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complex No. (2 Viewers)

dr baby beanie

*is so happeeeeee!!!*
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z1 = r1 (cos@ + isineB) which then becomes z1 = r1 cis@
z2 = r2 (cosB + isineB) which then becomes z2 = r2 cisB

now by DT's theorum: z1/z2 = r1/r2 cis(@-B)

which then when you expand out the cis(@-B) part becomes
z1/z2 = r1/r2 [cos(@-B) + isine(@-B)]

which is want you want isn't it?

However the other ppl who answered your questions early probably know more about it, and their's is probably more right than mine:confused:

Hope I'ved helped anway:)
 

chousta

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k, its just that the questions is a proof of the therm. so i dont think you can use the therm itself to prove it, coz thats wat i had originally(and that makes it easy) but the question is to prove the therm.
 

chousta

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dr Baby I GOT IT!@!!!

and it was really really easy!! i feel (ashamed)


 

kony

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1/Z2 = 1/r2 * cis (-B)

therefore, Z1.1/Z2 = r1/r2 * (cosA+isinA)(cosB-isinB)

=r1/r2 * (cosAcosB - sinBcosAi + cosAsinBi + sinAsinB)
=r1/r2 * [cos(A-B) - (sinBcosAi - cosAsinBi)]
=r1/r2 * [cos(A-B) - isin(A-B)]
 

adgala

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kony's answer was 4 minutes later than your answer, he was probably writing it when you posted.
 
P

pLuvia

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chousta in the future please don't delete your threads, because people may want to look at the solutions or people might ask the same questions thank you :)
 

haque

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chousta said:
dr Baby I GOT IT!@!!!

and it was really really easy!! i feel (ashamed)


Yep that's what i said(maybe i should have mentioned properties of sine as well )
 

Slidey

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On top of what pLuvia said, it might be a good idea in future to post new questions of the same focus in a single thread. For example, all these complex number questions could have been contained within one thread. That way you needn't delete a thread because it seems trivial. :)

(The posts may seem out of order - your three complex number threads, all bearing the same name, have been merged.)
 
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