What is De Movre's Theorem

de Moivre's formula, named after Abraham de Moivre, states that for any complex number (and, in particular, for any real number) x and any integer n it holds that

(cosx+isinx)^n = cos(nx)+ isin(nx)

Proof:(By mathematical induction)
Assume (cosx+isinx)^k = cos(kx)+isin(kx) is true
when n=k+1
LHS= (cosx+isinx)^k+1 (cos+isinx)^k X (cos+isinx)
= [cos(kx)+isin(kx)](cosx+isinx)
= (cosx)[cos(kx)]+[i^2(sinx)[sin(kx)]+ i[cos(kx)sinx+sin(kx)cosx]
= cos[(k+1)x]+sin[(k+1)x] <---because cosAcosB-sinAsinB= cos(A+B)
cosAsinB+ sinAcosB = sin (A+B)
so (cosx+isinx)^n = cos(nx)+ isin(nx) is true for n>= 0
for n<0
we let n=-m
so (cosx+isinx)^n= (cosx+isinx)^(-m)
= 1/ (cosx+isinx)^m
= 1/[cos(mx)+isin(mx)]
= [cos(mx)- isin(mx)]/{[cos(mx)+isin(mx)][cos(mx)-isin(mx)]}
= [cos(mx)- isin(mx)]/{[cos(mx)]^2- [isin(mx)]^2} because (a+b)(a-b)= a^2- b^2
= [cos(mx)- isin(mx)]/{[cos(mx)]^2+ [sin(mx)]^2}
=cos(mx)- isin(mx) because (cosA)^2+ (sinA)^2 = 1
= cos(-nx)- isin(-nx) because n=-m, so m=-n
= cos(nx) + isin(nx) because cos(-A)=cosA, sin(-A)= -sinA
so (cosx+isinx)^n = cos(nx)+ isin(nx) is true for n<0

詳情請看http://en.wikipedia.org/wiki/De_Moivre%27s_formula

呢個Theorem 是一套用來幫你計數的工具,原本好複雜的數可因此簡單左
你去呢度check下就知http://www.netsoc.tcd.ie/~jgilbert/maths_site/applets/complex_numbers/de_moivre_s_theorem.html