(t)uck.

    I've gotten as far as:

    a = gx - dy
    b = dx +gy

    from
    (a+ib) / (g + id)

    where

    a+ib = (g + id)(x+iy)

    I've managed to get

    [tex]x={\frac {b-{\it gy}}{d}}[/tex]

    and done the same for y (latex is a bit hard for me right now so I can't really show both and I'd like to have a go at it myself when I get what is done for x)

    But the book says I'm mean to get:
    [tex]x={\frac {{\it ag}+{\it bd}}{{g}^{2}+{d}^{2}}}[/tex]
    Help?
     
  2. jcsd
  3. Mar 6, 2007 #2

    AKG

    User Avatar
    Science Advisor
    Homework Helper

    You want to solve for x and y. If you have an expression for x that involves y, you haven't really solved for x. You have deduced two equations

    a = gx - dy
    b = dx +gy

    You used the second one to isolate x and got

    x = (b-gy)/d

    Now plug this into the first equation, solve for y. Then sub your solution for y back in to your expression for x, and solve for x. Note, how do you know you're allowed to divide by d, as in x = (b-gy)/d? What if d = 0?
     
  4. Mar 6, 2007 #3
    Oh bloody hell, I had that at the start but didn't see the division by d continued to g. Thanks.

    And about d = 0, the book says we're to assume that d isn't zero, than it just becomes normal division.
     
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