Key Concepts and Formulas

• Complex Cube Roots of Unity: The solutions to $z^3 = 1$ are $1, \omega, \omega^2$, where $\omega = e^{2\pi i/3} = -\frac{1}{2} + i\frac{\sqrt{3}}{2}$ and $\omega^2 = e^{4\pi i/3} = -\frac{1}{2} - i\frac{\sqrt{3}}{2}$.
• Properties of $\omega$: $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$. These are essential for simplifying expressions with $\omega$.
• Polynomial Factorization: Identifying and factoring out common factors allows us to find the roots of polynomials.

Step-by-Step Solution

Step 1: Find the roots of the second equation We are given the equation $z^3 + 2z^2 + 2z + 1 = 0$. Our goal is to find the roots of this equation, as these will be the candidates for common roots with the first equation.

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

We can try to factor this cubic polynomial. Notice that if $z = -1$, the equation becomes $(-1)^3 + 2(-1)^2 + 2(-1) + 1 = -1 + 2 - 2 + 1 = 0$. Thus, $z = -1$ is a root, and $(z+1)$ is a factor. We can rewrite the polynomial as: $z^3 + 1 + 2z^2 + 2z = 0$ Using the sum of cubes factorization, $z^3 + 1 = (z+1)(z^2 - z + 1)$. Also, $2z^2 + 2z = 2z(z+1)$. Substituting these back into the equation, we get: $(z+1)(z^2 - z + 1) + 2z(z+1) = 0$ Now, factor out $(z+1)$: $(z+1)(z^2 - z + 1 + 2z) = 0$ $(z+1)(z^2 + z + 1) = 0$ The roots are the solutions to $z+1 = 0$ and $z^2 + z + 1 = 0$.

• $z+1 = 0 \implies z = -1$
• $z^2 + z + 1 = 0$. This quadratic equation has roots $\omega$ and $\omega^2$, the non-real cube roots of unity.

So, the roots of the second equation are $-1, \omega, \omega^2$.

Step 2: Check if $z=-1$ satisfies the first equation The first equation is $z^{1985} + z^{100} + 1 = 0$. We need to check if any of the roots of the second equation also satisfy this equation. Let's start with $z = -1$. $(-1)^{1985} + (-1)^{100} + 1 = -1 + 1 + 1 = 1 \neq 0$ Since substituting $z = -1$ into the first equation does not result in 0, $z = -1$ is not a root of the first equation, and therefore not a common root.

Step 3: Check if $z=\omega$ satisfies the first equation Now, let's check $z = \omega$. $\omega^{1985} + \omega^{100} + 1$ Since $\omega^3 = 1$, we can simplify the exponents by finding their remainders when divided by 3. $1985 \div 3 = 661$ with a remainder of 2. So, $\omega^{1985} = \omega^2$. $100 \div 3 = 33$ with a remainder of 1. So, $\omega^{100} = \omega$. Substituting these back into the equation: $\omega^2 + \omega + 1 = 0$ This is a fundamental property of cube roots of unity. Therefore, $z = \omega$ is a root of the first equation, and hence a common root.

Step 4: Check if $z=\omega^2$ satisfies the first equation Next, check $z = \omega^2$. $(\omega^2)^{1985} + (\omega^2)^{100} + 1 = \omega^{3970} + \omega^{200} + 1$ Again, we simplify using $\omega^3 = 1$. $3970 \div 3 = 1323$ with a remainder of 1. So, $\omega^{3970} = \omega$. $200 \div 3 = 66$ with a remainder of 2. So, $\omega^{200} = \omega^2$. Substituting these back into the equation: $\omega + \omega^2 + 1 = 0$ This is also a fundamental property of cube roots of unity. Therefore, $z = \omega^2$ is a root of the first equation, and hence a common root.

Step 5: Count the common roots We found that $\omega$ and $\omega^2$ are common roots, while $-1$ is not. Therefore, there are two common roots.

Common Mistakes & Tips

• Incorrect factorization: Make sure to factor polynomials correctly. Grouping terms strategically is important.
• Forgetting properties of $\omega$: Remember that $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$. These are crucial for simplifying expressions.
• Not checking all roots: Always verify all potential common roots in both equations.

Summary

We found the roots of the second equation to be $-1$, $\omega$, and $\omega^2$. By substituting each of these into the first equation, we determined that $\omega$ and $\omega^2$ are also roots of the first equation, while $-1$ is not. Therefore, there are two common roots.

The final answer is $\boxed{2}$, which corresponds to option (B).