What is De Moivre's theorem?

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

(cos(x) + i(sinx))^n = cos(nx) + isin(nx) 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^(ixn) = cos(nx) + isin(nx)

resulting to

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

BS
Answered by BARUN S. Further Mathematics tutor

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