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

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SUMMARY

The discussion focuses on the proof involving complex numbers, specifically the manipulation of the expression \(\frac{a+ib}{c+id}\) using its complex conjugate. The correct approach involves multiplying by the complex conjugate of the denominator, resulting in the expression \(\frac{(ac+bd) - i(ad-bc)}{c^{2}+d^{2}}\). The participants emphasize the importance of combining terms over a common denominator and utilizing the identity \(c^2 + d^2 = (c + id)(c - id)\) to facilitate the proof. The goal is to demonstrate that \(\left(\frac{a+ib}{c+id}\right) \equiv \frac{a-ib}{c-id}\).

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Complex Numbers Proof

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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.
 
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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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