If z is a complex number, then the number of common roots of the...

Given,

${z}^{1985} + {z}^{100} + 1 = 0 \text{\text{ and } }$ {z}^{3} + 2{z}^{2} + 2z + 1 = 0

Now solving,

${z}^{3} + 2{z}^{2} + 2z + 1 = 0$

$⇒ \left(z + 1\right)\left({z}^{2} - z + 1\right) + 2z\left(z + 1\right) = 0$

$⇒ \left(z + 1\right)\left({z}^{2} + z + 1\right) = 0$

$⇒ z = - 1, z = \omega, {\omega}^{2}$

Now putting $z = \omega$ in {z}^{1985} + {z}^{100} + 1 we get,

$⇒ {\omega}^{1985} + {\omega}^{100} + 1$

$⇒ {\omega}^{2} + \omega + 1 = 0 \left\{\text{as}{\omega}^{3n} = 1\right\}$

Also, $z = {\omega}^{2}$

$⇒ {\omega}^{3970} + {\omega}^{200} + 1$

$⇒ \omega + {\omega}^{2} + 1 = 0$

Also, $z = - 1$ will not satisfy the equation ${z}^{1985} + {z}^{100} + 1$

Hence, there are two common root