JEE Advanced 2011 Paper 1Question 1
Let $\omega$ be a solution of $x^3-1=0$ with $\mathrm{Im}(\omega)>0$. If $a=2$ with $b$ and $c$ satisfying (E), then the value of $$\frac{3}{\omega^a}+\frac{1}{\omega^b}+\frac{3}{\omega^c}$$ is equal to
Complex Numbers
Solution
The columns give $a+b+c=0$, $a+8b+7c=0$, $9a+2b+3c=0$. With $a=2$: $b=12$, $c=-14$. Then $\frac{3}{\omega^2}+\frac1{\omega^{12}}+\frac3{\omega^{-14}}=3\omega+1+3\omega^{2}=1+3(\omega+\omega^2)=-2$.