22 CHAPTER 2. ALGEBRA AND NOTATION
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 2.5.2 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) .
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 +2lπk
)+ isin
(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 2.5.3 Find the three cube roots of i.
First note that i = 1(cos(
π
2
)+ isin
(π
2
)). Using the formula in the proof of the above
corollary, the cube roots of i are
1(
cos((π/2)+2lπ
3
)+ isin
((π/2)+2lπ
3
))where l = 0,1,2. Therefore, the roots are
cos(
π
6
)+ isin
(π
6
),cos
(56
π
)+ isin
(56
π
),cos
(32
π
)+ isin
(32
π
).
Thus the cube roots of i are
√3
2+ i(
12
),−√
32
+ i(
12
), and −i.
The ability to find kth roots can also be used to factor some polynomials.