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

H2instinct

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.

lanedance

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)