<s>[INST] Summation over roots of unity Find the value of $\displaystyle\sum_{r=1}^{4} \frac{1}{2-\alpha^r} $
where $ \alpha^k (k=0,1,2,3,4,5) $ are fifth roots of unity.
My approach:- 
As we know that $ \alpha^k (k=0,1,2,3,4,5) $ are fifth roots of unity, then $ \alpha^k - 1$ should be equal to zero. Therefore the final answer to the summation $\displaystyle\sum_{r=1}^{4} \frac{1}{2-\alpha^r} $ should be $\displaystyle\sum_{r=1}^{4}  1 = 4  $
But the answer given is $ \dfrac{ 49}{31} $ 
Any help or hint will be much appreciated! 
 [/INST] Since $\alpha_0,\alpha_1,\alpha_2 \dots \alpha_{4}$ are roots of the equation
$$x^5-1=0  \tag1$$ 
You can apply Transformation of Roots to find a equation whose roots are$$\frac{1}{2-\alpha_0} , \frac{1}{2-\alpha_1},\dots \frac{1}{2-\alpha_{4}}$$
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 $(1)$
$$\Bigg(\frac{2y-1}{y}\Bigg)^5-1=0 \implies (2y-1)^{5}-y^{5}=0$$
Use Binomial Theorem to find coefficient of $y^5$ and $y^{4}$.You will get sum of the roots using Vieta's Formulas.
Hope it helps!
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