Use De Moivre's Theorem to show that if z = cos(q)+isin(q), then (z^n)+(z^-n) = 2cos(nq) and (z^n)-(z^-n)=2isin(nq).

De Moivre's Theorem states that if z = cos(q)+isin(q), then

zn = (cos(q)+isin(q))n = cos(nq)+isin(nq)

But then 

z-n = cos(-nq)+isin(-nq).

Now, cos(-p)=cos(p), as cosine is a symmetric (even) function, and sin(-p)=-sin(p), as sine is an anti-symmetric (odd) fuction. Thus,

z-n  = cos(nq)-isin(nq).

The rest is just algebra:

zn+z-n = [cos(nq)+isin(nq)]+[cos(nq)-isin(nq)] = 2cos(nq).

zn-z-n = [cos(nq)+isin(nq)]-[cos(nq)-isin(nq)] = 2isin(nq).

DA
Answered by Dorian A. Further Mathematics tutor

15978 Views

See similar Further Mathematics A Level tutors

Related Further Mathematics A Level answers

All answers ▸

z = -2 + (2root3)i. Find the modulus and argument of z.


How do I integrate (sin x)^6?


Solve the following inequality: 2x^2 < x+3


You have three keys in your pocket which you extract in a random way to unlock a lock. Assume that exactly one key opens the door when you pick it out of your pocket. Find the expectation value of the number of times you need to pick out a key to unlock.


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences