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Given that $1,w,w^2,w^3,w^4.....w^{n-1}$ are nth roots of unity. find the sum of $$\sum_{i=1}^{n-1}\frac{1}{2-w^i}$$ The method I used was to write the equation of $w\;\text{as}\,z^n-1=0$ and further are writing this as $$z^n-1=(z-1)(z-w)(z-w^2)(z-w^3)....(z-w^{n-1})$$ and then replacing z here by 2. From this I was able to find out the denominator of the series in question but I couldn't figure out a way to simplify the numerator

Siddharth Jha
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    See https://math.stackexchange.com/questions/2209034/finding-sum-k-0n-1-dfrac-alpha-k2-alpha-k-where-alpha-k-is-prim OR https://math.stackexchange.com/questions/1909362/product-of-one-minus-the-tenth-roots-of-unity – lab bhattacharjee Dec 20 '17 at 05:12

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Here's a method different from those in the linked solutions. Use the geometric series: \begin{align} \sum_{i=0}^{n-1}\frac1{2-w^i}&=\frac12\sum_{i=0}^{n-1}\frac1{1-w^i/2} =\frac12\sum_{i,j=0}^\infty\frac{w^{ij}}{2^j}=\frac n2\sum_{k=0}^\infty \frac{1}{2^{kn}}=\frac{n2^{n-1}}{2^n-1}. \end{align} Therefore $$\sum_{i=0}^{n-1}\frac1{2-w^i}=\frac{n2^{n-1}}{2^n-1}-1$$ etc.

Angina Seng
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There is a simple trick. Assume that $$ p(z)=\prod_{k=1}^{n}(z-\zeta_k). $$ By applying $\frac{d}{dz}\log(\cdot)$ to both sides we get $$ \sum_{k=1}^{n}\frac{1}{z-\zeta_k} = \frac{p'(z)}{p(z)}.$$ In your case $p(z)=z^n-1=\prod_{k=1}^{n}(z-\omega^k)$ and $$ \sum_{k=1}^{n}\frac{1}{2-\omega^k} = \frac{p'(2)}{p(2)} = \frac{n2^{n-1}}{2^n-1}.$$

Jack D'Aurizio
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