Linear Algebra, Theory And Applications, 2012a

1.5. THE COMPLEX NUMBERS 17
and so (
)
x
√
x2 + y , y
√ 2 x2 + y 2
is a point on the unit circle. Therefore, there exists a unique angle, θ ∈ [0, 2π) such that
cos θ =
The polar form of the complex number is then
x
√
x2 + y , sin θ = y
√ 2 x2 + y . 2
r (cos θ + i sin θ)
where θ is this angle just described and r = √ x 2 + y 2 .
A fundamental identity is the formula of De Moivre which follows.
Theorem 1.5.4 Let r>0 be given. Then if n is a positive integer,
[r (cos t + i sin t)] n = r n (cos nt + i sin nt) .
Proof: It is clear the formula holds if n =1. Suppose it is true for n.
which by induction equals
[r (cos t + i sin t)] n+1 =[r (cos t + i sin t)] n [r (cos t + i sin t)]
= r n+1 (cos nt + i sin nt)(cost + i sin t)
= r n+1 ((cos nt cos t − sin nt sin t)+i (sin nt cos t +cosnt sin t))
= r n+1 (cos (n +1)t + i sin (n +1)t)
by the formulas for the cosine and sine of the sum of two angles.
Corollary 1.5.5 Let z be a non zero complex number. Then there are always exactly kk th
roots of z in C.
Proof: Let z = x + iy and let z = |z| (cos t + i sin t) be the polar form of the complex
number. By De Moivre’s theorem, a complex number,
is a k th root of z if and only if
r (cos α + i sin α) ,
r k (cos kα + i sin kα) =|z| (cos t + i sin t) .
This requires r k = |z| and so r = |z| 1/k and also both cos (kα) =cost and sin (kα) =sint.
This can only happen if
kα = t +2lπ
for l an integer. Thus
α = t +2lπ ,l ∈ Z
k
and so the k th roots of z are of the form
( ) ( ))
|z|
(cos
1/k t +2lπ t +2lπ
+ i sin
,l∈ Z.
k
k
Since the cosine and sine are periodic of period 2π, there are exactly k distinct numbers
which result from this formula.