state and prove De movrie's theorem

Asked by spsberry8 | 29th Dec, 2009, 08:27: AM

Expert Answer:

De Moivre's theorem states that for any real number x and any integer n, (cosx + isinx)n = cos(nx) + isin(nx).

It can be proved using the principle of mathematical induction

for n =1

(cosx + isinx) = cos(x) + isin(x)  is trivially true

Let the result be true for n =k

(cosx + isinx)k = cos(kx) + isin(kx)

To prove for n =k+1

i.e to prove P(k+1):(cosx + isinx)k+1 = cos((k+1)x) + isin((k+1)x)

(cosx + isinx)k (cosx + isinx) = (cos(kx) + isin(kx))(cosx + isinx)  using P(k)

= (coskxcosx- sinkxsinx)+i(coskxsinx-+sinkxcosx)

=cos((k+1)x) + isin((k+1)x)

So P(k+1) is true

Hence result holds for all natural numbers k

 

                                    

 

 

 

Answered by  | 29th Dec, 2009, 09:32: AM

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