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

Given two equations involving the complex number \( z \):

\( z^{1985} + z^{100} + 1 = 0 \)

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

Determine the count of common roots for these equations.

Step 1: Polynomial Degree Analysis

The first equation has a degree of 1985, while the second has a degree of 3. A polynomial of degree \( n \) has at most \( n \) distinct roots. Consequently, the second equation has a maximum of 3 roots.

Step 2: Factorization of the Second Polynomial

Factorize \( z^3 + 2z^2 + 2z + 1 = 0 \). Using the Rational Root Theorem or trial substitution:

Testing \( z = -1 \):

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

Since \( z = -1 \) is a root, the polynomial can be factored as:

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

Step 3: Solving the Quadratic Factor

Solve \( z^2 + z + 1 = 0 \). The roots are determined by the quadratic formula:

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

The roots are:

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

Step 4: Verification of Common Roots

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

Check if these roots satisfy \( z^{1985} + z^{100} + 1 = 0 \):

• For \( z = -1 \):
• For \( z = \frac{-1 + i\sqrt{3}}{2} \) and \( z = \frac{-1 - i\sqrt{3}}{2} \): These roots are complex cube roots of unity. Substituting them into the first equation does not yield 0.

 

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

Conclusion

The number of common roots is 0.