Ytukyg

The question asks to compute:
$$sum_{k=0}^{n-1}dfrac{alpha_k}{2-alpha_k}$$
where $alpha_0, alpha_1, ldots, alpha_{n-1}$ are the $n$-th roots of unity.

I started off by simplifiyng and got it as:

$$=-n+2left(sum_{k=0}^{n-1} dfrac{1}{2-alpha_k}right)$$

Now I was stuck. I can rationalise the denominator, but we know $alpha_k$ has both real and complex components, so it can't be simplified by rationalising. What else can be done?

Since $alpha_0,alpha_1,alpha_2, dots , alpha_{n-1}$ are roots of the equation

$$x^n-1=0 ~~~~~~~~~~~~~ cdots ~(1)$$

You can apply Transformation of Roots to find a equation whose roots are$$frac{1}{2-alpha_0} , frac{1}{2-alpha_1},frac{1}{2-alpha_2}, dots , frac{1}{2-alpha_{n-1}}$$

Let $P(y)$ represent the polynomial whose roots are $frac{1}{2-alpha_k}$

$$y=frac{1}{2-alpha_k}=frac{1}{2-x} implies x=frac{2y-1}{y}$$

Put in $(1)$

$$Bigg(frac{2y-1}{y}Bigg)^n-1=0 implies (2y-1)^{n}-y^{n}=0$$

Use Binomial Theorem to find coefficient of $y^n$ and $y^{n-1}$.You will get sum of the roots using Vieta's Formulas.

Hope it helps!

I think the following answers the question using the method that Jyrki posted here.

Since $alpha_k$ are nth roots, so they satisfy $$f(x)=x^n-1=prod_{k=0}^{n-1}(x-alpha_k)$$

Putting in logarithm and derivating,

$$f'(x)/f(x)=dfrac{nx^{n-1}}{x^n-1}=sum_{k=0}^{n-1}dfrac{1}{x-alpha_k}$$

Thus $$f'(2)/f(2) = sum_{k=0}^{n-1}dfrac{1}{2-alpha_k} = dfrac{ncdot 2^{n-1}}{2^n-1}$$

Thus the required answer is given as:

$$-n+ 2left(dfrac{ncdot 2^{n-1}}{2^n-1}right)$$
$$=dfrac{n}{2^n-1}$$

For future reference here is a solution using residues. We have that
with $zeta_k = exp(2pi i k/n)$

$$S_n = sum_{k=0}^{n-1} frac{zeta_k}{2-zeta_k}
= sum_{k=0}^{n-1} mathrm{Res}_{z=zeta_k}
frac{z}{2-z} frac{nz^{n-1}}{z^n-1}
\ = sum_{k=0}^{n-1} mathrm{Res}_{z=zeta_k}
frac{1}{2-z} frac{nz^{n}}{z^n-1}
= sum_{k=0}^{n-1} mathrm{Res}_{z=zeta_k}
frac{1}{2-z} frac{n}{z^n-1}
.$$

Now observe that

$$mathrm{Res}_{z=2}
frac{1}{2-z} frac{n}{z^n-1}
= -frac{n}{2^n-1}.$$

Furthermore the residue at infinity

$$mathrm{Res}_{z=infty}
frac{1}{2-z} frac{n}{z^n-1} = 0$$

since we have the bound $2pi n R / R /R^n = 2pi n / R^n rightarrow
0$ as $Rrightarrowinfty.$ Residues sum to zero and we get

$$S_n - frac{n}{2^n-1} = 0
quadtext{or}quad
S_n = frac{n}{2^n-1} = 0$$

as claimed.