Question

If $z$ is a complex number, then the number of common roots of the equation $z^{1985} + z^{100} + 1 = 0$ and $z^3 + 2z^2 + 2z + 1 = 0$, is equal to:

• A. 1
• B. 0
• C. 3
• D. 4

Answer 1

Answer: (B)

0

Answer 2

Solve $z^2 + z + 1 = 0$: The roots are the non-real cube roots of unity:

$z = \omega \quad \text{and} \quad z = \omega^2,$

where $\omega = e^{2\pi i/3}$ and $\omega^2 = e^{-2\pi i/3}$. These roots satisfy $\omega^3 = 1$ and $\omega^2 + \omega + 1 = 0$.

Check if $\omega$ and $\omega^2$ satisfy $z^{1985} + z^{100} + 1 = 0$: For $z = \omega$:

$\omega^{1985} = \omega, \quad \omega^{100} = \omega.$

Substituting into $z^{1985} + z^{100} + 1 = 0$:

$\omega + \omega + 1 = 2\omega + 1 \neq 0.$

For $z = \omega^2$:

$(\omega^2)^{1985} = \omega^2, \quad (\omega^2)^{100} = \omega^2.$

Substituting into $z^{1985} + z^{100} + 1 = 0$:

$\omega^2 + \omega^2 + 1 = 2\omega^2 + 1 \neq 0.$

Conclusion: Neither $\omega$ nor $\omega^2$ satisfies both equations. Therefore, there are no common roots.