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Question

State true or false:
$$A_1, A_2, A_3... A_n$$ are the vertices of a regular polygon of $$n$$ sides. $$|OA_1| = 1$$. Then,
$$\displaystyle \left | A_{1}A_{2} \right |\left | A_{1}A_{3} \right |....\left | A_{1}A_{n} \right |=n$$. 


Solution

From part (i), $$\displaystyle \left | A_{1}A_{r} \right |= \left | z_{1} \right |\left | 1-e^{2\left ( r-1 \right )\pi i/n} \right |$$
$$\displaystyle = \left | 1-e^{2\left ( r-1 \right )\pi i/n} \right | \left [ \because \left | z_{1} \right |= 1 \right ]$$
Hence $$\displaystyle \left | A_{1}A_{2} \right |.\left | A_{1}A_{3} \right |......\left | A_{1}A_{n} \right |$$
$$\displaystyle = \left | 1-e^{2\pi i/n} \right |\left | 1-e^{4\pi i/n} \right |.......\left | 1-e^{2\left ( n-1 \right )\pi i/n} \right |$$ ...(1)
Since $$\displaystyle e^{2\pi i/n}, e^{4\pi i/n}, .....e^{2\left ( n-1 \right )\pi i/n}$$ are the n-1 imaginary, $$\displaystyle n^{th}$$ roots of unity, we have the identity
$$\displaystyle z^{n}-1\equiv \left ( z-1 \right )\left ( z-e^{2\pi i/n} \right )\left ( z-e^{4\pi i/n} \right )......\left ( z-e^{2\left ( n-1 \right )\pi i/n} \right )$$
or $$\displaystyle \frac{z^{n}-1}{z-1}\equiv \left ( z-e^{2\pi i/n} \right )\left ( z-e^{4\pi i/n} \right )......\left ( z-e^{2\left ( n-1 \right )\pi i/n} \right )$$
or $$\displaystyle 1+z+z^{2}+....+z^{n-1}\equiv \left ( z-e^{2\pi i/n} \right )......\left ( z-e^{2\left ( n-1\pi i/n \right )} \right )$$
Putting $$z=1$$ in the above identity, we get
$$\displaystyle n= \left ( 1-e^{2\pi i/n} \right )\left ( 1-e^{4\pi i/n} \right ).....\left ( 1-e^{2\left ( n-1 \right )\pi i/n} \right )$$
Hence $$\displaystyle n= \left | n \right |= \left | 1-e^{2\pi i/n} \right |\left | 1-e^{4\pi i/n} \right |......\left | 1-e^{2\left ( n-1 \right )\pi i/n} \right |$$ ...(2)
$$\displaystyle \therefore $$ From (1) and (2), we get
$$\displaystyle \therefore \left | A_{1}A_{2} \right |.\left | A_{2}A_{3} \right |\cdots \left | A_{1}A_{n} \right |= n.$$

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