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:

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

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!

  • 1
    $\begingroup$ Thank you that was very useful! $\endgroup$ – SJ. Mar 30 '17 at 3:07

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,


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

Thus the required answer is given as:

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


For future reference here is a solution using residues. We have that with $\zeta_k = \exp(2\pi i k/n)$ so that $\zeta_k^n = 1$

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

which is our sum $S_n.$

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 $2\pi n R / R /R^n = 2\pi n / R^n \rightarrow 0$ as $R\rightarrow\infty.$ Residues sum to zero and we get

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

as claimed.


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