恩樂 asked in 科學及數學數學 · 1 decade ago

What is De Movre's Theorem

What is "De Movre's Theorem" ?

i saw a A.maths question in HKCEE 1997 paper ...

Using De Movre's Theorem ,show that : XXXXX

just tell me what is De Movre's Theorem is enough

thx !

Update:

oh , sorry . It is Moivre's Theorem

2 Answers

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  • 1 decade ago
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    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/appl...

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