state and prove De movrie's theorem

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