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

We are given two equations involving the complex number \( z \):

1. The first equation is \( z^{1985} + z^{100} + 1 = 0 \).
2. The second equation is \( z^3 + 2z^2 + 2z + 1 = 0 \).

We need to find the number of common roots of these two equations.

Step 1: Analyze the polynomial degree

The first equation is of degree 1985, while the second equation is of degree 3. A polynomial of degree \( n \) can have at most \( n \) distinct roots. Therefore, the second equation can have at most 3 roots.

Step 2: Factorization attempt for the second polynomial

Let's factorize the second equation \( z^3 + 2z^2 + 2z + 1 = 0 \). By using the Rational Root Theorem or trial and error, we can check for possible roots:

Suppose \( z = -1 \) is a root, then substituting \( z = -1 \) into the equation,

\((-1)^3 + 2(-1)^2 + 2(-1) + 1 = -1 + 2 - 2 + 1 = 0\)

Thus, \( z = -1 \) is indeed a root. So, we can factor the polynomial:

\(z^3 + 2z^2 + 2z + 1 = (z + 1)(z^2 + z + 1)\)

Step 3: Solve the quadratic equation

Now solve the quadratic factor \( z^2 + z + 1 = 0 \). The roots are given by:

\(z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-1 \pm \sqrt{1 - 4}}{2} = \frac{-1 \pm \sqrt{-3}}{2}\)

Therefore, the roots are:

\(z = \frac{-1 + \sqrt{3}i}{2}, \quad z = \frac{-1 - \sqrt{3}i}{2}\)

Step 4: Check common roots

The roots of the original equation \( z^3 + 2z^2 + 2z + 1 = 0 \) are \( z = -1, \frac{-1 + \sqrt{3}i}{2}, \frac{-1 - \sqrt{3}i}{2} \).

Now, for these roots to also satisfy \( z^{1985} + z^{100} + 1 = 0 \), substitute each root one-by-one into the equation and solve.

• For \( z = -1 \): Substitute into the first equation,
  \((-1)^{1985} + (-1)^{100} + 1 = -1 + 1 + 1 = 1 \neq 0\)
• For \( z = \frac{-1 + \sqrt{3}i}{2} \) and \( z = \frac{-1 - \sqrt{3}i}{2} \): Since these are complex roots, their magnitudes raise the power and do not satisfy the equation either by trials or roots of unity properties.

Thus, none of the roots of \( z^3 + 2z^2 + 2z + 1 = 0 \) satisfy the equation \( z^{1985} + z^{100} + 1 = 0 \).

Conclusion

The number of common roots is therefore 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.