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! 
 A: 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!
A: If $\alpha^5 = 1$, $\frac{1}{2-\alpha} = c_0 + c_1 \alpha + \ldots + c_4 \alpha^4$ where 
$$\eqalign{ 2 c_0 - c_4 &= 1\cr
            2 c_1 - c_0 &= 0\cr
            2 c_2 - c_1 &= 0\cr
            2 c_3 - c_2 &= 0\cr
            2 c_4 - c_3 &= 0\cr}$$
The solution of this is $$c_0 = 16/31,\; c_1 = 8/31,\; c_2 = 4/31,\; c_3 = 2/31,\; c_4 = 1/31$$
Then $$ \eqalign{\sum_{r=1}^4 \frac{1}{2-\alpha^r} &= \frac{16}{31}\sum_{r=1}^4 1 + 
\frac{8}{31}\sum_{r=1}^4 \alpha^r + \frac{4}{31}\sum_{r=1}^4 \alpha^{2r} +
\frac{2}{31}\sum_{r=1}^4 \alpha^{3r} + \frac{1}{31}\sum_{r=1}^4 \alpha^{4r}\cr &= \frac{16}{31} \cdot 4 - \frac{8}{31} - \frac{4}{31} - \frac{2}{31} - \frac{1}{31} = \frac{49}{31}}$$
A: Another proof:
$$P(x)=x^5-1=\prod_{r=0}^4(x-\alpha^r)$$
Hence 
$$\frac{P^{\prime}(x)}{P(x)}=\frac{5x^4}{x^5-1}=\sum_{r=0}^4\frac{1}{x-\alpha^r}$$
Now put $x=2$ in this formula.
A: Note that
$$\newcommand{\al}{\alpha}\frac1{2-\al^r}
=\sum_{k=0}^\infty\frac{\alpha^{kr}}{2^{k+1}}$$
and so
$$\sum_{r=0}^4\frac1{2-\al^r}
=\sum_{k=0}^\infty\frac{1}{2^{k+1}}\sum_{r=0}^4\al^{kr}.$$
The inner sum is zero, unless $5\mid r$. So
$$\sum_{r=0}^4\frac1{2-\al^r}=5\sum_{s=0}^\infty\frac1{2^{5s+1}}
=\frac{5}{2(1-1/32)}=\frac{80}{31}.$$
Now subtract $1/(2-1)=1$ to get $49/31$.
