12 CHAPTER 1. SOME PREREQUISITE TOPICS
1.7 Roots of Complex NumbersA fundamental identity is the formula of De Moivre which follows.
Theorem 1.7.1 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 1.7.2 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 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 and so the kth roots of z are of the form|z|1/k (cos
( t+2lπk
)+ isin
( t+2lπk
)), l ∈ Z. Since the cosine and sine are periodic of period
2π, there are exactly k distinct numbers which result from this formula. ■
Example 1.7.3 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.
Example 1.7.4 Factor the polynomial x3−27.