State true or false:
Assume that $A_{i} (i = 1, 2, . . . . n)$ are the vertices of a regular n-sided polygon inscribed in a circle of radius unity. Then,
${| A_{1} A_{2} |}_{1}^{2} + {| A_{1} A_{3} |}_{1}^{2} + \dots + {| A_{1} A_{n} |}_{1}^{2} = - 1$ . Type 1 for true and 0 for false

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Solution

With origin as the centre of the circle of redius unity, let $z_{1}, z_{2}, . . . . ., z_{n}$ represent the vertices $A_{1}, A_{2}, . . . ., A_{n}$ of the n-gon.Then we easily get
$z_{2} = z_{1} e^{2 π i / n}, z_{3} = z_{1} e^{4 π i / n}, . . . . .,$
$z_{n} =_{1} e^{2 (n - 1) π i / 2 n}$
Now ${| A_{1} A_{r} |}_{1}^{2} = {| z_{1} - z_{r} |}_{1}^{2}$
$= {∣ ∣ z_{1} - z_{1} e^{2 (r - 1) π i / n} ∣ ∣}_{1}^{2}$
$= {| z_{1} |}_{1}^{2} {∣ ∣ 1 - e^{2 (r - 1) π i / n} ∣ ∣}^{2} .$
$= {∣ ∣ ∣ 1 - cos \frac{2 (r - 1) π}{n} - i sin \frac{2 (r - 1) π}{n} ∣ ∣ ∣}^{2}$
[ $∵ | z_{1} | =$ radius of the circle=1]
$= {(1 - cos \frac{2 (r - 1) π}{n})}^{2} + {sin}^{2} \frac{2 (r - 1) π}{n}$
$= 2 - 2 cos \frac{2 (r - 1) π}{n}$ ...(A)
Hence $n \sum r = 2 {| A_{1} A_{r} |}_{1}^{2} = 2 (n - 1) - 2 n \sum r = 2 [cos \frac{2 (r - 1) π}{n}]$ ...(1)
Let $S = n \sum r = 2 cos \frac{2 (r - 1) π}{n}$ ...(2)
$= cos \frac{2 π}{n} + cos \frac{4 π}{n} + cos \frac{6 π}{n} + \dots + \dots + cos \frac{2 (n - 1)}{n} π$
Formula:
$cos A + cos (A + B) + cos (A + 2 B) + . . . . . . + cos (A + ¯ ¯¯¯¯¯¯¯¯¯¯¯ ¯ n - 1 B)$
$sin A + sin (A + B) + sin (A + 2 B) + . . . . + sin (A + ¯ ¯¯¯¯¯¯¯¯¯¯¯ ¯ n - 1 B)$
There are n angles in A.P. of common diff. B.
$S = \frac{sin n . \frac{B}{2}}{sin \frac{B}{2}} cos$ or $sin {\frac{1 s t a n g . + l a s t a n g .}{2}}$
In (2) there are (n-1) terms and angles are in A.P. of common difference $\frac{2 π}{n} .$
$S = \frac{sin [(n - 1) . π / n]}{sin (π / n)} cos ⎧ ⎪ ⎨ ⎪ ⎩ \frac{\frac{2 π}{n} + \frac{2 (n - 1) π}{n}}{2} ⎫ ⎪ ⎬ ⎪ ⎭$
$= \frac{sin (π - π / n)}{sin (π / n)} cos π$
$= \frac{sin (π / n)}{sin (π / n)} (- 1) = - 1$
Hence $n \sum r = 2 {| A_{1} A_{r} |}_{1}^{2} = 2 (n - 1) - 2 S,$ by (1)
$= 2 n - 2 + 2 = 2 n$ as $S = - 1$

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