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#1 2006-03-22 22:12:40

coolwind
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How to use DeMoivre's formula?

Use it to evaluate
(3/5+j4/5)^100 ?

Thaks~~
big_smile

#2 2006-03-22 23:40:34

George,Y
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Re: How to use DeMoivre's formula?

if your question is (3/5+i4/5)100=?the solution is words below

Key theories:

(1) (cosa+i sina)(cosb+i sinb)
= cosa cosb - sina sinb + (sina cosb + cosa sinb)i

= cos(a+b)+ sin(a+b)i

(cosa+i sina)(cosa+i sina)(cosa+i sina) = (cos2a+isin2a)(cosa+i sina) =cos3a+isin3a

Similarly, (cosx+i sinx)k=cos(kx)+i sin(kx)......(1b)

cos(x)+sin(x)=1    (3)

any a+bi can be expressed as
a+bi = (a+b)(a+bi)/(a+b) = (a+b) (a/(a+b)+b/(a+b) i)
a/(a+b) and b/(a+b) satisfy cos(x) and sin(x) in equation(3) . Actually we need  not point out angle x explicitly sometimes, besides, i forgot the formula! tongue

(a+bi)k= (a+b)k(...)  (4)

As for this case,

(3/5)+(4/5)=1 angle x could be arcsin(3/5)

the answer is cos[100arcsin(3/5)]+i sin[100arcsin(3/5)]

As far as i can reach what

Last edited by George,Y (2006-03-22 23:41:55)


X'(y-Xβ)=0

#3 2006-03-23 02:17:43

ganesh
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Re: How to use DeMoivre's formula?

Yes, I guess the question is (3/5 + 4/5i).
This is of the form a+bi. We got to convert this to polar form. That is r(Cosθ +iSinθ).
Equating the real and imaginary parts,
rCosθ=3/5, rSinθ=4/5.
Squaring both,
rCosθ  = 9/25, rSinθ = 16/25.
Adding the two,
r=25/25=1, r=1.
Therefore, Cosθ =3/5, Sinθ =4/5.
[r(Cosθ +iSinθ )]^100=r^100[Cosθ +iSinθ ]^100
Since r=1, r^100=1.
[Cosθ +iSinθ ]^100=Cos100θ +iSin100θ
θ=ArcCos3/5 = ArcSin 4/5


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