If $ \alpha_i, i=0,1,2...n-1 $ be the nth roots of unity, the $\sum_{i=0}^{n-1} \frac{\alpha_i}{3- \alpha_i}$ is equal to?

Question

If $ \alpha_i, i=0,1,2...n-1 $ be the nth roots of unity, the $\sum_{i=0}^{n-1} \frac{\alpha_i}{3- \alpha_i}$ is equal to?

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

Attempt: I know that $(3- \alpha_0)(3- \alpha_1)....(3-\alpha_{n-1})= (3^n-1)/2 $

But I have no clue about the numerator. Adding 3 and subracting 3 from the numberator would make the fraction simpler, but I would still have to sum up $ 1/(3- \alpha_i) $

Answer 1

Let $\dfrac{a_i}{3-a_i}=b_i\iff a_i=\dfrac{3b_i}{1+b_i}$

As $a_i^n=1,$

$$\left(\dfrac{3b_i}{1+b_i}\right)^n=1\iff(3^n-1)b_i^n-\binom n1b_i^{n-1}+\cdots=0$$

By Vieta's formula,

$$\displaystyle\sum_{i=0}^{n-1}b_i=\dfrac{\binom n1}{3^n-1}$$

Answer 2

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

So your sum is:

$$-n+3\sum \frac{1}{3-a_i}$$

Letting $p(x)=x^n-1$, show that $\frac{p'(x)}{p(x)}=\sum \frac{1}{x-a_i}$.

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More generally, if $p(x)=(x-b_1)(x-b_2)\cdots (x-b_n)$ then $$\sum \frac{b_i}{b-b_i} = -n + b\frac{p'(b)}{p(b)}$$

Answer 3

(For future reference.) Introducing

$$f(z) = \frac{z}{3-z} \frac{nz^{n-1}}{z^n-1}$$

we get

$$S_n = \sum_{q=0}^{n-1} \frac{\exp(2\pi i q/n)}{3-\exp(2\pi i q/n)} = \sum_{q=0}^{n-1} \mathrm{Res}_{z=\exp(2\pi i q/n)} f(z).$$

Residues sum to zero so we have

$$S_n + \mathrm{Res}_{z=3} f(z) + \mathrm{Res}_{z=\infty} f(z) = 0$$

Observe that

$$\mathrm{Res}_{z=3} f(z) = - \frac{n3^n}{3^n-1}$$

and

$$\mathrm{Res}_{z=\infty} f(z) = -\mathrm{Res}_{z=0} \frac{1}{z^2} f(1/z) \\ = -\mathrm{Res}_{z=0} \frac{1}{z^2} \frac{1/z}{3-1/z} \frac{n/z^{n-1}}{1/z^n-1} \\ = -\mathrm{Res}_{z=0} \frac{1}{z^2} \frac{1}{3z-1} \frac{nz}{1-z^n} \\ = -\mathrm{Res}_{z=0} \frac{1}{z} \frac{1}{3z-1} \frac{n}{1-z^n} = n.$$

Hence

$$S_n - \frac{n3^n}{3^n-1} + n = 0$$

or

$$S_n = \frac{n3^n-n(3^n-1)}{3^n-1} = \frac{n}{3^n-1}.$$