Mathematics - Complex Number Question with Solution | TestHub

Let $x$ be the (2009) ${}^{\text{th }}$ root of unity $\ne 1$, then

$x^{2009}-1=(x-1)(x-w)\dots ..\left(x-w^{2008}\right)$

Taking log on both sides, we get

$\ln \left(x^{2009}-1\right)=\ln (x-1)+\ln (x-w)+\ln \left(x-w^2\right)\dots ..+\ln \left(x-w^{2008}\right)$

∴ On differentiate both the side w.r.t. x , we get

$\frac{(2009)x^{2008}}{x^{2009}-1}=\frac{1}{x-1}+{\sum }_{r=1}^{2008}\frac{1}{x-w^r}\dots \dots .\left(2\right)$

Putting $x=2$ in equation (2), we get

$\Rightarrow 1+{\sum }_{r=1}^{2008}\frac{1}{2-{\mathrm{w}}^{\mathrm{r}}}=\frac{2009\left(2^{2008}\right)}{2^{2009}-1}$

Multiplying both sides of above equation by $\left(2^{2009}-1\right)$, we get

$\begin{array}{rlrlr}& \therefore \left(2^{2009}-1\right)\sum _{r=1}^{2008}\frac{1}{2-{\mathrm{w}}^{\mathrm{r}}}=2009\cdot 2^{2008}-2^{2009}+1& & & \end{array}$

$=2^{2008}(2009-2)+1=2^{2008}\cdot 2007+1=\left[(a)\left(2^b\right)+c\right]$

$\therefore a=2007,b=2008,c=1$

$Hencea+b+c=4016$