If 1, alpha_1, alpha_2, alpha_3, alpha_4 are | vedclass

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(C) The given equation is $z^5-1=0$,which has roots $1, \alpha_1, \alpha_2, \alpha_3, \alpha_4$. Thus,we can write $z^5-1=(z-1)(z-\alpha_1)(z-\alpha_2

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$-2$

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(C) The given equation is $z^5-1=0$,which has roots $1, \alpha_1, \alpha_2, \alpha_3, \alpha_4$. Thus,we can write $z^5-1=(z-1)(z-\alpha_1)(z-\alpha_2)(z-\alpha_3)(z-\alpha_4)$. Substituting $z=\omega$,we get $\omega^5-1=(\omega-1)(\omega-\alpha_1)(\omega-\alpha_2)(\omega-\alpha_3)(\omega-\alpha_4)$. Now,the expression becomes $(\omega-1)(\omega-\alpha_1)(\omega-\alpha_2)(\omega-\alpha_3)(\omega-\alpha_4)+\omega = \omega^5-1+\omega$. Since $\omega^3=1$,we have $\omega^5 = \omega^3 	imes \omega^2 = \omega^2$. So,the expression is $\omega^2+\omega-1$. Using the property of cube roots of unity,$1+\omega+\omega^2=0$,we have $\omega^2+\omega=-1$. Therefore,the value is $-1-1=-2$.

If $1, \alpha_1, \alpha_2, \alpha_3, \alpha_4$ are the roots of $z^5-1=0$ and $\omega$ is a cube root of unity,then $(\omega-1)(\omega-\alpha_1)(\omega-\alpha_2)(\omega-\alpha_3)(\omega-\alpha_4)+\omega$ is equal to

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