Complexity! (4)

If $z^{50} + z^{49} + z^{48} + ... + 1=0$ , then $z^{51} =1$ or $z$ is the 51st root of unity and $z_r = z^r$ .

$\begin{aligned} S & = \sum_{r=1}^{50} \frac 1{z_r -1} \\ & = \sum_{r=1}^{50} \frac 1{z^r -1} & \small \color{#3D99F6}{\text{Adding 1st \& 50th, 2nd \& 49th, 3rd \& 48th, ... together.}} \\ & = \sum_{r=1}^{25} \left(\frac 1{z^r -1} + \frac 1{z^{51-r} -1}\right) \\ & = \sum_{r=1}^{25} \frac {z^{51-r}+z^r-2}{\color{#3D99F6}{z^{51}} -z^{51-r} -z^r +1} & \small \color{#3D99F6}{\text{Note that }z^{51} = 1} \\ & = \sum_{r=1}^{25} \frac {z^{51-r}+z^r-2}{-z^{51-r} -z^r +2} \\ & = \sum_{r=1}^{25} -1 = - 25 \end{aligned}$

Therefore, $|S| = 25 = 5^2$ $\implies a = \boxed{5}$ .

As discussed here , I have proved earlier that a polynomial $p(x)$ of degree $n$ with real coefficients whose roots are given by $x_i\ne 1$ , $i\in[1,50]$ satisfies the identity, $\displaystyle \sum_{k=1}^{n} |\frac{1}{x_k-1}|=\frac{f'(1)}{f(1)}$

So here clearly $1$ is not a root for $\displaystyle p(z)=\sum_{r=0}^{50}z^r$ , and $p(1)=51,p'(1)=1275$ and so $\displaystyle |\sum_{k=1}^{50}\frac{1}{z_k-1}|=\frac{p'(1)}{p(1)}=\color{#D61F06}{5}^2$

Use the transformation $y$ $=$ $1/(x-1)$ .Use Veita formula to find the sum of roots as -25.Hence,we get the answer as 5.