Question

$1,z_1,z_2,z_3,...,z_{n-1}$ 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. none of these

Answer 1

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

If ${\alpha }_1,{\alpha }_2,...,{\alpha }_m$ are the roots of polynomial equation
$f\left(x\right)=a_0{x}^m+a_1{x}^{m-1}+...+a_{m-1}x+a_m=0$
Then,
$f\left(x\right)=a_0(x-{\alpha }_1)...(x-{\alpha }_m).$

and $\frac{f^{\prime }\left(x\right)}{f\left(x\right)}=\frac{1}{x-{\alpha }_1}+...+\frac{1}{x-{\alpha }_m}$
The equation in question is $x^n-1=0$

$f\left(x\right)=x^n-1=(x-1)(x-z_1)...(x-z_{n-1})$

Thus, $\frac{f^{\prime }\left(x\right)}{f\left(x\right)}=\frac{n{x}^{n-1}}{x^n-1}=\frac{1}{x-1}+\frac{1}{x-z_1}+...+\frac{1}{x-z_{n-1}}$

Substituting $x=3$:

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

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

Hence, (A) is correct.