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

Related Further Mathematics A Level answers

All answers ▸

How do I solve x^2 + x - 6 > 0 ?


Prove, by induction, that 4^(n+1) + 5^(2n-1) is always divisible by 21


Differentiate arctan(x) with respect to x


Given the equation x^3-12x^2+ax-48=0 has roots p, 2p and 3p, find p and a.