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I have the series
$$\sum_{k=1}^{N-1}\frac{1}{1-w_k} $$ where $w_k=e^{\frac{2\pi i k}{N}}$, how can I find the summation of this , another question related to this sum
$$\sum_{k=1}^{N-1}\frac{1}{z-w_k} $$

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3 Answers 3

$$\frac{1}{1-w^{r}}+\frac{1}{1-w^{n-r}}=\frac{2-w^{r}-w^{n-r}}{2-w^{r}-w^{n-r}}=1$$ where $w=e^{i2\pi/n}$

If you continue cancelling in pairs the sum will turn out to be = $\dfrac{N-1}{2}$

In case you have a middle term ,the term = $\dfrac{1}{1-e^{i\pi/2}}=\dfrac{1}{2}$

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Let $P(z)=z^{N}-1$. Then $$P(z)=\prod_{k=0}^{N-1}(z-w_{k})$$ where $w_{0}=1$. From here, it is easy to check that $$ \sum_{k=0}^{N-1}\frac{1}{z-w_{k}}=\frac{P^{\prime}(z)}{P(z)}% $$ hence $$ \sum_{k=1}^{N-1}\frac{1}{z-w_{k}}=\frac{P^{\prime}(z)}{P(z)}-\frac{1}{z-1}=\frac{Nz^{N-1}}{z^{N}-1}-\frac{1}{z-1}=\frac{(N-1)z^{N}-Nz^{N-1}+1}{(z^{N}-1)(z-1)}\text{.} $$ In particular, for $z=1$: $$ S:=\sum_{k=1}^{N-1}\frac{1}{1-w_{k}}=\lim_{z\rightarrow1}\frac{(N-1)z^{N}-Nz^{N-1}+1}{(z^{N}-1)(z-1)}. $$ In order to compute the limit, we write is as a limit at $0$, i.e., let $z=w+1$, $w\rightarrow0$, hence $$ S=\lim_{w\rightarrow0}\frac{(N-1)(w+1)^{N}-N(w+1)^{N-1}+1}{w((w+1)^{N}-1)}. $$ Next, recall that $$ \lim_{w\rightarrow0}\frac{(w+1)^{N}-1}{w}=N $$ hence $$ S=\frac{1}{N}\cdot\lim_{w\rightarrow0}\frac{(N-1)(w+1)^{N}-N(w+1)^{N-1}+1}{w^{2}}. $$ The last step is to apply l'Hopital's rule, hence \begin{align*} S & =\frac{1}{N}\cdot\lim_{w\rightarrow0}\frac{\left( (N-1)(w+1)^{N}-N(w+1)^{N-1}+1\right) ^{\prime}}{\left( w^{2}\right) ^{\prime}}=\frac{1}{N}\cdot\lim _{w\rightarrow0}\frac{N(N-1)(w+1)^{N-2}w}{2w}\\ & =\frac{1}{N}\cdot\lim_{w\rightarrow0}\frac{N(N-1)}{2}(w+1)^{N-2}=\frac{N-1}{2}. \end{align*}

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The answer for the first sum is $$\sum_{k=1}^{N-1}\frac{1}{1-w_k}=\frac{N-1}{2}$$

And for the second: $$\sum_{k=1}^{N-1}\frac{1}{z-w_k} = \sum_{k=0}^{N-1}\frac{1}{z-w_k}- \frac{1}{z-1}= \frac{N z^{N-1}}{z^N-1} -\frac{1}{z-1}$$

I believe series expansion in $w_k$, sum over $k$ and then summing back the series wll give you the above results.

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