2016.3.7 Question 7

For $\omega =\text{exp}<!--\; FUNCTION\; APPLICATION\; -->\frac{2\mathit{\pi i}}{n}$, we have for $k=0,1,2,\dots ,n-1$, that ${\omega }^k=\text{exp}<!--\; FUNCTION\; APPLICATION\; -->\frac{2\mathit{\pi ik}}{n}$. Therefore,

Also, notice that $\text{arg}<!--\; FUNCTION\; APPLICATION\; -->{\omega }^k=\frac{2\mathit{k\pi }}{n}$, which means that all ${\omega }^k$s are different.

This means that ${\omega }^0=1,{\omega }^1=1,{\omega }^2,\dots ,{\omega }^{n-1}$ are exactly the $n$ roots to the polynomial $z^n-1$, which has leading coefficient 1.

Therefore, we must have

as desired.

For the following parts, W.L.O.G. let the orientation of the polygon be such that $X_k={\omega }^k$.

1. Let $z$ represent the complex number for $P$, we have

   $$\begin{array}{llll}\hfill \prod _{k=0}^{n-1}\left|P{X}_k\right|& =\prod _{k=0}^{n-1}\left|z-{\omega }^k\right|& \hfill & \\ \hfill & =\left|\prod _{k=0}^{n-1}(z-{\omega }^k)\right|& \hfill & \\ \hfill & =\left|z^n-1\right|.& \hfill & \end{array}$$

   Since $P$ is equidistant from $X_0$ and $X_1$, we must have that $P=r\text{exp}<!--\; FUNCTION\; APPLICATION\; -->\left(\frac{\mathit{\pi i}}{n}\right)$ for some $r\in ℝ$, where $\left|r\right|=\left|\mathit{OP}\right|$. Therefore, we have

   $$\begin{array}{llll}\hfill \prod _{k=0}^{n-1}\left|P{X}_k\right|& =\left|z^n-1\right|& \hfill & \\ \hfill & =\left|r^n\text{exp}<!--\; FUNCTION\; APPLICATION\; -->\left(\frac{\mathit{\pi i}}{2}\right)-1\right|& \hfill & \\ \hfill & =\left|-r^n-1\right|& \hfill & \\ \hfill & =\left|r^n+1\right|.& \hfill & \end{array}$$

   If $n$ is even, then $r^n=|r{|}^n>0$, and therefore $\left|r^n+1\right|=r^n+1={\left|r\right|}^n+1={\left|\mathit{OP}\right|}^n+1$ as desired.

   If $n$ is odd, and $r>0$, then $r^n=|r{|}^n>0$, and

   $$\begin{array}{llll}\hfill \text{LHS}& =\left|r^n+1\right|& \hfill & \\ \hfill & =r^n+1& \hfill & \\ \hfill & =|r{|}^n+1& \hfill & \\ \hfill & ={\left|\mathit{OP}\right|}^n+1.& \hfill & \end{array}$$

   When $-1\le r<0$, we have $-1\le r^n=-|r{|}^n<0$, and

   $$\begin{array}{llll}\hfill \text{LHS}& =\left|r^n+1\right|& \hfill & \\ \hfill & =r^n+1& \hfill & \\ \hfill & =-|r{|}^n+1& \hfill & \\ \hfill & =-{\left|\mathit{OP}\right|}^n+1.& \hfill & \end{array}$$

   When $r<-1$, we have $r^n=-|r{|}^n<-1$, and

   $$\begin{array}{llll}\hfill \text{LHS}& =\left|r^n+1\right|& \hfill & \\ \hfill & =-r^n-1& \hfill & \\ \hfill & =|r{|}^n-1& \hfill & \\ \hfill & ={\left|\mathit{OP}\right|}^n-1.& \hfill & \end{array}$$

   In summary, when $n$ is odd, we have

   $$\prod _{k=0}^{n-1}\left|P{X}_k\right|=\{\begin{array}{cc}{\left|\mathit{OP}\right|}^n+1,\hfill & P\text{ is in the first quadrant},\hfill \\ -{\left|\mathit{OP}\right|}^n+1,\hfill & P\text{ is in the third quadrant and }|\mathit{OP}|\le 1,\hfill \end{array}$$

2. Notice that for a general point $P$ whose complex number is $z$, we have

   $$\begin{array}{llll}\hfill \prod _{k=1}^{n-1}\left|P{X}_k\right|& =(z-\omega )(z-{\omega }^2)\cdots \left(z-{\omega }^{n-1}\right)& \hfill & \\ \hfill & =\frac{z^n-1}{z-1}& \hfill & \\ \hfill & =1+z+z^2+\cdots +z^{n-1}.& \hfill & \end{array}$$

   If we let $P=X_0$, $z=1$, and $\text{RHS}=n$, just as we desired.