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

Alternate Solution:
$1,z_1,z_2,...,z_{n-1}$ are the roots of $z^n-1=0$
$\Rightarrow (z-1)(z-z_1)(z-z_2)...(z-z_{n-1})=0$

$\Rightarrow 1+z+z^2+...+z^{n-1}=0$

Replace $\frac{1}{3-z}=y\Rightarrow z=\frac{3y-1}{y}$

$1+\frac{3y-1}{y}+{\left(\frac{3y-1}{y}\right)}^2+...+{\left(\frac{3y-1}{y}\right)}^{n-1}=0$

Multiply by $y^{n-1}$
$\Rightarrow y^{n-1}+y^{n-2}(3y-1)+y^{n-3}(3y-1{)}^2+...+(3y-1{)}^{n-1}=0$

$\Rightarrow y^{n-1}(1+3+3^2+...+3^{n-1})-y^{n-2}(1+2\cdot 3+3\cdot 3^2+4\cdot 3^3+...+(n-1\left)3^{n-2}\right)+...+(-1{)}^{n-1}=0$

We know that, sum of roots $=-\frac{b}{a}$
$\therefore \frac{1}{3-z_1}+\frac{1}{3-z_2}+\frac{1}{3-z_3}+...+\frac{1}{3-z_{n-1}}=\frac{1+2\cdot 3+3\cdot 3^2+4\cdot 3^3+...+(n-1)3^{n-2}}{1+3+3^2+...+3^{n-1}}$
$=\frac{\sum _{r=1}^{n-1}r\cdot 3^{r-1}}{\sum _{r=1}^n{3}^{r-1}}$

Let $S=\sum _{r=1}^{n-1}r\cdot 3^{r-1}$
$S=1+2\cdot 3+3\cdot 3^2+4\cdot 3^3+...+(n-1)3^{n-2}$
$3S=3+2\cdot 3^2+3\cdot 3^3+...+(n-2)3^{n-2}+(n-1)3^{n-1}$
Subtracting above equations, we get
$-2S=\frac{3^{n-1}-1}{2}-(n-1)3^{n-1}$
$\Rightarrow S=\frac{n\cdot 3^{n-1}}{2}-\frac{3^n-1}{4}$

$S^{\prime }=\sum _{r=1}^n{3}^{r-1}=\frac{3^n-1}{2}$

Required Answer=
$=\frac{n\cdot 3^{n-1}}{3^n-1}-\frac{1}{2}$