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

11683 Views

See similar Further Mathematics A Level tutors

Related Further Mathematics A Level answers

All answers ▸

Find the general solution to: d^(2)x/dt^(2) + 7 dx/dt + 12x = 2e^(-t)


Why is the argument of a+bi equal to arctan(b/a)?


Integrate (x+4)/(x^2+2x+2)


How do I find and plot the roots of a polynomial with complex roots on an Argand diagram? e.g. f(z) =z^3 -3z^2 + z + 5 where one of the roots is known to be 2+i


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:

© 2025 by IXL Learning