Question

If $\omega =\cos \left(\frac{2\pi }{n}\right)+i\sin \left(\frac{2\pi }{n}\right)$, then numerical value of
$1+\frac{1}{2-\omega }+\frac{1}{2-{\omega }^2}+...+\frac{1}{2-{\omega }^{n-1}}$
equals

• A. $\frac{n\left(2^{n-1}\right)}{2^n+1}$
• B. $\frac{n\left(2^{n-1}\right)}{2^n-1}$
• C. $0$
• D. $1$

Answer 1

The correct option is B $\frac{n\left(2^{n-1}\right)}{2^n-1}$

The equation in question is $x^n-1=0$

$f\left(x\right)=x^n-1=(x-1)(x-\omega )...(x-{\omega }^{n-1})$

Thus, $\frac{f^{\prime }\left(x\right)}{f\left(x\right)}=\frac{n{x}^{n-1}}{x^n-1}=\frac{1}{x-1}+\frac{1}{x-\omega }+...+\frac{1}{x-{\omega }^{n-1}}$

Substituting $x=2$:
$\frac{1}{2-1}+\frac{1}{2-\omega }+...+\frac{1}{2-{\omega }^{n-1}}=\frac{n{2}^{n-1}}{2^n-1}$