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

11244 Views

See similar Further Mathematics A Level tutors

Related Further Mathematics A Level answers

All answers ▸

Integral of ln x


Evaluate the following product of two complex numbers: (3+4i)*(2-5i)


What is the modulus of 3+4i?


Prove that 1+4+9+...+n^2 = n(n+1)(2n+1)/6.


We're here to help

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

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences