Condition that a polygon is regular in the complex plane

complex numbersgeometrypolygonsroots-of-unity

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?

Best Answer

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.

enter image description here

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;
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