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.
Best Answer
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 $\lambda_1+\cdots+\lambda_n=1$.