# Calculation using unit roots of 5

If $1, \alpha_1,\alpha_2,\alpha_3,\alpha_4$ be the roots of $x^5-1 = 0$, then prove that $$\frac{\omega -\alpha_1}{\omega^2 -\alpha_1}.{\frac{\omega -\alpha_2}{\omega^2 -\alpha_2}.\frac{\omega -\alpha_3}{\omega^2 -\alpha_3}.\frac{\omega -\alpha_4}{\omega^2 -\alpha_4}}=\omega$$

For this question, i am trying to elimate the roots term and trying to bring $\omega$, I dont to expand the terms and make it complicated , is their any short method i can use.

• What is $\omega$? Aug 24 '17 at 8:26
• It is not mention but what I presume that $1+\omega + \omega^2 = 0$ or cube root of unity Aug 24 '17 at 8:28
• That doesn't answer my question. Aug 24 '17 at 8:30
• I cross checked the question nothing is mentioned about $\omega$ so i presume it as cube root of unity Aug 24 '17 at 8:33
• Why should it be? Aug 24 '17 at 8:44

$$\forall \omega, \omega^5 - 1 = (\omega-1)(\omega-\alpha_1)(\omega-\alpha_2)(\omega-\alpha_3)(\omega-\alpha_4)$$ thus $$\omega^{10}-1 = (\omega^{2})^5-1= (\omega^2-1)(\omega^2-\alpha_1)(\omega^2-\alpha_2)(\omega^2-\alpha_3)(\omega^2-\alpha_4)$$

Now, the initial product $\mathcal{P}$ is equal to: $$\mathcal{P} = \frac{\omega^5-1}{\omega-1}\times \frac{\omega^2-1}{\omega^{10}-1}= \frac{\omega+1}{\omega^5+1}$$

As $\omega$ is assumed to be a cube root of $1$,then,

$$\mathcal{P} = \frac{\omega+1}{\omega^3\omega^2+1} =\frac{\omega+1}{\omega^2+1} =\frac{-\omega^2}{-\omega} = \omega$$

• I also thought of this approach but my only concern is $\omega$ is cube root of unity, your approach is correct but as answer is not coming hence we have to consider $\omega$ as cube root of unity Aug 24 '17 at 8:44
• @SamarImamZaidi As you stated, $\omega^2 = -\omega-1$. You can simplify the denominator by expressing $\omega^5$ only using $\omega$'s and $1$'s and see for yourself if the identity is valid or not. Aug 24 '17 at 8:47
• Prove $\omega+1=-\omega^2$ and $\omega^5 +1 = -\omega$ Aug 24 '17 at 8:50
• It is not difficult to verify that when you multiply the denominator by $\omega$ you get the numerator. After all $\omega^6=(\omega^3)^2$. IOW. This answer is correct, +1. Aug 24 '17 at 8:50
• @Wyllich Thanks for the answer, apply, cube root property, $\omega^5=\omega^2$, and $1+ \omega+\omega^2=0$, please add this step Aug 24 '17 at 8:52