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Q.
If $1,\alpha,\alpha^2,........,\alpha^{n-1} $ are the nth roots of unity, then $\displaystyle\sum_{i=1}^{n-1}\frac{1}{2-a^i}$ is equal to
Complex Numbers and Quadratic Equations
Solution:
Since $1, \alpha, \alpha^{2}, ...\alpha^{n-1}$ are the nth roots of unity
$\therefore x^{n}-1=\left(x-1\right)\left(x-\alpha\right)\left(x-\alpha\right)^{2}\ldots\ldots\left(x-\alpha^{n-1}\right)$
$\therefore log\left(x^{n}-1\right)=log\left(x-1\right)+log\left(x-\alpha\right)$
$+log\left(x-\alpha\right)^{2}+.....+log\left(x-\alpha^{n-1}\right)$
Diff. both sides w.r.t. $‘x’$, we get
$\frac{nx^{n-1}}{x^{n}-1}=\frac{1}{x-1}+\frac{1}{x+\alpha}$
$+\frac{1}{x-\alpha^{2}}+...+\frac{1}{x-\alpha^{n-1}}$
Putting $x = 2$, we get
$\frac{n2^{n-1}}{2^{n}-1}=\frac{1}{1}+\frac{1}{2-\alpha}+\frac{1}{2-\alpha^{2}}+...+\frac{1}{2-\alpha^{n-1}}$
$\therefore \frac{n.2^{n-1}}{2^{n}-1}-1$
$= \displaystyle \sum_{i=1}^{n=1}$$\frac{1}{2-\alpha}$.
Hence $\displaystyle \sum_{i=1}^{n=1}$$\frac{1}{2-\alpha^i}$
$=\frac{n.2^{n-1}-2^{n}+1}{2^{n}-1}$
$=\frac{\left(n-2\right)2^{n-1}+1}{2^{n}+1}$