1. We need the common roots of

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

So first, solve the cubic.

2. Factorize the cubic:

$z^3+2z^2+2z+1=(z+1)(z^2+z+1).$

Hence its roots are:

• $z=-1$
• roots of $z^2+z+1=0$

The roots of $z^2+z+1=0$ are the non-real cube roots of unity, say $\omega,\omega^2$, where

$\omega^3=1,\quad \omega\neq 1,\quad 1+\omega+\omega^2=0.$

So the possible common roots are among

$-1,\ \omega,\ \omega^2.$

3. Check each root in the first equation.

Case 1: $z=-1$

Substitute into $z^{1985}+z^{100}+1$:

$(-1)^{1985}+(-1)^{100}+1=-1+1+1=1\neq 0.$

So $z=-1$ is not a common root.

Case 2: $z=\omega$

Using $\omega^3=1$, reduce exponents modulo $3$.

For $1985$:

$1985 \equiv 2 \pmod{3}$ so $\omega^{1985}=\omega^2.$

For $100$:

$100 \equiv 1 \pmod{3}$ so $\omega^{100}=\omega.$

Therefore,

$\omega^{1985}+\omega^{100}+1=\omega^2+\omega+1=0.$

So $\omega$ is a common root.

Case 3: $z=\omega^2$

Again reduce exponents modulo $3$.

$\left(\omega^2\right)^{1985}=\omega^{3970}.$ Since

$3970 \equiv 1 \pmod{3},$ we get

$\left(\omega^2\right)^{1985}=\omega.$

Also,

$\left(\omega^2\right)^{100}=\omega^{200}.$ Since

$200 \equiv 2 \pmod{3},$ we get

$\left(\omega^2\right)^{100}=\omega^2.$

Thus,

$\left(\omega^2\right)^{1985}+\left(\omega^2\right)^{100}+1=\omega+\omega^2+1=0.$

So $\omega^2$ is also a common root.

4. Therefore the common roots are exactly

$\omega,\ \omega^2,$

and the number of common roots is

$\boxed{2}.$

5. Comparing with the stored correct answer:

• Derived answer: $2$
• Stored correct answer: B ($2$)

They agree.