Question

Let $A_1,A_2,\cdots ,A_n$ be the vertices of a regular polygon of $n$ sides in a circle of radius unity and
$a=|A_1{A}_2{|}^2+|A_1{A}_3{|}^2+\cdots |A_1{A}_n{|}^2,$
$b=|A_1{A}_2||A_1{A}_3|\cdots |A_1{A}_n|,$ then $\frac{a}{b}=$

Answer 1

Let us assume that $O$ is the centre of the regular polygon and $z_0,z_1,\cdots ,z_{n-1}$ represent the affixes of $A_1,A_2,\cdots ,A_n,$ such that
$z_0=1,z_1=\alpha ,z_2={\alpha }^2,\cdots ,z_{n-1}={\alpha }^{n-1},$ where $\alpha =e^{i2\pi /n}$
(Since all $n^{th}$ roots of unity can be represented as verticies of regular polygon inscribed in a circle of unit radius)
Now, $|A_1{A}_r{|}^2=|{\alpha }^r-1{|}^2=|1-{\alpha }^r{|}^2$
$={|1-\cos \frac{2r\pi }{n}-i\sin \frac{2r\pi }{n}|}^2$
$={(1-\cos \frac{2r\pi }{n})}^2+{\left(\sin \frac{2r\pi }{n}\right)}^2$
$=2-2\cos \frac{2r\pi }{n}$
$\therefore \sum _{r=2}^n|A_1{A}_2{|}^2=\sum _{r=1}^n(2-2\cos \frac{r\pi }{n})$
$=2(n-1)-2[\cos \frac{2\pi }{n}+\cos \frac{4\pi }{n}+\cdots \cos \frac{2(n-1)\pi }{n}]$
$=2(n-1)-2$ Real part of $(\alpha +{\alpha }^2+\cdots +{\alpha }^{n-1})$
$=2(n-1)-2(-1)=2n$
$[\because 1+\alpha +{\alpha }^2+\cdots +{\alpha }^{n-1}=0]$
$\therefore |A_1{A}_2{|}^2+|A_1{A}_3{|}^2+\cdots |A_1{A}_n{|}^2=2n$

Also, let $E=|A_1{A}_2||A_1{A}_3|\cdots |A_1{A}_n|$
$=|1-\alpha ||1-{\alpha }^2||1-{\alpha }^3|\cdots |1-{\alpha }^{n-1}|$
$=|(1-\alpha )(1-{\alpha }^2)(1-{\alpha }^3)\cdots (1-{\alpha }^{n-1})|$
Since, $1,\alpha ,{\alpha }^2,\cdots ,{\alpha }^{n-1}$ are the roots of $z^n-1=0$
$\Rightarrow (z-1)(z-\alpha )(z-{\alpha }^2)\cdots (z-{\alpha }^{n-1})=z^n-1$
$\Rightarrow (z-\alpha )(z-{\alpha }^2)\cdots (z-{\alpha }^{n-1})=\frac{z^n-1}{z-1}$
$=1+z+z^2+\cdots +z^{n-1}$
Substituting $z=1,$ we have
$=(1-\alpha )(1-{\alpha }^2)(1-{\alpha }^3)\cdots (1-{\alpha }^{n-1})=n$
$\therefore |1-\alpha ||1-{\alpha }^2||1-{\alpha }^3|\cdots |1-{\alpha }^{n-1}|=n$
Hence, the value of $\frac{a}{b}=\frac{2n}{n}=2$