Summation containing $n$-th roots of unity

Question

If $a_{0},a_{1},a_{2},a_{3}....a_{n-1}$ are the nth roots of unity, then the value of $$\sum_{i=0}^{n-1}{\left(\frac{a_{i}}{3-a_{i}}\right)}.$$

I tried expressing $a_{i}$ in euler form as $e^{\frac{2k\pi i}{n}}$ but I am stuck after it.

Edit: It is a question from JEE examination.

Answer 1

$$ \begin{align} \sum_{k=0}^{n-1}\frac{e^{\frac{2\pi ik}{n}}}{3-e^{\frac{2\pi ik}{n}}} &=\sum_{k=0}^{n-1}\frac{\frac13e^{\frac{2\pi ik}{n}}}{1-\frac13e^{\frac{2\pi ik}{n}}}\tag1\\ &=\sum_{m=1}^\infty\sum_{k=0}^{n-1}\frac1{3^m}e^{\frac{2\pi ikm}{n}}\tag2\\ &=\sum_{m=1}^\infty\frac1{3^m}n\,[n\mid m]\tag3\\ &=\sum_{j=1}^\infty\frac1{3^{jn}}n\tag4\\[6pt] &=\frac{n}{3^n-1}\tag5 \end{align} $$ Explanation:
$(1)$: cancel $3$ in numerator and denominator
$(2)$: use the Taylor series $\frac{x}{1-x}=\sum\limits_{k=1}^\infty x^k$
$(3)$: $\sum\limits_{k=0}^{n-1}e^{\frac{2\pi ikm}{n}}=n\,[n\mid m]$ (Iverson brackets)
$(4)$: select the terms where $m=jn$
$(5)$: sum the geometric series

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Explanation of step $\boldsymbol{(3)}$

Note that the sum of the following finite geometric series $$ \sum\limits_{k=0}^{n-1}e^{\frac{2\pi ikm}{n}}=\frac{e^{\frac{2\pi inm}{n}}-1}{e^{\frac{2\pi im}{n}}-1}=\frac0{e^{\frac{2\pi im}{n}}-1}\tag6 $$ shows that if $n\nmid m$, $$ \sum\limits_{k=0}^{n-1}e^{\frac{2\pi ikm}{n}}=0\tag7 $$ It is pretty simple to see that if $n\mid m$, $$ \sum\limits_{k=0}^{n-1}e^{\frac{2\pi ikm}{n}}=\sum\limits_{k=0}^{n-1}1=n\tag8 $$

Answer 2

This is only a hint,

$$ \begin{aligned} \sum_{i=0}^{n-1}{\left(\frac{a_{i}}{3-a_{i}}\right)}+n&=\sum_{i=0}^{n-1}{\left(\frac{1}{1-\frac{a_{i}}{3}}\right)}\\ &=\frac{\sum_{i=0}^{n-1}{\left(\prod_{j\neq i}{\left(1-\frac{a_{j}}{3}\right)}\right)}}{\prod_{i=0}^{n-1}{\left(1-\frac{a_{i}}{3}\right)}} \end{aligned} $$

Then You can use Vieta's formula.