Question

If $1,z_1,z_2,z_3,\cdots z_{n-1}$ are $n$ roots of unity then the value of $\frac{1}{3-z_1}+\frac{1}{3-z_2}+\cdots +\frac{1}{3-z_{n-1}}$ is equal to :

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

Answer 1

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

$x^n=1$
$\Rightarrow x^n-1=0$
$x^n-1=(x-1)(x-z_1)(x-z_2)\cdots (x-z_{n-1})$ where $1,z_2,\cdots ,z_{n-1}$ are the $n$ roots
Now taking $\log$ both sides
$\log (x^n-1)=\log (x-1)+\log (x-z_1)+\log (x-z_2)+\cdots +\log (x-z_{n-1})$
differentiating both side w.r.t $x$, we get
$\frac{n{x}^{n-1}}{x^n-1}=\frac{1}{x-1}+\frac{1}{x-z_1}+\frac{1}{x-z_2}+\cdots +\frac{1}{x-z_{n-1}}$
Putting $x=3$, we get
$\frac{1}{3-z_1}+\frac{1}{3-z_2}+\cdots +\frac{1}{3-z_{n-1}}=\frac{n\cdot 3^{n-1}}{3^n-1}-\frac{1}{2}$