The common roots of the equations z^3+2z^2+2z | vedclass

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Question

(A) Given equation $z^3+2z^2+2z+1=0$ can be factored as $(z+1)(z^2+z+1)=0$. Its roots are $-1, \omega, \omega^2$,where $\omega$ is a complex cube root

Vedclass · Accepted Answer

$\omega, \omega^2$

Vedclass · Answer

(A) Given equation $z^3+2z^2+2z+1=0$ can be factored as $(z+1)(z^2+z+1)=0$. Its roots are $-1, \omega, \omega^2$,where $\omega$ is a complex cube root of unity. Let $f(z) = z^{2014}+z^{2015}+1$. Testing $z=-1$: $f(-1) = (-1)^{2014}+(-1)^{2015}+1 = 1-1+1 = 1 
eq 0$. So,$-1$ is not a common root. Testing $z=\omega$: $f(\omega) = \omega^{2014}+\omega^{2015}+1 = (\omega^3)^{671} \cdot \omega + (\omega^3)^{671} \cdot \omega^2 + 1 = \omega + \omega^2 + 1 = 0$. So,$\omega$ is a common root. Testing $z=\omega^2$: $f(\omega^2) = (\omega^2)^{2014}+(\omega^2)^{2015}+1 = \omega^{4028}+\omega^{4030}+1 = (\omega^3)^{1342} \cdot \omega^2 + (\omega^3)^{1343} \cdot \omega + 1 = \omega^2 + \omega + 1 = 0$. So,$\omega^2$ is a common root. Thus,the common roots are $\omega$ and $\omega^2$.

The common roots of the equations $z^3+2z^2+2z+1=0$ and $z^{2014}+z^{2015}+1=0$ are

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