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For any two points inside a regular polygon on the complex plane, where the vertices of the polygon are on a unit circle, one of the vertices is at (1, 0), and the centre of the polygon is at (0,0). Can we prove that the multiplication of these two complex number is still inside this polygon (a rotated and shortened version of one of the two original complex numbers)?

I guess it's right but I haven't figured out how to prove it. Or maybe it's just not right.

Take the square as an example, with four vetices $v_1, \ldots v_4$, I am just wondering whether the multiplication of any $a_1a_2$ is still in this regular polygon.

enter image description here

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    $\begingroup$ Is one of the vertices at the point $1$ on the complex plane? $\endgroup$ Apr 3 at 2:57
  • $\begingroup$ Do you mean (1,0)? Yes, I would set one of the vertices at (1,0). $\endgroup$
    – tyrela
    Apr 3 at 3:05
  • $\begingroup$ You say polygon, is $n$-sided, or do you have an idea in mind? @tyrela $\endgroup$
    – Masd
    Apr 3 at 3:10
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    $\begingroup$ The question should also mention that the number $1$ ($1+0i$) is one of the vertices, since that does change the answer. $\endgroup$
    – aschepler
    Apr 3 at 3:26
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    $\begingroup$ I think you also need to add that the centre of the polygon is at 0, otherwise the answer is obviously no. $\endgroup$ Apr 3 at 20:40

5 Answers 5

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The vertices of your polygon are given by $1,\omega,\omega^2,\ldots,\omega^{n-1}$, where $\omega=e^{2\pi i/n}$ is a primitive $n$th root of unity. Since the polygon is convex, each point inside it is a convex linear combination* of the points $\{1,\omega,\omega^2,\ldots,\omega^{n-1}\}$. This means that, if $z_1$ and $z_2$ are both inside the polygon, each is a convex linear combination of $\{\omega^j\colon 0\leq j<n\}$, and so their product is a convex linear combination of $$\{\omega^{j+k}\colon 0\leq j,k<n\}.$$ (This is a special case of a more general fact: if $z_1$ is a convex linear combination of a set $S_1$ and $z_2$ is a convex linear combination of a set $S_2$, then $z_1z_2$ is a convex linear combination of the product set $\{s_1s_2\colon s_1\in S_1,s_2\in S_2\}$.) Now, since $\omega$ is an $n$th root of unity, each $\omega^{j+k}$ is equal to $\omega^\ell$ for some $0\leq\ell<n$ (take $\ell=j+k$ if $j+k<n$, and $\ell=j+k-n$ otherwise). So, $$\{\omega^{j+k}\colon 0\leq j,k<n\}=\{\omega^\ell\colon 0\leq \ell<n\}$$ is simply the set of vertices of the original polygon. So, $z_1z_2$ is a convex linear combination of the vertices of the polygon, and so it lies in the polygon, as desired.

*A convex linear combination of points $x_1,\ldots,x_n$ is an expression of the form $\lambda_1x_1+\cdots+\lambda_nx_n$ with $\lambda_1,\ldots,\lambda_n\geq 0$ and $\lambda_1+\cdots+\lambda_n=1$.

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    $\begingroup$ Really thanks for the answer! Sorry that I have to grab some food and I will get back to your answer asap. $\endgroup$
    – tyrela
    Apr 3 at 4:09
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    $\begingroup$ The OP only specified that the vertices are on the unit circle, so the points may be are different from the roots of unity, and in that case the statement is false. Your proof only works because you assumed that the point are the $n$-th roots of unity. $\endgroup$
    – jjagmath
    Apr 3 at 12:44
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    $\begingroup$ In particular this uses the fact that the product of two vertices is a vertex, which only holds if one of the vertices is at 1. Otherwise the product of two vertices is not a vertex and this proof doesn't go through, and in fact that's a counterexample. $\endgroup$ Apr 3 at 14:25
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    $\begingroup$ @jjagmath: the question states that the polygon is regular and cyclic, and that $(1,0)$ is one of the vertices. I believe that that implies that the vertices are roots of unity. $\endgroup$
    – robjohn
    Apr 3 at 15:11
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    $\begingroup$ @robjohn FYI, please see my comment, and its continuation, which deals with the apparent intent, but the question not being completely clear, regarding the issue of whether or not the $(1,0)$ point is meant to apply to just a square or to all regular polygons. $\endgroup$ Apr 3 at 15:16
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The statement is false. Consider the regular triangle with vertices $-1$, $\frac{1+i\sqrt 3}{2}$ and $\frac{1-i\sqrt 3}{2}$. Then $-.9$ is inside the triangle but $(-.9)^2$ is not.

Update: now that the OP specified that one of the vertex is $1$, this counterexample doesn't apply. The set of vertices are the $n$-th roots of unity and Carl Schildkraut's proof works.

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    $\begingroup$ @JohnOmielan The question ask in general for a "regular polygon on the complex plane, where the vertices of the polygon are on a unit circle". The part where $(1,0)$ is one of the vertex is one particular case ("Take the square as an example ...") $\endgroup$
    – jjagmath
    Apr 3 at 8:34
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    $\begingroup$ More generally, it is necessary that the set of vertices is closed under multiplication (otherwise take points sufficiently close to vertices whose product lands outside), which necessitates that $1$ is one of the vertices. $\endgroup$
    – ronno
    Apr 3 at 12:09
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    $\begingroup$ There's a later aschepler comment which states "The question should also mention that the number $1 (1+0i)$ is one of the vertices, since that does change the answer", and then from @tyrela there's this comment reply of "Got it, I will edit it here." Thus, the intent is for all cases, so the ... $\endgroup$ Apr 3 at 14:17
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    $\begingroup$ (cont.) clarification should have been in the first paragraph with the general statement. As such, as currently stated, it's at least ambiguous, and it's quite reasonable to have interpreted it as you did. $\endgroup$ Apr 3 at 14:18
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    $\begingroup$ @tyrela Ok. Now that you precise that, your conjecture is correct. But your notation is faulty. Either denote the points as complex numbers: $0,1,1+i,\ldots$ or as pairs of reals $(0,0),(1,0),(1,1)\ldots$, but mixing them by writing the complex number $a+bi$ (or the point $(a,b)$) as $(a,bi)$ is incorrect. $\endgroup$
    – jjagmath
    Apr 4 at 2:20
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Not an answer, just a couple of illustrative animations for posterity of heptagons ($n = 7$) that are "standard" (vertices are $n$th roots of unity) or not standard. Each shows a regular polygon $P$ in blue, and the rotated-and-scaled polygon $zP$ in green as $z$ moves along one edge of $P$.

An animation loop showing "the standard regular heptagon" multiplied by a point along the edge from unity to the first non-trivial seventh root of unity. The light green "product" heptagon is contained in the light blue original. An animation loop showing a "non-standard regular heptagon" multiplied by a point along the "first edge." The light green "product" heptagon is generally not contained in the light blue original.

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    $\begingroup$ Really thanks for the illustration. It visually verifies the answers. Nice and neat! $\endgroup$
    – tyrela
    Apr 5 at 2:35
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This is a nice problem -- here's an abstract way of thinking about it. This will also explain why the polygon must contain the vertex $1$.

Let $P \subseteq \mathbb C$ be a nonempty convex polygon, whose finite set of vertices $D$ lies the unit circle $\mathbb S^1$. Then we can reduce the question to classifying how the set $D$ must look:

Lemma 1: P is multiplicatively closed if and only if $D$ is

For the direction $\Rightarrow$, we have that the product of any two vertices must again be in $P$. By inspecting the absolute value, it must in fact land in $D$ again. The direction $\Leftarrow$ uses the convexity argument of Carl Schildkraut. If $D$ is multiplicatively closed, so is its convex hull, which is $P$.

Lemma 2: $D$ is multiplicatively closed if and only if it is the set of $n$-th roots of unity, for some $n$. In particular, it must contain $1$.

This is a standard algebraic argument applied to $\mathbb S^1$: A nonempty finite sub-semigroup of a group must be a subgroup.

Putting those lemmas together, we get the solution:

Claim: $P$ is multiplicatively closed if and only if it is a regular $n$-gon with vertex at $1$.

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Start with a simple case - the solutions to $z^{n}=z_{0}$ will lie on a circle centered at $z=z_{0}$ and the vertices will be $z=1, z=w, z=w^{2},...,z=w^{n-1}$

Take any two of these and you will achieve the complex number $z=w^{m}$ which will be a point on the same circle.

Or have I misunderstood your question? (It happens a lot...)

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    $\begingroup$ I think the two points are to be chosen from the polygon's "inside", not two of its vertices. $\endgroup$
    – aschepler
    Apr 3 at 3:18
  • $\begingroup$ Ah... so "in" here should be taken to mean "inside"..? I'll ponder some more. $\endgroup$
    – Red Five
    Apr 3 at 3:19
  • $\begingroup$ yes sorry, it's "inside". I will modify my question here. $\endgroup$
    – tyrela
    Apr 3 at 3:21

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