Mathematics - Complex Number Question with Solution | TestHub

$k=2$,

Let $e^{i\frac{2\pi }{n}}=\alpha$ then ${\sum }_{j=1}^{n-12}\frac{1}{1-e^j}=\frac{1}{1-\alpha }+\frac{1}{1-{\alpha }^2}+\dots \dots +\frac{1}{1-{\alpha }^{n-1}}$

Where $\alpha$ is a nth root of unity $\left(\alpha ,{\alpha }^2,{\alpha }^3,\dots \dots .{\alpha }^{n-1}\right)$ are the roots of $\frac{x^n-1}{x-1}=(x-\alpha )\left(x-{\alpha }^2\right)\dots \dots \left(x-{\alpha }^{n-1}\right)$

Taking log on both side

$\log \frac{x^n-1}{x-1}=\log (x-\alpha )+\log \left(x-{\alpha }^2\right)+\dots \dots +\log \left(x-{\alpha }^{n-1}\right)$

Diff w.r.t. x and use ${\lim }_{x\to 1}$

$\Rightarrow \frac{n-1}{2}=\frac{1}{1-\alpha }+\frac{1}{1-{\alpha }^2}+\dots \dots .+\frac{1}{1-{\alpha }^{n-1}}$