- #1
H2instinct
- 20
- 0
Complex Numbers Proof
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.
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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