What is De Moivre's theorem?

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In complex number ( especially for any real number) x and integer n it holds that

(cos(x) + i(sinx))^n = cos(n*x) + isin(n*x) where i is the imaginary unit representing as i*i = -1.

This is called  De Moivre's theorem.

This theorem can be proved by Euler's theorem which states 

e^(i*x) = cos(x) + isin(x)

then

(e^(i*x))^n = (cos(x) + isin(x))^n which equals to

e^(i*x*n) = cos(n*x) + isin(n*x)

resulting to

 (cos(x) + isin(x))^n = cos(n*x) + isin(n*x)

BARUN S. A Level Further Mathematics  tutor, GCSE Further Mathematics...

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