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  Discussion about math, puzzles, games and fun.   Useful symbols: √ ∞ ≠ ≤ ≥ ≈ ⇒ ∈ Δ θ ∴ ∑ ∫ π -

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#1 2008-11-23 02:47:21

Daniel123
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Complex numbers



smile

Last edited by Daniel123 (2008-11-23 02:49:43)

 

#2 2008-11-24 10:03:22

Daniel123
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Re: Complex numbers

Actually, it's true for all real n.

 

#3 2008-11-24 10:26:28

mathsyperson
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Re: Complex numbers

Fairly simple if you convert to modulus-argument form.


Why did the vector cross the road?
It wanted to be normal.
 

#4 2008-11-24 10:29:49

Daniel123
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Re: Complex numbers

Indeed smile

 

#5 2009-09-04 07:21:29

rzaidan
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Re: Complex numbers

Hi Daniel123
this is the proof for natural n:
√3+i=2(√3/2  + (1/2)i)=2(cos 30 +i sin(30 ))  r=2 ,  theta=30 then by demoivers theorem

(√3+i)^n=2^n(cos n(30) +i sin( n (30 ))  and similarly

√3-i=2(√3/2  - (1/2)i)=2(cos 330 +i sin(330 ))  r=2 ,  theta=30 then by demoivers theorem

(√3-i)^n=2^n(cos n(330) +i sin( n (330 ))   therefore

(√3+i)^n+(√3-i)^n=2^n(cos n(30) +i sin( n (30 ))+ 2^n(cos n(330)+i sin( n (330 )) but
sin( n (30))=- sin( n (330 )) therefore the imaginary part cancells and the result is in R or
(√3+i)^n+(√3-i)^n=2^n(cos n(30)+2^n(cos n(330) but cos n(30)=(cos n(330) i.e
(√3+i)^n+(√3-i)^n=2(2^n(cos n(30))=2^(n+1)   (cos n(30) ∈R
Best Regards
Riad Zaidan

 

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