If $z$ is a complex number, then the number of common roots of the equations $z^{1985}+z^{100}+1=0$ and $z^3+2z^2+2z+1=0$ is equal to:
Step-by-Step Solution
Key Concept: Factor $z^3+2z^2+2z+1=(z+1)(z^2+z+1)=0$, giving roots $z=-1$ and $z=\omega,\omega^2$ (cube roots of unity). Check each root in $z^{1985}+z^{100}+1=0$.
Factor: $z^3+2z^2+2z+1=(z+1)(z^2+z+1)=0$. Roots: $z=-1,\omega,\omega^2$.
Test $z=-1$: $(-1)^{1985}+(-1)^{100}+1=-1+1+1=1\ne0$. ✗
Test $z=\omega$: $\omega^{1985}+\omega^{100}+1=\omega^{3\cdot661+2}+\omega^{3\cdot33+1}+1=\omega^2+\omega+1=0$. ✓
Test $z=\omega^2$: $(\omega^2)^{1985}+(\omega^2)^{100}+1=\omega^{3970}+\omega^{200}+1=\omega+\omega^2+1=0$. ✓
Two common roots.
Correct Answer: 2