1.5. ROOTS OF COMPLEX NUMBERS 7

and so (x√

x2 + y2,

y√x2 + y2

)is a point on the unit circle. Therefore, there exists a unique angle θ ∈ [0,2π) such that

cosθ =x√

x2 + y2, sinθ =

y√x2 + y2

.

The polar form of the complex number is then r (cosθ + isinθ) where θ is this angle justdescribed and r =

√x2 + y2 ≡ |z|.

θ

x+ iy = r(cos(θ)+ isin(θ))r =√

x2 + y2r

1.5 Roots Of Complex NumbersA fundamental identity is the formula of De Moivre which follows.

Theorem 1.5.1 Let r > 0 be given. Then if n is a positive integer,

[r (cos t + isin t)]n = rn (cosnt + isinnt) .

Proof: It is clear the formula holds if n = 1. Suppose it is true for n.

[r (cos t + isin t)]n+1 = [r (cos t + isin t)]n [r (cos t + isin t)]

which by induction equals

= rn+1 (cosnt + isinnt)(cos t + isin t)

= rn+1 ((cosnt cos t− sinnt sin t)+ i(sinnt cos t + cosnt sin t))

= rn+1 (cos(n+1) t + isin(n+1) t)

by the formulas for the cosine and sine of the sum of two angles. ■

Corollary 1.5.2 Let z be a non zero complex number. Then there are always exactly k kth

roots of z in C.

Proof: Let z = x+ iy and let z = |z|(cos t + isin t) be the polar form of the complexnumber. By De Moivre’s theorem, a complex number

r (cosα + isinα) ,

is a kth root of z if and only if

rk (coskα + isinkα) = |z|(cos t + isin t) .