Complex number weird sum. Question given in red. My working in black. $$\color{red}{\sum_{r=0}^{50}z^r=0}\iff z_k=\exp\underbrace{\left(\frac{{\cal i}2\pi k}{51}\right)}_{\theta_k},k\in\{n\mid n\le50,n\in\mathbb N\},i:=\sqrt{-1}$$
I proceed: $$\color{red}{S=\sum_{k=1}^{50}\frac1{1-z_k}}=\sum_{k=1}^{50}\frac1{1-\cos\theta_k-i\sin\theta_k}=\sum_{k=1}^{50}\frac1{2\cos^2(\theta_k/2)-2i\sin(\theta_k/2)\cos(\theta_k/2)}$$
What to do now?
$$S=\frac12\sum_{k=1}^{50}\frac{\sec(\theta_k/2)}{e^{-\theta_k/2}}=??$$
Help!

According to help in comments:
$$S=\sum_{k=1}^{50}\sum_{r=0}^{\infty}z_k^r=\sum_{r=0}^{\infty}\sum_{k=1}^{50}z_k^r=\sum_{r=0}^{\infty}\sum_{k=1}^{50}\exp\displaystyle\left(\frac{i2\pi kr}{51}\right)=\sum_{r=0}^{\infty}\frac{\exp(i2\pi r/51)-\exp(i2\pi r)}{1-\exp(i2\pi r/51)}=??$$
 A: $$\begin{array}{rcl}
\prod\limits_{k=1}^{50} (z - z_k) = \sum\limits_{s=0}^{50} z^s
& \stackrel{\color{blue}{\text{ take log and diff }}}{\implies} 
& \displaystyle\;\sum\limits_{k=1}^{50}\frac{1}{z-z_k} = \frac{\sum\limits_{s=0}^{50} sz^{s-1}}{\sum\limits_{s=0}^{50} z^s}\\
& 
\stackrel{\color{blue}{\text{ subst } z \text{ by } 1}}{\implies} 
&
\displaystyle
\sum\limits_{k=1}^{50}\frac{1}{1-z_k} = \frac{\sum\limits_{s=0}^{50} s}{\sum\limits_{s=0}^{50} 1}
= \frac{\frac{50(50+1)}{2}}{51} = 25
\end{array}
$$
A: Hint 1: Using the formula for the sum of a geometric series, the corrected sum is
$$
\sum_{r=0}^{50}z^r=\frac{1-z^{51}}{1-z}
$$

Hint 2:
$$
\begin{align}
\sum_{k=1}^{50}\frac1{1-e^{2\pi ki/51}}
&=\sum_{k=1}^{50}\frac{e^{-\pi ki/51}}{e^{-\pi ki/51}-e^{\pi ki/51}}\tag{1}\\
&=-\sum_{k=1}^{50}\frac{e^{\pi ki/51}}{e^{-\pi ki/51}-e^{\pi ki/51}}\tag{2}
\end{align}
$$
Explanation:
$(1)$: multiply each term by $\frac{e^{-\pi ki/51}}{e^{-\pi ki/51}}$
$(2)$: substitute $k\mapsto51-k$
Try averaging $(1)$ and $(2)$.
