Question

If $1,z_1,z_2,z_{n-1}$ are the $n^{th}$ roots of unity then the value of $\frac{1}{3-z_1}+\frac{1}{3-z_2}+.......+\frac{1}{3-z_{n-1}}$ is equal to

• A. $\frac{{n.3}^{n-1}}{3^n-1}+\frac{1}{2}$
• B. $\frac{{n.3}^{n-1}}{3^n-1}-1$
• C. $\frac{{n.3}^{n-1}}{3^n-1}+1$
• D. $\frac{{n.3}^{n-1}}{3^n-1}-\frac{1}{2}$

Answer 1

The correct option is D $\frac{{n.3}^{n-1}}{3^n-1}-\frac{1}{2}$

$z_1,z_2,z_3........z_{n-1}$ are normal $n^{th}$

$x^n-1=(x-1)(x-z_1)(x-z_2)(x-z_3).......(x-z_{n-1})$

ln $(x^n-1)=ln(x-1)(x-z_2)(x-z_3).........(x-z_{n-1})$

$\frac{n{x}^{n-1}}{x^n-1}=\frac{1}{x-1}+\frac{1}{x-z_1}+\frac{1}{x{-}_2}......\frac{1}{x-z_{n-1}}$

Putting $x=3$

$\frac{n{3}^{n-1}}{3^{n-1}}=\frac{1}{2}+\frac{1}{3-z_1}+\frac{1}{3-z_2}........\frac{1}{3-z_{n-1}}$

$\frac{1}{3-z_1}+\frac{1}{3-z_2}..............\frac{1}{3-z_{n-1}}=\frac{n(3{)}^{n-1}}{3^n-1}\frac{1}{2}$