Question

If $1, \alpha_1, \alpha_2, \alpha_3, \alpha_4$ be the roots of $x^5-1=0$, then the value of $\frac{\omega-\alpha_1}{\omega^2-\alpha_1} \frac{\omega-\alpha_2}{\omega^2-\alpha_2} \frac{\omega-\alpha_3}{\omega^2-\alpha_3} \frac{\omega-\alpha_4}{\omega^2-\alpha_4} \ldots$ is . (where $\omega$ is imaginary cube root of unity.)

• A. $ \omega$
• B. $\omega^2$
• C. $2 \omega$
• D. $2 \omega^2$

Answer 1

Answer: (A)

$(z-1)\left(z-\alpha_1\right) \ldots \ldots . .\left(z-\alpha_4\right)=z^5-1$
Put $ z=\omega, z=\omega^2$ and divide
$ \frac{(\omega-1)\left(\omega-\alpha_1\right)\left(\omega-\alpha_2\right)\left(\omega-\alpha_3\right)\left(\omega-\alpha_4\right)}{\left(\omega^2-1\right)\left(\omega^2-\alpha_1\right)\left(\omega^2-\alpha_2\right)\left(\omega^2-\alpha_3\right)\left(\omega^2-\alpha_4\right)}=\frac{\omega^5-1}{\omega^{10}-1} $
$ \frac{\left(\omega-\alpha_1\right)\left(\omega-\alpha_2\right)\left(\omega-\alpha_3\right)\left(\omega-\alpha_4\right)}{\left(\omega^2-\alpha_1\right)\left(\omega^2-\alpha_2\right)\left(\omega^2-\alpha_3\right)\left(\omega^2-\alpha_4\right)} $
$ =\frac{\left(\omega^2-1\right)^2}{(\omega-1)^2}=(\omega+1)^2=\omega^4=\omega$