Prove De Moivre's by induction for the positive integers

First we check the formula for our first case, n=1 where (cosx + isinx)n =cosnx + isinnx (cosx + isinx)1 = cos(1x) + isin(1x) - holds true for n=1 next we assume true for our case n=k(cosx + isinx)k = coskx + isinkxnext we show that if the case for n=k is true, then the case n=k+1 is true.by inputting n=k+1 we get (cosx+isinx)(k+1)=(cosx+isinx)(cosx+isinx)know as we assumed for n=k ,this equals (cosx+isinx)(coskx+isinkx)this gives cosxcoskx +icosxsinkx +isinxcoskx -sinxsinkxsimplifying to (coskxcosx -sinxsinkx) + i(cosxsinkx + sinxcoskx)finally through the compound angle formulae we reach our desired result: (cosx+isinx)k+1 = cos((k+1)x) + isin((k+1)x) conclusion: If n=k holds true, then n=k+1 holds true. Since n=1 holds true we have now shown that De Moivre's theorem holds true for all positive integers.

HS
Answered by Harry S. Further Mathematics tutor

3211 Views

See similar Further Mathematics A Level tutors

Related Further Mathematics A Level answers

All answers ▸

A spring with a spring constant k is connected to the ceiling. First a weight of mass m is connected to the spring. Deduce the new equilibrium position of the spring, find its equation of motion and hence deduce its frequency f.


Prove by induction the sum of the natural numbers from 1 to n is n(n+1)/2


Find the solution the the differential equation d^2y/dx^2 + (3/2)dy/dx + y = 22e^(-4x)


Cube roots of 8?


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