1.5. THE COMPLEX NUMBERS 15

Theorem 1.5.4 Let r ≥ 0 be given. Then if n is a positive integer,

[r (cos t+ i sin t)]n= rn (cosnt+ i sinnt) .

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

[r (cos t+ i sin t)]n+1

= [r (cos t+ i sin t)]n[r (cos t+ i sin t)]

which by induction equals

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

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

= rn+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 k kth

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 complexnumber. By De Moivre’s theorem, a complex number,

r (cosα+ i sinα) ,

is a kth root of z if and only if

rk (cos kα+ i sin kα) = |z| (cos t+ i sin 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

and so the kth roots of z are of the form

|z|1/k(cos

(t+ 2lπ

k

)+ i sin

(t+ 2lπ

k

)), l ∈ Z.

Since the cosine and sine are periodic of period 2π, there are exactly k distinct numberswhich result from this formula. ■

Example 1.5.6 Find the three cube roots of i.

First note that i = 1(cos(π2

)+ i sin

(π2

)). Using the formula in the proof of the above

corollary, the cube roots of i are

1

(cos

((π/2) + 2lπ

3

)+ i sin

((π/2) + 2lπ

3

))where l = 0, 1, 2. Therefore, the roots are

cos(π6

)+ i sin

(π6

), cos

(5

6π

)+ i sin

(5

6π

),

and

cos

(3

2π

)+ i sin

(3

2π

).

Thus the cube roots of i are√32 + i

(12

), −

√3

2 + i(12

), and −i.

The ability to find kth roots can also be used to factor some polynomials.