Condition that a polygon is regular in the complex plane

Question

In Generalized formula for third point to form an equilateral triangle I deduced that for $A(z_1)$ and $B(z_2)$ the formula for $C(z_3)$ such that $\triangle ABC$ is equilateral has the formula:

$z_3=z_1+\omega(z_1-z_2),~\omega^3=1,~\omega\in\mathbb{C}-\mathbb{R}$

Rearranging we get

$(\omega+1)z_1-z_3-\omega z_2=0$

Multiply by $\omega$

$(\omega^2+\omega)z_1-\omega z_3-\omega^2z_2=0$

$\omega^3-1=0\Rightarrow(\omega-1)(\omega^2+\omega+1)=0$ and since $\omega\ne1,~\omega^2+\omega+1=0\Rightarrow\omega^2+\omega=-1$

$-z_1-\omega z_3-\omega^2z_2=0$

Multiply by -1

$z_1+\omega z_3+\omega^2z_2=0$

Generally we have this equation:

$z_1+\omega z_2+\omega^2z_3=0$

Is this valid for n-th sided regular polygons? That means,

A polygon is regular if and only if

$\mathbf{\sum\limits_{k=1}^n z_k\omega_k=0}$ ($\omega$ is the n-th root of unity, $\omega_k$ is the k-th such root)

where the vertices of the polygon are $\mathbf{A_k(z_k)}$.

I attempted to prove it by translating the polygon such that the center coincides with the origin of the orthogonal axis but I got stuck.

EDIT:

I've dwelt on this problem last night and based on @Jean Marie's demonstration I thought of a more practical proof for the implication regular polygon $\Rightarrow$ sum $= 0$.

Let $A_1A_2A_3(...)A_n$ be a regular polygon with $A_k(z_k)$ in trigonometrical order and $G(z_G)$ be the center of gravity of the polygon.

We will prove $\sum\limits_{k=1}^n z_k\omega_k=0$ by subtracting $(z_G\sum\limits_{k=1}^n \omega_k)$.

$LHS=\sum\limits_{k=1}^n \big((z_k-z_G)\omega_k\big)$

$RHS=-z_G\sum\limits_{k=1}^n \omega_k=0$, since $\sum\limits_{k=1}^n \omega_k=0$

We are left to prove

$\sum\limits_{k=1}^n \big((z_k-z_G)\omega_k\big)=0$

Consider $C_k(z_k-z_G)$.

Regular polygon $C_1C_2C_3(...)C_n$ is obtained by translating $A_1A_2A_3(...)A_k$ such that the center of gravity coincides with the origin of the orthogonal axis.

$C_1,~C_2,~C_3,~(...)~C_n$ will be points on the circle with radius $=$ the circumradius of the polygon.

Also consider $C_k'\big((z_k-z_G)\omega_k\big)$.

$C_k'$ is obtained from $C_k$ by circular rotations.

For n - odd : $C_1'C_2'C_3'(...)C_n'$ is regular polygon (1) $\Rightarrow$ the relation is obvious.

For n - even : $C_1'C_2'C_3'(...)C_{n/2}'$ and $C_{n/2+1}'C_{n/2+2}'(...)C_n'$ are regular polygons (2) $\Rightarrow$

$\sum\limits_{k=1}^{n/2}\big((z_k-z_G)\omega_k\big)=0$

$\sum\limits_{k=n/2+1}^n\big((z_k-z_G)\omega_k\big)=0$

Summing up the equations we get the equation that was left to prove.

We need to prove that

(1) For a set $S$ with n elements, n odd, if for every index $k$ (counting from $0$) we "mark" the element with index $2k+1~mod~n$, then every element of $S$ is going to be marked.

(2) For a set $S$ with n elements, n even, if for every index $k$ (counting from $0$) we "mark" the element with index $2k+1~mod~n$, then elements with odd indexes will be marked twice.

How can I prove these statements using modular mathematics?

Answer 1

It isn't a necessary and sufficient condition.

• The implication : "points $z_k$s are the vertices of a regular polygon" (assumed indexed in the trigonometric order) implies "sum = $0$" is true.

Indeed the vertices of a regular $n$-gon are the images of the "standard" regular $n$-gon using this "similitude"

$$s(z)=A(z-z_G) \ \ \text{with} \ \ A=re^{i \theta } \ \ \text{and} \ \ z_G=\frac1n \sum z_k \tag{1}$$

where $\theta$ is the rotation angle, $r$ is the enlargment/shrinking factor and $z_G$ is the barycenter of points $z_k$.

Relationship (1) can be expressed in a slightly different way:

$$s(z)=Az+B$$

Therefore

$$\sum_{k=1}^n z_k \omega_n^k=\sum_{k=1}^n s(\omega_n^k) \omega_n^k=\sum_{k=1}^n (A\omega_n^k +B) \omega_n^k$$

$$=A\underbrace{\sum_{k=1}^n \omega_n^{2k}}_0 +B \underbrace{\sum_{k=1}^n \omega_n^k}_0=0\tag{2}$$

The fact that the first summation in (2) is zero is a consequence of a more general result one can find here. A specific proof could also be given by distinguishing the cases $n$ odd/$n$ even.

• Whereas, in the other direction, "sum=$0$" doesn't imply "points $z_k$ constitute a regular polygon".

Here is a counterexample for $n=4$ with $w_4=i$ :

$$z_1=0+0i,\ \ z_2=3-2i, \ \ z_3=-1+2i, \ \ z_4=1-3i$$

their sum

$$\sum_{k=1}^4 z_k \omega_4^k=(0)i+(3-2i)i^2+(-1+2i)i^3+(1-3i)i^4$$

$$=-(3-2i)-i(-1+2i)+(1-3i)=0$$

whereas the $z_k$ aren't the vertices of a square.

Fig 1: Case $n=8$: A "continuous" mapping of the regular octagon made by the $w_8^k$s onto the regular octagon of the $z_k$s. The "continuous" aspect is of course unimportant ; I have used it because of its aesthetical appeal.

Matlab program of the figure:

    n=8;
    z=exp((i*2*pi/n)*(0:n));
    A=@(t)(t);B=@(t)(1-exp(i*(t-1)));
    for t=1:0.1:4
       plot(A(t)*z+B(t));
    end;

Answer 2

Working in the complex plane throughout ...

Let $R_{n,k}$ be the "standard" regular $\{n/k\}$-gon (that is, the regular polygon with Schläfli symbol $\{n/k\}$ (where we leave the fraction un-reduced, and we allow $k=0$), whose vertices are at $\omega_n^{mk}$ for $m=0,1,\ldots,n-1$, and where $\omega_n$ is the principal $n$-th root of unity. (These vertices follow Grünbaum's convention, not depicted in Wikipedia, when $n$ and $k$ are not relatively prime. For instance, wikipedia shows $\{8/2\}$ as two squares overlapping at an angle; Grünbaum's convention considers $\{8/2\}$ to be a single $\{4/1\}$ square, traversed twice.) Here are the $\{5/k\}$-gons:

Both $R_{5,1}$ and $R_{5,4}$ are "regular convex pentagons" (and both $R_{5,2}$ and $R_{5,3}$ are "regular pentagrams"), but their opposing orientations are important here. Note that $R_{5,0}$'s vertices all coincide at $1+0i$, but there are still five of them.

For an $n$-gon $P$, let $[P]$ be the vector of its vertices (taken in order around the polygon) $$[P]:=(p_0, p_1, p_2,\ldots, p_{n-1})$$ As it happens, we can write uniquely $$[P] = a_0 [R_{n,0}] + a_1 [R_{n,1}] + \cdots + a_{n-1} [R_{n,n-1}] \tag{1}$$ for appropriate complex coefficients $a_k$. (Proof: Equating individual vector components gives a linear system of $n$ equations in the $n$ unknowns $a_k$.) Observe that each $a_k [R_{n,k}]$ is the vertex-vector of a scaled-and-rotated copy of $R_{n,k}$; this says that

Any $n$-gon is a "vertex sum" of $n$ regular $\{n/k\}$-gons (one for each $k$).

Because the centroid of polygons $R_{n,k}$ for $k\neq 0$ are at the origin, the centroid of $P$ (ie, the polygon's location in the plane) is determined by the $R_{n,0}$ term. On the other hand, the shape of $P$ is determined by those $k\neq 0$ terms, and we can say

$P$ is a regular $\{n/k\}$-gon, for $k\neq 0$, if and only if $a_k\neq 0$, and $a_m=0$ for $0<m\neq k$.

If we only care about non-oriented shapes, we can test whether $P$ is an $\{n/k\}$-gon or an $\{n/(n-k)\}$-gon; ie,

$P$ is a non-oriented regular $\{n/k\}$-gon, for $k\neq 0$, if and only if $a_m\neq0$ for $m=k$ or $m=n-k$ (but not both, unless $k=n/2$), and $a_m=0$ for all other $m>0$.

Now, OP's summation with roots of unity, call it $S_1(P)$, amounts to the dot product of a polygon's vertex-vector with that of $R_{n,1}$. Considering more-generally the dot product with vertex-vector of $R_{n,k}$, we have, invoking the decomposition $(1)$, $$S_k(P) := [P]\cdot [R_{n,k}] = \sum_{j=0}^{n-1} a_j [R_{n,j}]\cdot [R_{n,k}] \tag{2}$$ But $$[R_{n,j}]\cdot [R_{n,k}]=\sum_{m=0}^{n-1} \omega_n^{mj} \omega_n^{mk}=\sum_{m=0}^{n-1}\left(\omega_n^{j+k}\right)^m=\begin{cases} n, & j+k=0 \pmod{n} \\ 0, & \text{otherwise}\end{cases} \tag{3}$$ Therefore, $(2)$ reduces to $$S_k(P) = n a_{n-k} \tag{4}$$ (In the case of $k=0$, we take $a_{n-0}=a_0$.) Therefore, the test for (oriented) regularity becomes

$P$ is a regular $\{n/k\}$-gon if and only if $S_{n-k}(P)\neq 0$, and $S_m(P)=0$ for $0<m\neq n-k$.

For non-oriented regularity:

$P$ is a non-oriented regular $\{n/k\}$-gon if and only if $S_m(P)\neq0$ for $m=k$ or $m=n-k$ (but not both, unless $k=n/2$), and $S_m(P)=0$ for all other $m>0$.

In the case of a triangle $T$, decomposition $(1)$ is simply $$[T] = a_0[R_{3,0}] + a_1[R_{3,1}]+ a_2[R_{3,2}] \tag{1.3}$$ OP's test for regularity (that is, equilaterality) amounts to checking whether $S_1(T)=0$, which corresponds to whether the $R_{3,2}$ component of $(1)$ vanishes. This is consistent with the oriented test given above. For a non-oriented test, we'd check that $S_1(T)=0$ or $S_2(T)=0$ (but not both).

So, for general polygons, OP was correct to suspect a connection (congratulations on the insight!), but simply didn't allow for enough tests.

---

The ideas here are captured in a modified theorem of Barlotti. Barlotti observed that any $n$-gon is a "vertex sum" of affinely-regular non-oriented $\{n/k\}$-gons (with $0\leq k\leq n/2$). And it happens that, for $k\neq0$ and $k\neq n/2$, an affinely-regular $\{n/k\}$-gon is itself the vertex sum of oppositely-oriented actually-regular ones, so that we get a sum ($(1)$ above) with an $\{n/k\}$ component for each $0\leq k\leq n-1$.

(One can also write any polyhedron as a "vertex sum" of combinatorially-equivalent "symmetric" polyhedra. The nature of the "symmetric" components, and how many there are, depend upon eigen-analysis of the adjacency matrix of the polyhedron's vertex-and-edge skeleton. Likewise in higher dimensions. But I digress ...)

An interesting related result is a generalization of Napoleon's Theorem:

Given an oriented $n$-gon, erect an $\{n/k\}$-gon, for some $k$, on each edge, and connect their centroids in order to create a new oriented polygon. (Note that if $k=0$, the new polygon matches the old, so we could ignore it.) On the sides of that polygon, erect $\{n/k\}$-gons, for a different $k$; iterate.

After $n-1$ steps (or $n-2$ steps, if we're ignoring $k=0$), the resulting polygon will be a regular $\{n/k\}$-gon, for the leftover value of $k$!

As it happens, the order of the $k$s used in the iteration doesn't matter.

Napoleon's Theorem for an oriented triangle (ignoring $k=0$) erects $\{3/1\}$-triangles on the sides to get a regular $\{3/2\}$-triangle; and erects $\{3/2\}$-triangles on the sides to get a regular $\{3/1\}$-triangle. (The two results are the "inner" and "outer" Napoleon triangles. Which is which depends upon how the original triangle is oriented.) Likewise, the iterative process above associates with every $n$-gon a regular "leftover" $\{n/k\}$-gon for every $k$ ... kinda like the decomposition $(1)$. However, these Napoleonic polygons do not typically match the size and/or rotation of the components in $(1)$.