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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: $$S=\sum_{k=1}^{50}\sum_{r=0}^{\infty}z_k^r=\sum_{r=0}^{\infty}\sum_{k=1}^{50}z_k^r=\sum_{r=0}^{\infty}\sum_{k=1}^{50}\exp\displaystyle\left(\frac{i2\pi kr}{51}\right)=\sum_{r=0}^{\infty}\frac{\exp(i2\pi r/51)-\exp(i2\pi r)}{1-\exp(i2\pi r/51)}=??$$

Yogeshwar Bala
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RE60K
  • 17,716

2 Answers2

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$$\begin{array}{rcl} \prod\limits_{k=1}^{50} (z - z_k) = \sum\limits_{s=0}^{50} z^s & \stackrel{\color{blue}{\text{ take log and diff }}}{\implies} & \displaystyle\;\sum\limits_{k=1}^{50}\frac{1}{z-z_k} = \frac{\sum\limits_{s=0}^{50} sz^{s-1}}{\sum\limits_{s=0}^{50} z^s}\\ & \stackrel{\color{blue}{\text{ subst } z \text{ by } 1}}{\implies} & \displaystyle \sum\limits_{k=1}^{50}\frac{1}{1-z_k} = \frac{\sum\limits_{s=0}^{50} s}{\sum\limits_{s=0}^{50} 1} = \frac{\frac{50(50+1)}{2}}{51} = 25 \end{array} $$

achille hui
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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 $(2)$.

robjohn
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