This document discusses using complex numbers to solve combinatorial problems involving integer divisibility in an elegant way. It shows that the sum of the kth powers of the roots of unity is 0 if k is not divisible by the prime p, allowing one to determine divisibility of k by p. As an example, it uses this to count the number of subsets of an n-element set with cardinalities divisible by 3 by substituting roots of unity into the binomial theorem. Complex numbers allow expressing this concisely and performing algebraic manipulations.

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