Question

$1,z_1,z_2,...z_{n-1}$ are the n 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. $\frac{n{3}^{n-1}}{3^n-1}+1$

Answer 1

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

$z^n-1=(z-1)(z-z_1)(z-z_2)...(z-z_{n-1})$ ...(1)
Taking logarithm on both sides and differentiate w.r.t.z
$\Rightarrow \frac{n{z}^{n-1}}{z^n-1}=\frac{1}{z-1}+\frac{1}{z-z_1}+\frac{1}{z-z_2}+....+\frac{1}{z-z_{n-1}}$
put z=3 we get
$\frac{1}{3-z_1}+\frac{1}{3-z_2}+...\frac{1}{3-z_{n-1}}=\frac{n{3}^{n-1}}{3^n-1}-\frac{1}{2}$.