Question

If $a_0,a_1,a_2,............,a_{n-1}$ are nth roots of unity, then $a_k$ = cos $\frac{2k\pi }{n}$ + i sin $\frac{2k\pi }{n}$ where $0\le k\le n-1$.

Also then Value of $2^{n-1}$sin$\left(\frac{\pi }{n}\right)$sin$\left(\frac{2\pi }{n}\right)$.........sin$\left(\frac{n-1}{n}\pi \right)$is

• A. n
• B. -1
• C. n-1
• D. n/3

Answer 1

The correct option is A

n

$a_0,a_1,a_2,.............a_{n-1}$ are nth roots of unity, then they are roots of

$x^n$ - 1 = $(x-a_0)(x-a_1)..............(x-a_{n-1})$ and $a_0$ = 1

$(x-a_1)(x-a_2)..............(x-a_{n-1})$ = $\frac{x^n-1}{x-1}$ = 1 + x + x ........+ $x^{n-1}$

Put x = 1 then $(1-a_1)(1-a_2).........(1-a_{n-1})$ = n

We have for $1\le k\le n-1$

$|1-a_k{|}^2$ =${[1-cos\frac{2k\pi }{n}]}^2$ + $si{n}^2\frac{2k\pi }{n}$

= 2$[1-cos\frac{2k\pi }{n}]$ = $4si{n}^2\frac{k\pi }{n}$

$|(1-a_1)(1-a_2).............(1-a_{n-1})|$ = n

$\Rightarrow$ $2^{n-1}$sin$\left(\frac{\pi }{n}\right)$sin$\left(\frac{2\pi }{n}\right)$............sin$\left(\frac{n-1}{n}\pi \right)$ = n