Question

lf 1, $a_1,a_2,...,a_{n-1}$ are $n^{th}$ roots of unity then $\frac{1}{1-a_1}+\frac{1}{1-a_2}+\dots +\frac{1}{1-a_{n-1}}$ equals?

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

Answer 1

Answer: (B)

$\frac{n-1}{2}$

We know that $w,w^2,1$ are the cube roots of unity

Hence,

$\frac{1}{1-w}+\frac{1}{1-w^2}$

$=\frac{1}{1-w}+\frac{1}{(1-w)(1+w)}$

Now,

$=\frac{1}{1-w}(1+\frac{1}{1+w})$

$=\frac{1}{1-w}\left(\frac{2+w}{1+w}\right)$

$=\frac{2+w}{1-w^2}$

We know

$1+w+w^2=0$

$2+w+w^2=1$ ...(by adding 1 on both sides).

$2+w=1-w^2$ substituting we get

$=\frac{2+w}{2+w}$

$=1$

$=\frac{3-1}{2}$

Hence, the correct alternative is $\frac{n-1}{2}$