If z is a complex number, then the number of common roots of the equation z 1985 + z 100 + 1 = 0 and z 3 + 2…

Given,

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

Now solving,

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

$\Rightarrow \left(z+1\right)\left(z^2-z+1\right)+2z\left(z+1\right)=0$

$\Rightarrow \left(z+1\right)\left(z^2+z+1\right)=0$

$\Rightarrow z=-1, z=\omega , {\omega }^2$

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

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

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

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

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

$\Rightarrow \omega +{\omega }^2+1=0$

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

Hence, there are two common root