1Complex numbers
IA Vectors and Matrices
1.5 De Moivre’s theorem
Theorem (De Moivre’s theorem).
cos nθ + i sin nθ = (cos θ + i sin θ)
n
.
Proof.
First prove for the
n ≥
0 case by induction. The
n
= 0 case is true since
it merely reads 1 = 1. We then have
(cos θ + i sin θ)
n+1
= (cos θ + i sin θ)
n
(cos θ + i sin θ)
= (cos nθ + i sin nθ)(cos θ + i sin θ)
= cos(n + 1)θ + i sin(n + 1)θ
If n < 0, let m = −n. Then m > 0 and
(cosθ + i sin θ)
−m
= (cos mθ + i sin mθ)
−1
=
cos mθ − i sin mθ
(cos mθ + i sin mθ)(cos mθ − i sin mθ)
=
cos(−mθ) + i sin(−mθ)
cos
2
mθ + sin
2
mθ
= cos(−mθ) + i sin(−mθ)
= cos nθ + i sin nθ
Note that “
cos nθ
+
i sin nθ
=
e
inθ
= (
e
iθ
)
n
= (
cos θ
+
i sin θ
)
n
” is not a valid
proof of De Moivre’s theorem, since we do not know yet that
e
inθ
= (
e
iθ
)
n
. In
fact, De Moivre’s theorem tells us that this is a valid rule to apply.
Example.
We have
cos
5
θ
+
i sin
5
θ
= (
cos θ
+
i sin θ
)
5
. By binomial expansion
of the RHS and taking real and imaginary parts, we have
cos 5θ = 5 cos θ − 20 cos
3
θ + 16 cos
5
θ
sin 5θ = 5 sin θ − 20 sin
3
θ + 16 sin
5
θ