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

12571 Views

See similar Further Mathematics A Level tutors

Related Further Mathematics A Level answers

All answers ▸

write the sum cos(x)+cos(2x)+...+cos(nx) as a quotient only involving sine and cosine functions


Given y=arctan(3e^2x). Show dy/dx= 3/(5cosh(2x) + 4sinh(2x))


Find all of the roots of unity, Zn, in the case that (Zn)^6=1


Find the integrating factor of the following first order ODE: dx/dt = -2tx +t.


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

MyTutor is part of the IXL family of brands:

© 2026 by IXL Learning