# Sums of roots of unity

This note, which I’ll probably expand upon later, is about roots of
unity. I was reading Shafarevich’s excellent
*Discourses on
Algebra*,
and wondered what the identity at the end of section 2.4 meant for
the polynomial `x`^{n} − 1, the defining polynomial for the `n`th roots
of unity (where `n` > 1).

The reason I wondered was that the roots of unity are very
symmetric: they lie evenly along the unit circle in the complex
plane. As an example, the ninth roots of unity (i.e., the roots of
the polynomial when `n` is 9) are represented in figure
1.

Really, I started thinking about the identity (as discussed below), but then I wondered what other ways there are to show that the sum of all of the roots of unity is 0. This note has a number of ways of showing this fact.

## 1. The intuitive way

The `n`th roots of unity lie evenly on the unit circle, so their
center of mass better be at the origin. So, the sum of the complex
numbers as vectors is zero.

## 2. The direct way

The most direct way to find the sum of the `n`th roots of unity is
as follows. Let `x` = `ω`_{0} + ⋯ + `ω`_{n − 1} be the sum of all
`n` of the roots of unity. Since roots of unity have unit length,
since `ω`_{i}^{n} = 1 for all `i`, and since 1/`ω`_{i} is also an
`n`th root of unity, `ω`_{i}`x` must be exactly `x` (`ω`_{i}
just permutes the components of the sum). That is,
(1 − `ω`_{i})`x` = 0, which means either `x` = 0 or `ω`_{i} = 1, and
because we can choose an `i` such that `ω`_{i} ≠ 1, we conclude
that `x` = 0.

## 3. By representation theory

The direct way reminded me of one-dimensional representations of the
cyclic group `C`_{n}. Take an irreducible complex one-dimensional
representation `ρ` of `C`_{n} = ⟨`α`⟩, and first note that the
value of `ρ`_{α} completely characterizes `ρ` since
`ρ`_{αk} = (`ρ`_{α})^{k}.

One can show that all of the irreducible representations of `C`_{n}
are one-dimensional from the fact that `C`_{n} is abelian, and
furthermore, using the fact that the sum of the squares of the
dimensions of the irreducible representations is equal to |`C`_{n}|,
we can conclude that each root of unity corresponds to a different
irreducible representation of `C`_{n}.

Next, we compute `x` = 1 + `α` + `α`^{2} + ⋯ + `α`^{n − 1}. Since
`ρ`_{x} is an endomorphism of complex representations from
irreducible `ρ`, it must be some constant in **C**. We can see,
as before, that 1 − `α` is in the kernel of `ρ`_{x}, so either
`ρ` is the trivial representation or `x` = 0, and since we can
choose `ρ` not to be trivial, this completes the proof.

## 4. By polynomial theory

This kind of method should be well known to anyone who did
high-school competition math. We take the polynomials `f`(`x`) = `x`^{n} − 1
and `g`(`x`) = (`x` − `ω`_{0})(`x` − `ω`_{1})⋯(`x` − `ω`_{n − 1}) and note
that they are the same because they are both degree `n`, have the
same `n` roots, and are monic. We use the fact they are monic by
seeing that dividing each polynomial through by `x` − `ω`_{i} results
in a monic polynomial, and so dividing through by all such monomials
will result in 1 for each of them, proving `f` = `g`. Now, the
`x`^{n − 1} term has 0 as a coefficient according to `f` and
−`ω`_{0} − `ω`_{1} − ⋯ − `ω`_{n − 1} according to `g`, which
completes the proof.

## 5. By Shafarevich’s identity

This method is the reason I thought about any of this to begin with.
It’s not very direct, but I think it is kind of interesting. First,
we need the identity from the end of section 2.4 of
*Discourses on Algebra*, which I will derive here. To do
this, we will develop the theory of the interpolating polynomial.
The idea is that, given a set of `n` + 1 values `v`_{i} at distinct
values `x`_{i}, what is the `n`th degree polynomial `f` with
`f`(`x`_{i}) = `v`_{i} for all `i`?

Let `F`(`x`) = (`x` − `x`_{1})(`x` − `x`_{2})⋯(`x` − `x`_{n + 1}), and let `F`_{i}(`x`) be the
polynomial obtained by dividing `F` through by `x` − `x`_{i}. We define

`f`

_{i}(

`x`) =

1 |

F_{i}(x_{i}) |

`F`

_{i}(

`x`)

so that `f`_{i}(`x`_{i}) = 1 and `f`_{i}(`x`_{j}) = 0 when `i` ≠ `j`. This provides
a convenient basis for polynomials: observe that

`f`(

`x`) =

`v`

_{1}

`f`

_{1}(

`x`) +

`v`

_{2}

`f`

_{2}(

`x`) + ⋯ +

`v`

_{n + 1}

`f`

_{n + 1}(

`x`)

is the interpolating polynomial as discussed.

If `k` is a natural number not exceeding `n`, if we have
`v`_{i} = `x`_{i}^{k}, then since `f`(`x`) must be `x`^{k}, we have

`x`

^{k}=

x_{1}^{k} |

F_{1}(x_{1}) |

`F`

_{1}(

`x`) + ⋯ +

x_{n + 1}^{k} |

F_{n + 1}(x_{n + 1}) |

`F`

_{n + 1}(

`x`)

and since each `F`_{i} is a monic degree-`n` polynomial, this means
that if `k` < `n`,

x_{1}^{k} |

F_{1}(x_{1}) |

x_{n + 1}^{k} |

F_{n + 1}(x_{n + 1}) |

and if `k` = `n`, then

x_{1}^{n} |

F_{1}(x_{1}) |

x_{n + 1}^{n} |

F_{n + 1}(x_{n + 1}) |

These are the identities at the end of section 2.4.

We won’t be using these identities directly, but rather in
spirit^{[1]}. We let `x`_{i} = `ω`_{i} and `v`_{i} = 1, where
`ω`_{i} is an `n`th root of unity (with `ω`_{1} = 1). Notice
that each `F`_{i} is an (`n` − 1)-degree polynomial, so since both
`f`(`x`) and `g`(`x`) = 1 are at-most-(`n` − 1)-degree polynomials which
agree at `n` points, `f` = `g`. By taking the leading `x`^{n − 1}
coefficient, this means

1 |

F_{1}(ω_{1}) |

1 |

F_{n}(ω_{n}) |

Notice that `F`_{i}(`ω`_{i}) is the product of all `ω`_{i} − `ω`_{j} for `j` ≠ `i`, and
`ω`_{i} − `ω`_{j} = `ω`_{i}(1 − `ω`_{i}^{−1}`ω`_{j}). Then, we can
see `F`_{i}(`ω`_{i}) = `ω`_{i}^{n − 1}`F`_{1}(`ω`_{1}) since
`ω`_{i}^{−1} is just permuting the roots of unity. Because
(`ω`_{i}^{n − 1})^{−1} = `ω`_{i}, we have

ω_{1} |

F_{1}(ω_{1}) |

ω_{n} |

F_{1}(ω_{1}) |

and so `ω`_{1} + ⋯ + `ω`_{n} = 0.

^{[1]}Truthfully, this is because I forgot the exact formulation of the identities while considering their application to roots of unity.