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

The Correct Option is B

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.