## Dirichlet convolution

Let $f$ and $g$ be two functions defined on the positive integers. Then the Dirichlet convolution of $f$ and $g$, written $f \ast g$, is another function on the positive integers, defined as follows:

$\displaystyle (f \ast g)(n) = \sum_{ab = n} f(a) g(b)$

The sum is taken over all possible factorizations of $n$ into a product $ab$ of positive integers. For example, suppose $f(n) = n$ and $g(n) = n^2$. Then

$\displaystyle \begin{array}{rcl}(f \ast g)(6) &=& \displaystyle \sum_{ab = 6} f(a) g(b) \\[1.5em] &=& f(1)g(6) + f(2)g(3) + f(3)g(2) + f(6)g(1) \\[1em] &=& 1 \cdot 6^2 + 2 \cdot 3^2 + 3 \cdot 2^2 + 6 \cdot 1^2 \\[1em] &=& 36 + 18 + 12 + 6 = 72.\end{array}$

At this point this may seem somewhat arbitrary, but over the next few posts we’ll see that it’s not arbitrary at all—this operation is deeply connected with the Möbius function $\mu$ and hence with primes and factorization.

So what properties does $\ast$ have?

• It’s commutative, that is, $f \ast g = g \ast f$. This follows directly from the commutativity of multiplication:

$\displaystyle (f \ast g)(n) = \sum_{ab = n} f(a) g(b) = \sum_{ba = n} g(b) f(a) = (g \ast f)(n).$

• It’s also associative, that is, $f \ast (g \ast h) = (f \ast g) \ast h$. This also follows from the associativity of multiplication, and distributivity of multiplication over addition. The proof is kind of fiddly but not really difficult:

$\displaystyle \begin{array}{rcl} (f \ast (g \ast h))(n) &=& \displaystyle \sum_{ab = n} f(a) (g \ast h)(b) \\[1.5em] &=& \displaystyle \sum_{ab = n} f(a) \left( \sum_{cd = b} g(c) h(d) \right) \\[1.5em] &=& \displaystyle \sum_{ab = n} \sum_{cd = b} f(a) \left( g(c) h(d) \right) \\[1.5em] &=& \displaystyle \sum_{a(cd) = n} f(a) (g(c) h(d)) \\[1.5em] &=& \displaystyle \sum_{(ac)d = n} (f(a) g(c)) h(d) \\[1.5em] &=& \displaystyle \sum_{ed = n} \sum_{ac=e} (f(a) g(c)) h(d) \\[1.5em] &=& \displaystyle \sum_{ed = n} \left(\sum_{ac=e} f(a)g(c)\right) h(d) \\[1.5em] &=& \displaystyle \sum_{ed = n} (f \ast g)(e) h(d) \\[1.5em] &=& ((f \ast g) \ast h)(n) \end{array}$

• Define the function $\varepsilon$ as follows:

$\displaystyle \varepsilon(n) = \begin{cases} 1 & \text{if } n=1 \\ 0 & \text{otherwise.} \end{cases}$

Then I claim that $\varepsilon$ is an identity element for $\ast$, that is, for any function $f$ we have $f \ast \varepsilon = \varepsilon \ast f = f$. By definition,

$\displaystyle \displaystyle (f \ast \varepsilon)(n) = \sum_{ab = n}f(a) \varepsilon(b)$

But $\varepsilon(b) = 0$ for every $b$ other than $1$, which cancels out most of the terms of the sum. The only term that does not become zero is $f(n) \varepsilon(1) = f(n)$. So the right-hand sum reduces to just $f(n)$, that is, $f \ast \varepsilon = f$. Since we already know $\ast$ is commutative this proves that $\varepsilon \ast f = f$ too.

So, to sum up1, we have defined the Dirichlet convolution $\ast$, and shown that it is associative, commutative, and has $\varepsilon$ as an identity element. That is, to use some fancy math words, we have shown that $(\ast, \varepsilon)$ is a commutative monoid over the set of functions defined on the positive integers.2 Next time we will see what this has to do with the Möbius function $\mu$.

1. Haha.

2. Technically we should also specify the codomain of the functions, but it doesn’t matter much: typically it is taken to be the complex numbers, but really anything with a commutative, associative multiplication that distributes over addition will do.

Posted in combinatorics, proof | Tagged , , , | 4 Comments

## The Möbius function proof, part 2 (the subset parity lemma)

Continuing from my previous post, we are in the middle of proving that $\mu$ satisfies the same equation as $S$, that is,

$\displaystyle \sum_{d \mid n} \mu(d) = \begin{cases}1 \qquad n=1 \\ 0 \qquad n>1,\end{cases}$

and that therefore $\mu(n) = S(n)$ for all $n \geq 1$, that is, $\mu(n)$ is the sum of all the $n$th primitive roots of unity.

We had gotten as far as the following lemma:

Subset parity lemma: any nonempty finite set has an equal number of even-sized and odd-sized subsets.

In fact, this lemma is the only remaining piece of the proof: if we can prove this then we will have a complete proof that $S(n) = \mu(n)$. Just for fun, we’re actually going to consider two different proofs of this lemma today.

## The first proof

The first proof will be by a combinatorial or counting argument, and was suggested in a comment by Andrey Mokhov. The idea is to match up the subsets in such a way that:

• Each subset is matched with exactly one other subset, that is, the matching is 1-1.
• Each even subset is matched with an odd subset, and vice versa.

Since the subsets are in pairs and each pair has one even and one odd subset, the inescapable conclusion will be that there are the same number of even and odd subsets. For example, suppose there are a bunch of people in a room and you tell them to get into pairs where each pair has one person with a hat and one person without a hat. If they are able to do this exactly, with no unpaired people left over, then you know (without actually having to count people or hats!) that there must be an equal number of hatted and hatless people.

Call the entire set $T$. Let’s first consider the case where $T$ has an odd number of elements. How will we match up its subsets? That is, given some subset $U \subseteq T$, which other subset should $U$ be matched with? A natural idea—which turns out to be the right one in this case—is to match $U$ with its complement $T - U$, that is, the subset of elements from $T$ which are not in $U$. Since the complement of $U$’s complement is $U$ again (and also because $U$ cannot be its own complement), this does indeed put the subsets of $T$ into pairs. Also, if $|U|$ is even, then $|T - U| = |T| - |U| = \text{odd} - \text{even} = \text{odd}$ (and vice versa if $|U|$ is odd). So each pair has one even and one odd subset.

A picture will probably help. Here’s an example where $|T| = 5$. As you can check, each row in the picture is a pair of subsets which are each other’s complement, with one even subset and one odd subset:

Now, what about if $|T|$ is even? Matching $U$ with $T - U$ doesn’t work anymore, since they have the same parity—both are even or both are odd. But we can use a cute trick to fix things. Pick some particular element $t \in T$ (note, this is the place in the proof where we depend on the assumption that $T$ is nonempty!)—for example, we might pick the smallest element if they can be ordered, but it really doesn’t matter which one we pick. Previously, when complementing a subset $U$, we “flipped” the membership of each element of $T$: if $x$ is element of $U$, then $x$ is not an element of $T - U$, and vice versa. Now we do the same, except that we don’t flip $t$: if $t \in U$ we keep it; if $t \not\in U$ we leave it out. But we still flip all the other elements. Here’s an example when $|T| = 4$:

Note how we match all the sets that don’t have the smallest (blue) element with each other, and likewise all the sets that do have the smallest element are matched with each other. But within each pair, all the elements other than the smallest one are flipped.

You can convince yourself that this does indeed match up all the subsets into pairs. Moreover, each pair has one even and one odd subset, because the sizes of the paired sets add up to an odd number: $n-1$ if they don’t contain $t$, or $n + 1$ if they do. QED.

And that’s it for the first proof! We’ve shown that any nonempty finite set has the same number of even and odd subsets, because we can always match them up perfectly.

## Binomial coefficient sums

And now for a small tangent. Remember that a set of size $n$ has $\binom{n}{k}$ subsets of size $k$, where $\binom n k$ is a binomial coefficient, or an entry in Pascal’s Triangle. Since we now know there are the same number of even and odd subsets, if we add $\binom{n}{k}$ when $k$ is even and subtract when $k$ is odd, we must get zero:

$\displaystyle \sum_{0 \leq k \leq n} (-1)^k \binom n k = 0$

That is, the alternating sum of each row of Pascal’s Triangle (besides the zeroth row) is 0:

$\displaystyle \begin{array}{ccccccccccc} &&&1&-&1&&&&=&0 \\ &&1&-&2&+&1&&&=&0 \\ &1&-&3&+&3&-&1&&=&0 \\ 1&-&4&+&6&-&4&+&1&=&0 \end{array}$

and so on. As you can see, this is already really obvious for odd rows, for example, $1-3+3-1$, since equal numbers cancel out. This is why in some sense the proof was so easy for odd $n$: we just took the complement of each $U$. However, for even rows, such as $1-4+6-4+1$, it is not as obvious: why should we have $1+6+1 = 4+4$? So the proof for even $n$ required a bit more cleverness.

In any case, this discussion of binomial coefficiencts brings us to:

## The second proof

The second proof was hinted at in a comment by decourse. Recall the Binomial Theorem:

$\displaystyle (x+y)^n = \sum_{0 \leq k \leq n} \binom n k x^{n-k} y^k$

Set $x = 1$ and $y = -1$. Then the Binomial Theorem says that

$\displaystyle 0 = 0^n = (1-1)^n = \sum_{0 \leq k \leq n} \binom n k (-1)^k.$

And voila! The fact that this sum is zero is the same as saying that there are the same number of even and odd subsets. So we can see that the subset parity lemma is really just a special case of the Binomial Theorem. QED.

## Onward

Which proof is better? I like them both. The second proof is really short, and would be your best bet to quickly communicate the idea of the proof to someone who already knows the Binomial Theorem. On the other hand, although it’s longer, the first proof is really cute and feels simpler somehow—it doesn’t depend on the Binomial Theorem and doesn’t even use any actual numbers or equations.

So, where are we? We’ve proved the subset parity lemma. But this means that

$\displaystyle \sum_{d \mid n} \mu(d) = \begin{cases}1 \qquad n=1 \\ 0 \qquad n>1,\end{cases}$

since we saw that this amounts to looking at subsets of the set of unique prime factors of $n$, and adding $1$ for even subsets and $-1$ for odd subsets.

This proves that $\mu(n) = S(n)$ (the sum of primitive $n$th roots of unity) since $S(n)$ satisfies the same equation. But it’s also a useful fact in its own right, will play an important role in our continuing exploration of the Möbius function $\mu$.

Posted in arithmetic, combinatorics, complex numbers, primes, proof | Tagged , , , , , , , , , | 2 Comments

## The Möbius function proof, part 1

In my last post, I introduced the Möbius function $\mu(n)$, which is defined in terms of the prime factorization of $n$:

• $\mu(n) = 0$ if $n$ has any repeated prime factors, that is, if $n$ is divisible by a perfect square.
• Otherwise, if $n$ has $k$ distinct prime factors, $\mu(n) = (-1)^k$. In other words, $\mu(n) = 1$ if $n$ has an even number of distinct prime factors, and $-1$ if the number of prime factors is odd.

Previously, we also considered the sum $S(n)$ of all the primitive $n$th roots of unity. Today, we will begin proving they are the same: $S(n) = \mu(n)$ for all $n \geq 1$! This is surprising since it is not immediately obvious what the two functions have to do with each other: one is about sums of complex numbers, and the other is about prime factorization. On the other hand, perhaps it is not all that surprising after all: we’ve already seen that primitive roots of unity have a lot to do with divisibility and prime factorization. In any case, this connection is the key that cracks open a lot of really cool mathematics.

So, how will we do it? Recall that we proved $S(n)$ satisfies this equation:

$\displaystyle \sum_{d \mid n} S(d) = \begin{cases}1 \qquad n=1 \\ 0 \qquad n>1.\end{cases}$

In the case of $S(n)$ this was fairly “obvious”: we proved it essentially by staring at some pictures. But we will show that $\mu(n)$ also satisfies this equation, that is,

$\displaystyle \sum_{d \mid n} \mu(d) = \begin{cases}1 \qquad n=1 \\ 0 \qquad n>1.\end{cases}$

(In contrast to $S(n)$, this is not so obvious! But it’s true, and we will prove it.) Given a starting value for $n = 1$, we have seen that this equation uniquely determines all values of a function for $n > 1$: for example, given $S(1)$, we can use the equation to uniquely determine $S(2)$; given $S(1)$ and $S(2)$, we can then uniquely determine $S(3)$; and so on. So, if $S(n)$ and $\mu(n)$ both satisfy this equation, they must be the same for all $n$ as long as they have the same starting value—and they do, since $S(1) = \mu(1) = 1$.

I have to say that this is one of my favorite proofs—it rests on a beautiful little combinatorial argument, and, as we’ll see, it gives some insight into the Principle of Inclusion-Exclusion as well. We’ll start the proof today and then see the really beautiful part in my next post. So, without further ado:

Theorem. $\displaystyle \sum_{d \mid n} \mu(d) = \begin{cases}1 \qquad n=1 \\ 0 \qquad n>1.\end{cases}$

Corollary. $\mu(n) = S(n)$ for all $n \geq 1$.

Proof. Since $\mu(1) = 1$, we can already see that $\sum_{d \mid 1} \mu(d) = \mu(1) = 1$ as desired. So now consider $\sum_{d \mid n} \mu(d)$ for some $n > 1$; we wish to show this sum is zero.

Since $\mu(d) = 0$ for values of $d$ which are divisble by a perfect square, those values of $d$ contribute nothing to the sum. We can therefore concentrate on only those $d$ which are not divisible by a perfect square (we call such numbers squarefree). Each squarefree $d$ must be a product of distinct prime factors, since if any prime factor was repeated, then $d$ would be divisible by a perfect square. Since $d \mid n$, the prime factorization of each squarefree $d$ is therefore a subset of the set of distinct prime factors of $n$, and conversely, each subset of the distinct prime factors of $n$ yields a distinct squarefree divisor $d$.

For example, consider $n = 360 = 2^3 \cdot 3^2 \cdot 5$. Each squarefree divisor of $n$ can have at most one copy of each of $2$, $3$, and $5$. There are therefore $2^3 = 8$ squarefree divisors of $n$, one corresponding to each subset of $\{2, 3, 5\}$:

$\displaystyle \begin{array}{rcr} d & \text{factorization} & \mu(d) \\ 1 & & 1 \\ 2 & 2 & -1 \\ 3 & 3 & -1 \\ 5 & 5 & -1 \\ 6 & 2 \cdot 3 & 1 \\ 10 & 2 \cdot 5 & 1 \\ 15 & 3 \cdot 5 & 1 \\ 30 & 2 \cdot 3 \cdot 5 & -1 \end{array}$

Adding up all the values of $\mu(d)$ listed in the right column, we can see that there are four $1$’s and four $-1$’s, which cancel to yield $0$ as claimed.

As another example, for $n = 21 = 3 \cdot 7$, all of the divisors of $n$ are squarefree: we have

$\displaystyle \mu(1) + \mu(3) + \mu(7) + \mu(3 \cdot 7) = 1 - 1 - 1 + 1 = 0.$

Our goal is to show that this sum is always zero, and the only way for it to be zero is if there are always an equal number of $1$’s and $-1$’s. Given the definition of $\mu$, we know that the $1$’s come from even-sized subsets of the distinct prime factors of $n$, and the $-1$’s come from odd-sized subsets. So in other words, we need to show that there are the same number of even-sized subsets as there are odd-sized subsets.

The fact that these are sets of distinct prime factors really makes no difference, since we are just counting them. What we really need, in essence, is the following lemma:

Subset parity lemma: any nonempty finite set has an equal number of even-sized and odd-sized subsets.

Can you see how to prove this? (And as an aside, do you see why it is false if we exclude the word “nonempty”? Do you see what this has to do with the $n=1$ versus $n > 1$ cases?) My next post will explain a nice proof of the lemma and then conclude the proof that $\mu(n) = S(n)$!

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## The Möbius function

Time to pull back the curtain a bit! My recent series of posts on complex roots of unity may seem somewhat random and unmotivated so far, but the fact is that I definitely have a destination in mind—we are slowly and inexorably heading towards some really deep and beautiful mathematics. It has just taken a while to build up all the requisite concepts!

Last time, we defined $S(n)$ as the sum of all the primitive $n$th roots of unity. We noted that

$\displaystyle \sum_{d \mid n} S(d) = \begin{cases}1 \qquad n=1 \\ 0 \qquad n>1\end{cases}$

and used this to deduce values of $S(n)$. In principle we could carry this as far as we want to compute $S(n)$ for any $n$, but this doesn’t necessarily give us any insight into the nature of $S$. For example, is $S(n)$ always either $-1$, $0$, or $1$? Is there a nicer way to characterize $S(n)$, and/or a quicker way to compute it?

It turns out that $S(n)$ is more than just an idle curiosity. In fact, it is famous enough that it has a name: it is known as the Möbius function, and it is usually written $\mu(n)$. There are indeed nicer ways to characterize and compute $\mu(n)$. Here’s how it is most commonly defined:

• $\mu(n) = 0$ if $n$ has any repeated prime factors, that is, if $n$ is divisible by a perfect square.
• Otherwise, if $n$ has $k$ distinct prime factors, $\mu(n) = (-1)^k$. In other words, $\mu(n) = 1$ if $n$ has an even number of distinct prime factors, and $-1$ if the number of prime factors is odd.

For example:

• $\mu(8) = 0$ since $8$ is divisible by a perfect square (namely, $4$).
• $\mu(5) = -1$ since $5$ has an odd number of prime factors (namely, one).
• $\mu(10) = 1$ since $10$ has two distinct prime factors.
• $\mu(30) = -1$ since $30$ has three distinct prime factors.
• $\mu(1) = 1$ since $1$ has zero prime factors, and zero is even.

So far, this seems to match up with what we had already deduced about $S(n)$, but it is not at all obvious whether $S(n)$ and $\mu(n)$ are the same. Why should adding up a particular set of complex numbers have anything to do with the number of prime factors of $n$? In my next post, we’ll prove that they are in fact the same. The proof is a nice exercise in combinatorics and relates to Pascal’s Triangle. (Michael Paul Goldenberg hinted at such a proof in a comment on my previous post.)

As you might guess, $\mu$ is deeply related to prime numbers and the Riemann zeta function. It turns out it is also related to the Principle of Inclusion-Exclusion. All this and more to come!

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## Computing sums of primitive roots

Remember this picture?

It, and other pictures like it, express the fact that for a given $n$, if we take the primitive roots for each of the divisors of $n$, together they make up exactly the set of all $n$th roots of unity. The above picture is for the specific case of $n=12$: the $12$th roots of unity (the dots on the bottom circle) are composed of the primitive roots for $n=1$, $2$, $3$, $4$, $6$, and $12$ (the dots on each of the top circles). I proved this in another post.

Of course, if two sets of complex numbers are the same, then their sums must also be the same. Let’s write $S(n)$ for the sum of all the primitive $n$th roots: in my previous post we worked out $S(n)$ for certain $n$ but weren’t sure how to compute it in other cases. Well, today that’s going to change!

We already know by symmetry that the sum of all the $n$th roots of unity is zero, except when $n=1$ in which case the sum is $1$. Putting all of this together,

$\displaystyle \sum_{0 \leq k < n} \omega_n^k = \sum_{d \mid n} S(d) = \begin{cases}1 \qquad n=1 \\ 0 \qquad n>1\end{cases}$

That is, the sum of all $n$th roots of unity is the same as summing the primitive roots, $S(d)$, for each divisor $d$ of $n$. (The notation $d \mid n$ means $d$ evenly divides $n$, so the summation symbol with $d \mid n$ underneath means we are summing over all divisors of $n$.)

So what have we gained? Well, we can use this equation “backwards” to compute values for $S(d)$!

• We already know $S(1) = 1$.
• When $n = 2$, the equation tells us that $S(1) + S(2) = 0$, so it must be that $S(2) = -1$. (Of course we already knew that too.)
• When $n = 3$, we have $S(1) + S(3) = 0$ (note that $S(2)$ is not included since $2$ is not a divisor of $3$), so $S(3) = -1$ as well.
• $S(1) + S(2) + S(4) = 1 + (-1) + S(4) = 0$, so $S(4) = 0$, which also checks out with our previous knowledge.
• Recall that $S(10)$ was the first value we were unsure about. Well, $S(1) + S(2) + S(5) + S(10) = 1 + (-1) + (-1) + S(10) = 0$, so $S(10)$ must be $1$.

And so on. Since each value of $S(n)$ can be computed in this way once we know $S(d)$ for all the divisors of $n$ (which are smaller than $n$), we can continue filling in a table of values of $S(n)$ like this forever.

$\displaystyle \begin{array}{c|rrrrrrrrrrrrrrrrrrrr} n & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 & 19 & 20 \\ \hline S(n) & 1 & -1 & -1 & 0 & -1 & 1 & -1 & 0 & 0 & 1 & -1 & 0 & -1 & 1 & 1 & 0 & -1 & 0 & -1 & 0 \end{array}$

For example, to fill in $S(20)$ we compute $S(1) + S(2) + S(4) + S(5) + S(10) + S(20) = 1 - 1 + 0 - 1 + 1 + S(20) = 0$ and hence $S(20) = 0$.

Do you notice any patterns? It is not too hard to see that $S(n)$ must always be an integer (why?), but so far it has always been either $-1$, $0$, or $1$; will it always be one of those three values? Before my next post you might like to try extending the table of values for $S(n)$ further and exploring these questions yourself!

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## Sums of primitive roots

In my previous post, we saw that adding up all the complex $n$th roots of unity always yields zero (unless $n=1$, in which case the sum is $1$). Intuitively, this is because the roots are symmetrically distributed around the unit circle, so their contributions to the sum all cancel out.

Today I want to consider the question: what happens when we add up only the primitive $n$th roots of unity? Recall that a primitive $n$th root is one which is not also an $m$th root for some smaller $m < n$. As points on the unit circle, the primitive roots correspond to the spokes which are relatively prime to $n$.

In some cases we clearly still get zero, again because of symmetry. Shown below are $n = 4$, $8$, $9$, and $12$:

For $n=1$ we of course get $1$ again:

For $n=2$ the only primitive root is $-1$, so the sum is clearly $-1$. It’s not too hard to see that the sum of primitive roots for $n=3$ will also be $-1$:

Why is that? Well, when $n=3$, the only root that is not a primitive root is $1$, so if we subtract out $1$ from the sum of all roots, we are left with the sum of the remaining, primitive roots. But we know the sum of all roots is zero, and hence the sum of the remaining primitive roots must be $-1$. In fact, this is not specific to $n=3$; the same argument applies to any prime number. When $n$ is prime, every root is primitive except for $1$ itself, so in order for the sum of all the roots to be zero, the sum of all the primitive roots must be $-1$.

However, this still leaves other values of $n$ where it is not obvious what we will get as the sum of primitive roots. For example, here are $n=10$, $14$, and $15$:

Let’s look at $n=10$. The primitive roots have reflection symmetry across the $x$-axis, so their sum must lie somewhere on the $x$-axis (since their “up-down pulls” cancel out). Since they are leaning more in the positive direction, the sum will be bigger than zero. But how much bigger? Just thinking about the $x$ components, we are adding together irrational numbers involving things like the cosine of $\pi/5$, so maybe we just get some weird irrational sum like $0.8267834852\dots$? It seems plausible, but actually, we don’t: we instead get a sum of… exactly $1$! And the sums for $n=14$ and $n=15$… are also exactly $1$. Something strange is going on.

But you don’t have to take my word for it! Next time we’ll put together what we know about the sum of all roots and how the set of $n$th roots is made up of primitive roots for the divisors of $n$ to compute sums of primitive roots.

Posted in geometry, pictures | Tagged , , , , , , | 4 Comments

## Sums and symmetry

Let’s continue our exploration of roots of unity. Recall that for any positive integer $n$, there are $n$ complex numbers, evenly spaced around the unit circle, whose $n$th power is equal to $1$. These are called the $n$th roots of unity.

Today let’s answer a simple question: what happens if you add all the $n$th roots of unity?

For example, the third roots of unity are $1$ and $-1/2 \pm i\sqrt{3}/2$. If we sum them, we get

$\displaystyle 1 + (-1/2 + i\sqrt{3}/2) + (-1/2 - i \sqrt{3}/2) = 1 - 1 = 0$

Notice how the two $\pm i \sqrt{3}/2$ terms cancel, and then we are left with $1 - 1/2 - 1/2 = 0$.

What about the fourth roots of unity? That’s easy:

$\displaystyle 1 + i + (-1) + (-i) = 0.$

Hmm, zero again. Will we always get zero?

In fact, we will, because of symmetry. Here’s one way to think about it. Adding complex numbers is like adding vectors: we can think of a complex number as trying to “pull” the origin toward itself. Adding complex numbers means they “pull” the origin in multiple directions at the same time. Since the $n$th roots of unity are evenly spaced around the unit circle, all their “pulls” exactly cancel out. It’s like having a bunch of people evenly spaced around the edge of a circular parachute, all pulling at the same time; the parachute doesn’t go anywhere since it is being pulled equally in all directions. (Of course the parachute could rip, but metaphors can only take you so far…)

We can see this a bit more formally as follows. Consider the sum

$\displaystyle S = 1 + \omega_n^1 + \omega_n^2 + \dots + \omega_n^{n-1}.$

If we multiply both sides by $\omega_n$, we can distribute $\omega_n$ over the sum on the right side, and we get

$\displaystyle \omega_n S = \omega_n^1 + \omega_n^2 + \omega_n^3 + \dots + \omega_n^n.$

But $\omega_n^n = 1$, so the right-hand side is really the same as $S$. That is, $\omega_n S = S$. If we subtract $S$ from both sides and factor out $S$, we get $(\omega_n - 1) S = 0$. The only way for this to be true is if either $\omega_n - 1 = 0$ or $S = 0$. $S = 0$ is what we were trying to show, but what’s this $\omega_n - 1 = 0$ possibility? Well, $\omega_n = 1$ only happens when $n = 1$. When $n = 1$, we have only a single dot:

In this case, the sum is actually $1$, not zero. So when $n = 1$, the sum of all $n$th roots is $1$, and for any larger $n$ the sum is $0$:

$\displaystyle \sum_{k=0}^{n-1} \omega_n^k = \begin{cases} 1 \qquad n=1 \\ 0 \qquad n > 1 \end{cases}$

So what did this proof have to do with symmetry? Well, remember that complex multiplication corresponds to rotation. In particular, multiplying by $\omega_n$ is the same as rotating by $1/n$ of a full circle. The equation $\omega_n S = S$ then follows from the fact that the roots of unity have rotational symmetry: if we rotate all of them by $1/n$ of a turn, we end up with exactly the same set of points, so their sum must still be the same. And the only number that doesn’t change when you rotate it is—you guessed it—zero. Unless you rotate by a complete turn, that is, which is what happens when $n = 1$.

Next time, we’ll try adding up all the primitive $n$th roots of unity—something much more interesting happens!

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