Question

State true or false:

$A_1,A_2,A_3...A_n$ are the vertices of a regular polygon of $n$ sides. $|O{A}_1|=1$. Then,

$\left|A_1{A}_2\right|\left|A_1{A}_3\right|....\left|A_1{A}_n\right|=n$.

Answer 1

From part (i), $\left|A_1{A}_r\right|=\left|z_1\right||1-e^{2(r-1)\pi i/n}|$
$=|1-e^{2(r-1)\pi i/n}|[\because \left|z_1\right|=1]$
Hence $\left|A_1{A}_2\right|.\left|A_1{A}_3\right|......\left|A_1{A}_n\right|$
$=|1-e^{2\pi i/n}||1-e^{4\pi i/n}|.......|1-e^{2(n-1)\pi i/n}|$ ...(1)
Since $e^{2\pi i/n},e^{4\pi i/n},.....e^{2(n-1)\pi i/n}$ are the n-1 imaginary, $n^{th}$ roots 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 $n=\left|n\right|=|1-e^{2\pi i/n}||1-e^{4\pi i/n}|......|1-e^{2(n-1)\pi i/n}|$ ...(2)
$\therefore$ From (1) and (2), we get
$\therefore \left|A_1{A}_2\right|.\left|A_2{A}_3\right|\cdots \left|A_1{A}_n\right|=n.$