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**coolwind****Member**- Registered: 2005-10-30
- Posts: 30

Use it to evaluate

(3/5+j4/5)^100 ?

Thaks~~

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**George,Y****Member**- Registered: 2006-03-12
- Posts: 1,306

if your question is (3/5+i4/5)[sup]100[/sup]=?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)[sup]k[/sup]=cos(kx)+i sin(kx)......(1b)

cos(x)²+sin(x)²=1 (3)

any a+bi can be expressed as

a+bi = (a²+b²)(a+bi)/(a²+b²) = (a²+b²) (a/(a²+b²)+b/(a²+b²) i)

a/(a²+b²) and b/(a²+b²) satisfy cos(x) and sin(x) in equation(3) . Actually we need not point out angle x explicitly sometimes, besides, i forgot the formula!

(a+bi)[sup]k[/sup]= (a²+b²)[sup]k[/sup](...) (4)

As for this case,

(3/5)²+(4/5)²=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

*Last edited by George,Y (2006-03-22 00:41:55)*

**X'(y-Xβ)=0**

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 20,349

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,

r²Cos²θ = 9/25, r²Sin²θ = 16/25.

Adding the two,

r²=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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