Finding Roots of Complex Equations

The problem asks us to find which of the given options is a root of equation-II: \(z^{1985} + z^{100} + 1 = 0\). We are also given equation-I: \(z^3 + 2z^2 + 2z + 1 = 0\). Let's first analyze equation-I, as its roots might be related to the options provided.

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

This is a cubic equation. We can try to find simple roots by inspection. 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 \(z=-1\) makes the equation true, \(z=-1\) 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 or synthetic division to find the other factor:

The quotient is \(z^2 + z + 1\). So, equation-I can be written as:

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

The roots of this equation are \(z+1 = 0\) or \(z^2 + z + 1 = 0\).

• From \(z+1 = 0\), we get \(z = -1\).
• The equation \(z^2 + z + 1 = 0\) is significant in complex numbers. Its roots are the complex cube roots of unity, denoted by \(\omega\) and \(\omega^2\). These roots satisfy the property \(1 + \omega + \omega^2 = 0\) and \(\omega^3 = 1\).

So, the roots of equation-I are \(-1\), \(\omega\), and \(\omega^2\). Note that the options provided are related to these roots or their negatives.

Checking Roots for Equation-II: \(z^{1985} + z^{100} + 1 = 0\)

Now we need to check which of the given options satisfies equation-II. The options are \(-1\), \(-\omega\), \(-\omega^2\), and \(\omega\). We will substitute each option into equation-II and see if the expression evaluates to zero.

Testing Option 1: \(z = -1\)

Substitute \(z=-1\) into \(z^{1985} + z^{100} + 1\):
\((-1)^{1985} + (-1)^{100} + 1\)
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.

Testing Option 2: \(z = -\omega\)

Substitute \(z=-\omega\) into \(z^{1985} + z^{100} + 1\):
\((-\omega)^{1985} + (-\omega)^{100} + 1\)
\(= (-1)^{1985} \omega^{1985} + (-1)^{100} \omega^{100} + 1\)
\(= -1 \cdot \omega^{1985} + 1 \cdot \omega^{100} + 1\)
\(= -\omega^{1985} + \omega^{100} + 1\)

We use the property \(\omega^3 = 1\) to simplify the powers of \(\omega\). We find the remainder when the exponent is divided by 3.

• For \(\omega^{1985}\): \(1985 \div 3\). \(1985 = 3 \times 661 + 2\). So, \(\omega^{1985} = \omega^2\).
• For \(\omega^{100}\): \(100 \div 3\). \(100 = 3 \times 33 + 1\). So, \(\omega^{100} = \omega^1 = \omega\).

Substitute these back into the expression:
\(= -(\omega^2) + \omega + 1\)
\(= 1 + \omega - \omega^2\)

Using the property \(1 + \omega + \omega^2 = 0\), we know \(1 + \omega = -\omega^2\).
So, \(1 + \omega - \omega^2 = (-\omega^2) - \omega^2 = -2\omega^2\).
Since \(\omega^2 \neq 0\), \(-2\omega^2 \neq 0\). Thus, \(z=-\omega\) is not a root of equation-II.

Testing Option 3: \(z = -\omega^2\)

Substitute \(z=-\omega^2\) into \(z^{1985} + z^{100} + 1\):
\((-\omega^2)^{1985} + (-\omega^2)^{100} + 1\)
\(= (-1)^{1985} (\omega^2)^{1985} + (-1)^{100} (\omega^2)^{100} + 1\)
\(= -1 \cdot \omega^{3970} + 1 \cdot \omega^{200} + 1\)
\(= -\omega^{3970} + \omega^{200} + 1\)

Simplify powers of \(\omega\) using \(\omega^3 = 1\):

• For \(\omega^{3970}\): \(3970 \div 3\). \(3970 = 3 \times 1323 + 1\). So, \(\omega^{3970} = \omega^1 = \omega\).
• For \(\omega^{200}\): \(200 \div 3\). \(200 = 3 \times 66 + 2\). So, \(\omega^{200} = \omega^2\).

Substitute these back into the expression:
\(= -(\omega) + \omega^2 + 1\)
\(= 1 - \omega + \omega^2\)

Using \(1 + \omega + \omega^2 = 0\), we know \(1 + \omega^2 = -\omega\).
So, \(1 - \omega + \omega^2 = (-\omega) - \omega = -2\omega\).
Since \(\omega \neq 0\), \(-2\omega \neq 0\). Thus, \(z=-\omega^2\) is not a root of equation-II.

Testing Option 4: \(z = \omega\)

Substitute \(z=\omega\) into \(z^{1985} + z^{100} + 1\):
\(\omega^{1985} + \omega^{100} + 1\)

From our calculations in Option 2:

• \(\omega^{1985} = \omega^2\)
• \(\omega^{100} = \omega\)

Substitute these back:
\(= \omega^2 + \omega + 1\)

Using the fundamental property of cube roots of unity, \(1 + \omega + \omega^2 = 0\).
So, \(\omega^2 + \omega + 1 = 0\).
Since the expression evaluates to 0, \(z=\omega\) is a root of equation-II.

Conclusion

Based on the testing of all options, only \(z=\omega\) satisfies equation-II (\(z^{1985} + z^{100} + 1 = 0\)).

Revision Table: Complex Roots and Equations

Additional Information: Complex Cube Roots of Unity

The complex cube roots of unity are the solutions to the equation \(z^3 = 1\). These are \(1\), \(\omega\), and \(\omega^2\). Geometrically, they are points in the complex plane that form vertices of an equilateral triangle inscribed in the unit circle, with one vertex at \((1,0)\).

• The principal cube root is \(1\).
• The other two roots are \(\omega = e^{i2\pi/3} = \cos(2\pi/3) + i\sin(2\pi/3) = -\frac{1}{2} + i\frac{\sqrt{3}}{2}\).
• The third root is \(\omega^2 = e^{i4\pi/3} = \cos(4\pi/3) + i\sin(4\pi/3) = -\frac{1}{2} - i\frac{\sqrt{3}}{2}\). Note that \(\omega^2\) is the complex conjugate of \(\omega\).

Key properties that are frequently used in problems involving \(\omega\):

• \(\omega^3 = 1\)
• \(1 + \omega + \omega^2 = 0\)
• \(\omega^2 = \frac{1}{\omega}\)
• \(\omega = \frac{1}{\omega^2}\)
• \(\omega \cdot \omega^2 = \omega^3 = 1\)

These properties are crucial for simplifying expressions involving powers of \(\omega\), as demonstrated in solving this problem.