Understanding the Problem: Finding Common Roots

The question asks us to find the number of common roots between two given equations involving a complex variable \(z\):

• Equation I: \(z^3 + 2z^2 + 2z + 1 = 0\)
• Equation II: \(z^{1985} + z^{100} + 1 = 0\)

To find the common roots, we first need to determine the roots of Equation I and then check which of these roots also satisfy Equation II.

Solving Equation I: \(z^3 + 2z^2 + 2z + 1 = 0\)

This is a cubic equation. Let's try finding a simple root by inspection. Notice the coefficients are symmetric (1, 2, 2, 1). This suggests testing values like 1, -1, or perhaps relating it to reciprocal equations, although this specific form isn't a standard reciprocal equation.

Let's test \(z = -1\):

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

Since substituting \(z = -1\) results in 0, \(z = -1\) is a root of Equation I. This means \((z+1)\) is a factor of the polynomial \(z^3 + 2z^2 + 2z + 1\).

We can perform polynomial division to find the other factor:

Divide \(z^3 + 2z^2 + 2z + 1\) by \((z+1)\):

So, Equation I can be factored as \((z+1)(z^2 + z + 1) = 0\).

The roots are given by \(z+1=0\) or \(z^2+z+1=0\).

• From \(z+1=0\), we get the root \(z = -1\).
• From \(z^2+z+1=0\), we use the quadratic formula \(z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\): \[z = \frac{-1 \pm \sqrt{1^2 - 4(1)(1)}}{2(1)} = \frac{-1 \pm \sqrt{1 - 4}}{2} = \frac{-1 \pm \sqrt{-3}}{2} = \frac{-1 \pm i\sqrt{3}}{2}\]

These two complex roots are special; they are the non-real complex cube roots of unity. We often denote them as \(\omega\) and \(\omega^2\), where \(\omega = \frac{-1 + i\sqrt{3}}{2}\) and \(\omega^2 = \frac{-1 - i\sqrt{3}}{2}\). These roots satisfy the properties \(\omega^3 = 1\) and \(1 + \omega + \omega^2 = 0\).

So, the roots of Equation I are \(-1\), \(\omega\), and \(\omega^2\).

Let's summarise the roots of Equation I:

Checking Roots of Equation I in Equation II: \(z^{1985} + z^{100} + 1 = 0\)

Now, we test each root of Equation I to see if it satisfies Equation II.

Test Root 1: \(z = -1\)

Substitute \(z = -1\) into Equation II:

\((-1)^{1985} + (-1)^{100} + 1\)

Since 1985 is an odd number, \((-1)^{1985} = -1\).

Since 100 is an even number, \((-1)^{100} = 1\).

So, the expression becomes \(-1 + 1 + 1 = 1\).

Since \(1 \neq 0\), \(z = -1\) is not a root of Equation II.

Test Root 2: \(z = \omega\)

Substitute \(z = \omega\) into Equation II:

\(\omega^{1985} + \omega^{100} + 1\)

We use the property \(\omega^3 = 1\). We need to find the remainders of the exponents when divided by 3.

• For the exponent 1985: \(1985 = 3 \times 661 + 2\). So \(1985 \equiv 2 \pmod{3}\). Thus, \(\omega^{1985} = \omega^{3 \times 661 + 2} = (\omega^3)^{661} \cdot \omega^2 = 1^{661} \cdot \omega^2 = \omega^2\).
• For the exponent 100: \(100 = 3 \times 33 + 1\). So \(100 \equiv 1 \pmod{3}\). Thus, \(\omega^{100} = \omega^{3 \times 33 + 1} = (\omega^3)^{33} \cdot \omega^1 = 1^{33} \cdot \omega = \omega\).

Substituting these back into Equation II with \(z=\omega\):

\(\omega^{1985} + \omega^{100} + 1 = \omega^2 + \omega + 1\)

We know the property of complex cube roots of unity: \(1 + \omega + \omega^2 = 0\).

So, \(\omega^2 + \omega + 1 = 0\).

Thus, \(z = \omega\) is a root of Equation II.

Test Root 3: \(z = \omega^2\)

Substitute \(z = \omega^2\) into Equation II:

\((\omega^2)^{1985} + (\omega^2)^{100} + 1\)

Using the property \((a^m)^n = a^{mn}\):

• \((\omega^2)^{1985} = \omega^{2 \times 1985} = \omega^{3970}\). We find the remainder of 3970 when divided by 3: \(3970 = 3 \times 1323 + 1\). So \(3970 \equiv 1 \pmod{3}\). Thus, \(\omega^{3970} = \omega^{3 \times 1323 + 1} = (\omega^3)^{1323} \cdot \omega^1 = 1^{1323} \cdot \omega = \omega\).
• \((\omega^2)^{100} = \omega^{2 \times 100} = \omega^{200}\). We find the remainder of 200 when divided by 3: \(200 = 3 \times 66 + 2\). So \(200 \equiv 2 \pmod{3}\). Thus, \(\omega^{200} = \omega^{3 \times 66 + 2} = (\omega^3)^{66} \cdot \omega^2 = 1^{66} \cdot \omega^2 = \omega^2\).

Substituting these back into Equation II with \(z=\omega^2\):

\((\omega^2)^{1985} + (\omega^2)^{100} + 1 = \omega + \omega^2 + 1\)

Again, using the property \(1 + \omega + \omega^2 = 0\).

So, \(\omega + \omega^2 + 1 = 0\).

Thus, \(z = \omega^2\) is a root of Equation II.

Conclusion: Common Roots Found

The roots of Equation I are \(-1\), \(\omega\), and \(\omega^2\).

We tested these roots in Equation II:

• \(z = -1\) is NOT a root of Equation II.
• \(z = \omega\) IS a root of Equation II.
• \(z = \omega^2\) IS a root of Equation II.

The common roots of Equation I and Equation II are \(\omega\) and \(\omega^2\).

The number of common roots is 2.

Revision Table: Key Concepts for Common Roots

Additional Information: Complex Roots and Polynomials

When dealing with polynomials with real coefficients, complex roots always appear in conjugate pairs. In Equation I, we found roots \(-1\), \(\frac{-1 + i\sqrt{3}}{2}\), and \(\frac{-1 - i\sqrt{3}}{2}\). The two complex roots are indeed conjugates of each other, as expected.

The complex cube roots of unity, \(\omega\) and \(\omega^2\), are located on the unit circle in the complex plane, at angles \(120^\circ\) and \(240^\circ\) (or \(2\pi/3\) and \(4\pi/3\) radians) from the positive real axis, respectively. They satisfy the equation \(z^3 - 1 = 0\), which factors as \((z-1)(z^2+z+1)=0\). So \(\omega\) and \(\omega^2\) are precisely the roots of \(z^2+z+1=0\).

When working with high powers of \(\omega\) or \(\omega^2\), the property \(\omega^3 = 1\) is fundamental. Any power \(\omega^n\) can be reduced by finding the remainder of \(n\) when divided by 3. If \(n = 3q + r\), then \(\omega^n = \omega^{3q+r} = (\omega^3)^q \cdot \omega^r = 1^q \cdot \omega^r = \omega^r\), where \(r\) is the remainder (0, 1, or 2). Similarly for \(\omega^2\), we use \((\omega^2)^n = \omega^{2n}\) and then find the remainder of \(2n\) when divided by 3.

Equation II involves high powers, making this property of roots of unity indispensable. Without recognizing that the non-real roots of \(z^2+z+1=0\) are related to cube roots of unity, solving Equation II for these roots would be significantly more complex.

The number of roots of a polynomial equation is equal to its degree (counting multiplicity). Equation I is degree 3, and it has 3 roots: \(-1, \omega, \omega^2\). Equation II is degree 1985, and it has 1985 roots. We only needed to check if the roots of the lower-degree equation satisfy the higher-degree equation to find the common roots.