Question

Assume that $A_i(i=1,2,......,n)$ are the vertices of a regular n-gon inscribed in circle of radius unity.
Prove that : $|A_1{A}_2||A_1{A}_3|....|A_1{A}_n|=n$

Answer 1

From part(i),$|A_1{A}_r|=|z_1||1-e^{2(r-1)\pi i/n}|$
$=|1-e^{2(r-1)\pi i/n}|[\because |z_1|=1]$
Hence $|A_1{A}_2|\cdot |A_1{A}_3|.......|A_1{A}_n|$
$=|1-e^{2\pi i/n}||1-e^{4\pi i/n}|.......|1-e^{2(n-1)\pi i/n}|.....\left(1\right)$
$Since{e}^{2\pi i/n},e^{4\pi i/n}.....e^{2(n-1)\pi i/n}arethen-1$
imaginary, $n^{th}$ root of unity, we have the identity
$z^n-1\equiv (z-1)(z-e^{2\pi i/n})(z-e^{4\pi i/n})......(z-e^{2(n-1)\pi i/n})$
or $\frac{z^n-1}{z-1}\equiv (z-e^{2\pi i/n})(z-e^{4\pi i/n})......(z-e^{2(n-1)\pi i/n})$
or$1+z+z^2+....+z^{n-1}\equiv (z-e^{2\pi i/n})......(z-e^{2(n-1)\pi i/n})$
Putting z =1 in the above identity, we get
$n=(1-e^{2\pi i/n})(1-e^{4\pi i/n})......(1-e^{2(n-1)\pi i/n})$
Hence $n=|n|=|1-e^{2\pi i/n}||1-e^{4\pi i/n}|......|1-e^{2(n-1)\pi i/n}|....\left(2\right)$
From (1) and (2), we get
$\therefore |A_1{A}_2|\cdot |A_1{A}_3|.......|A_1{A}_n|=n$