de Moivre identity, proof of


To prove the de Moivre identityMathworldPlanetmath, we will first prove by inductionMathworldPlanetmath on n that the identity holds for all natural numbersMathworldPlanetmath.

For the case n=0, observe that

cos(0θ)+isin(0θ)=1+i0=(cos(θ)+isin(θ))0.

Assume that the identity holds for a certain value of n:

cos(nθ)+isin(nθ)=(cos(θ)+isin(θ))n.

Multiply both sides of this identity by cos(θ)+isin(θ) and expand the left side to obtain

cos(θ)cos(nθ)-sin(θ)sin(nθ)+icos(θ)sin(nθ)+isin(θ)cos(nθ) =(cos(θ)+isin(θ))(cos(nθ)+isin(nθ))
=(cos(θ)+isin(θ))n+1.

By the angle sum identities,

cos(θ)cos(nθ)-sin(θ)sin(nθ) =cos(nθ+θ)
cos(θ)sin(nθ)+sin(θ)cos(nθ) =sin(nθ+θ)

Therefore,

cos((n+1)θ)+isin((n+1)θ)=(cos(θ)+isin(θ))n+1.

Hence by induction de Moivre’s identity holds for all natural n.

Now let -n be any negative integer. Then using the fact that cos is an even and sin an odd function, we obtain that

cos(-nθ)+isin(-nθ) =cos(nθ)-isin(nθ)
=cos(nθ)-isin(nθ)cos2(nθ)+sin2(nθ)
=1cos(nθ)+isin(nθ)cos(nθ)-isin(nθ)cos(nθ)-isin(nθ)
=1cos(nθ)+isin(nθ),

the denominator of which is (cos(nθ)+isin(nθ))n. Hence

cos(-nθ)+isin(-nθ)=(cos(θ)+isin(θ))-n.
Title de Moivre identity, proof of
Canonical name DeMoivreIdentityProofOf
Date of creation 2013-03-22 14:34:08
Last modified on 2013-03-22 14:34:08
Owner rspuzio (6075)
Last modified by rspuzio (6075)
Numerical id 10
Author rspuzio (6075)
Entry type Proof
Classification msc 12E10
Synonym proof of de Moivre’s formulaMathworldPlanetmathPlanetmath
Synonym proof of de Moivre’s theorem
Related topic AngleSumIdentity