1.15. ROOTS OF COMPLEX NUMBERS 41

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

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

Theorem 1.15.1 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 1.15.2 Let z be a non zero complex number. Then there are always exactly kkth 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 zif and only if

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

This requires rk = |z| and so r = |z|1/k and also both cos(kα) = cos t and sin(kα) = sin t.This can only happen if kα = t +2lπ for l an integer. Thus

α =t +2lπ

k, l ∈ Z