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1. Complex #'s Proof, 2.Complex Particle movement, Magntitute of Acc. and Vel.

  1. Sep 22, 2009 #1
    Complex Numbers Proof

    167qx39.jpg

    Multiplying the top and bottom by the complex conj. of the bottom:

    [tex]\frac{a+ib}{c+id} * \frac{c-id}{c-id}[/tex]

    Gives me:

    [tex]\frac{(ac+bd) - i(ad-bc)}{c^{2}+d^{2}}[/tex]

    In form x+iy it is:

    [tex]\frac{(ac+bd)}{c^{2}+d^{2}} + (\frac{(bc-ad)}{c^{2}+d^{2}})*i [/tex]

    This is where I get stuck. In order to prove [tex]\left(\frac{a+ib}{c+id}\right)[/tex]*[tex]\equiv\frac{a-ib}{c-id}[/tex] I think that I am supposed to take the complex conjugate of my previous answer and then work backwards until I get to [tex]\frac{a-ib}{c-id}[/tex]. I have tried this with several different variations and I am not coming up with the proof at all. I need a bump in the right direction here.
     
    Last edited: Sep 23, 2009
  2. jcsd
  3. Sep 23, 2009 #2

    lanedance

    User Avatar
    Homework Helper

    i think you're heading right direction, first combine everything on the same denominator, then use the fact
    [tex] c^2+d^2 = (c+id)(c-id) [/tex]

    and previously you multiplied
    [tex] (a+ib)(c-id) = (ac+bd) + i(bc-ad)[/tex]

    so now use the fact
    [tex] (a+ib)(c+id) = (ac+bd) - i(bc-ad)[/tex]
    (check if you need)
     
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