1Complex numbers

IA Vectors and Matrices

1.5 De Moivre’s theorem
Theorem (De Moivre’s theorem).
cos + i sin = (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 + i sin )(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 + i sin )
1
=
cos i sin
(cos + i sin )(cos i sin )
=
cos() + i sin()
cos
2
+ sin
2
= cos() + i sin()
= cos + i sin
Note that
cos
+
i sin
=
e
inθ
= (
e
)
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
)
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