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#1 2008-11-22 03:47:21

Daniel123
Member
Registered: 2007-05-23
Posts: 663

Complex numbers

smile

Last edited by Daniel123 (2008-11-22 03:49:43)

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#2 2008-11-23 11:03:22

Daniel123
Member
Registered: 2007-05-23
Posts: 663

Re: Complex numbers

Actually, it's true for all real n.

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#3 2008-11-23 11:26:28

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

Re: Complex numbers

Fairly simple if you convert to modulus-argument form.


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

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#4 2008-11-23 11:29:49

Daniel123
Member
Registered: 2007-05-23
Posts: 663

Re: Complex numbers

Indeed smile

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#5 2009-09-03 09:21:29

rzaidan
Member
Registered: 2009-08-13
Posts: 59

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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