A cool trick involving roots of unity

1. This note is about a delightful application of complex numbers to combinatorics and to integer divisi-
bility problems in general. I learnt this from an excellent Math Stackexchange answer by Qiaochu Yuan.
(http://math.stackexchange.com/questions/4961/interesting-results-easily-achieved-using-complex-numbers)
Let 1, ω, ω 2 , . . . , ω p−1 be the p roots of unity for some prime p. Consider the sum
p−1
Sk = (ω i )k
i=0
of the kth powers of the roots of unity. It is easy to show that
p if p | k
Sk = (1)
0 else
This can be shown for instance by considering the equation xp − 1 = 0 that ω i satisfies and using the fact
p−1
that the sum of the roots of this polynomial must equal 0, i.e., i=0 ω i = 0. Since p is a prime, the claim
in equation (1) follows. This is a very useful property to have - we are able to ascertain divisibility of an
arbitrary integer k by a prime p simply by looking at the value that an algebraic expression Sk takes. The
fact itself isn’t very deep - there is nothing more fundamental going on here than the fact that cos(2πx) and
sin(2πx) have zeroes at half-integral and integral locations respectively. However, complex numbers provide
us a very succint way of expressing this fact and also give us lots of notational flexibility for algebraic
manipulation.
Lets look at a simple application of this fact. Consider the following question
Question: How many subsets of an n-element set have cardinalities divisible by 3? In other words,
evaluate n
3
n
3k
k=0
.
We start with the familiar binomial identity
n
n k
x = (1 + x)n
k
k=0
. Substituting the roots of unity one by one into this yields
n
n k
1 = 2n (2)
k
k=0
n
n k
ω = (1 + ω)n (3)
k
k=0
n
n 2k
ω = (1 + ω 2 )n (4)
k
k=0
Adding the equations together, we note that only those terms when k is a multiple of 3 survive. Thus,
n
3
n 1
= 3 2n + (1 + ω)n + (1 + ω 2 )n (5)
3k
k=0
1
= 3 2n + (−1)n ω n + (−1)n ω 2n (6)
1 nπ
= 3 2n + 2 cos 3 (7)

2. The question can be answered easily enough without applealing to complex numbers (for ex., by consid-
ering separate cases for each value of n mod 3) but this is very elegant.
There is more information to be gleaned from this method. Lets say 3 is replaced by an arbitrary integer
d and we are interested in asymptotics. From analysis similar to the one above, we can conclude that this
2n
asymptotic is roughly d but to do a second order analysis, the other complex terms matter. This is an
example of a case where complex numbers emerge organically from a problem that has nothing to do with
them initially.
2