JEE Main · 2024 · 30 Jan 2024 · Shift 2 · Complex Numbers

$z^3+2z^2+2z+1=0$ has $z=-1$ as a root (direct check), so it factors as $(z+1)(z^2+z+1)=0$. The remaining roots are the primitive cube roots of unity $\omega ,{\omega }^2$.

Check $z=-1$ in $z^{1985}+z^{100}+1$: $(-1{)}^{1985}+(-1{)}^{100}+1=-1+1+1=1\ne 0$, so $-1$ is not a common root.

Check $z=\omega$: since ${\omega }^3=1$, $1985=3(661)+2\Rightarrow {\omega }^{1985}={\omega }^2$, and $100=3(33)+1\Rightarrow {\omega }^{100}=\omega$. The expression becomes ${\omega }^2+\omega +1=0$ — a root.

Check $z={\omega }^2$: $3970=3(1323)+1\Rightarrow ({\omega }^2{)}^{1985}=\omega$, and $200=3(66)+2\Rightarrow ({\omega }^2{)}^{100}={\omega }^2$. The expression becomes $\omega +{\omega }^2+1=0$ — also a root.

Both $\omega$ and ${\omega }^2$ are common roots, so there are 2 common roots.