# What is De Movre's Theorem

What is &quot;De Movre&#39;s Theorem&quot; ?

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

Using De Movre&#39;s Theorem ,show that : XXXXX

just tell me what is De Movre&#39;s Theorem is enough

thx !

Update:

oh , sorry . It is Moivre's Theorem

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de Moivre&#39;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] &lt;---because cosAcosB-sinAsinB= cos(A+B)

cosAsinB+ sinAcosB = sin (A+B)

so (cosx+isinx)^n = cos(nx)+ isin(nx) is true for n&gt;= 0

for n&lt;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&lt;0

• 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