Finding the Roots of Equation-I: z³ + 2z² + 2z + 1 = 0

The question asks for the roots of the cubic equation, referred to as Equation-I: \( z^3 + 2z^2 + 2z + 1 = 0 \).

To find the roots of a cubic equation, we can first try to find any simple integer roots using the Rational Root Theorem. Possible rational roots are divisors of the constant term (1), which are ±1.

• Let's test \( z = 1 \):
  \( (1)^3 + 2(1)^2 + 2(1) + 1 = 1 + 2 + 2 + 1 = 6 \neq 0 \). So, \( z=1 \) is not a root.
• Let's test \( z = -1 \):
  \( (-1)^3 + 2(-1)^2 + 2(-1) + 1 = -1 + 2(1) - 2 + 1 = -1 + 2 - 2 + 1 = 0 \). So, \( z=-1 \) is a root.

Since \( z = -1 \) is a root, \( (z - (-1)) \), which is \( (z + 1) \), is a factor of the polynomial \( z^3 + 2z^2 + 2z + 1 \). We can perform polynomial division to find the other factor.

Polynomial Division to Find Factors

We divide \( z^3 + 2z^2 + 2z + 1 \) by \( z + 1 \):

\( \begin{array}{c|cc cc} \multicolumn{2}{r}{z^2} & +z & +1 \\ \cline{2-5} z+1 & z^3 & +2z^2 & +2z & +1 \\ \multicolumn{2}{r}{z^3} & +z^2 \\ \cline{2-3} \multicolumn{2}{r}{0} & z^2 & +2z \\ \multicolumn{2}{r}{} & z^2 & +z \\ \cline{3-4} \multicolumn{2}{r}{} & 0 & z & +1 \\ \multicolumn{2}{r}{} & & z & +1 \\ \cline{4-5} \multicolumn{2}{r}{} & & 0 & 0 \\ \end{array} \)

The quotient is \( z^2 + z + 1 \). Thus, the equation \( z^3 + 2z^2 + 2z + 1 = 0 \) can be factored as \( (z + 1)(z^2 + z + 1) = 0 \).

Finding the Remaining Roots

The roots of the original equation are the roots of the factors:

• From \( z + 1 = 0 \), we get the root \( z = -1 \).
• From \( z^2 + z + 1 = 0 \), we need to find the roots of this quadratic equation.

The quadratic equation \( z^2 + z + 1 = 0 \) is a special equation whose roots are the complex cube roots of unity, usually denoted by \( \omega \) and \( \omega^2 \).

Recall the properties of complex cube roots of unity:

• The roots of \( z^3 - 1 = 0 \) are \( 1, \omega, \omega^2 \).
• The equation \( z^3 - 1 = 0 \) can be factored as \( (z-1)(z^2 + z + 1) = 0 \).
• Therefore, the roots of \( z^2 + z + 1 = 0 \) are \( \omega \) and \( \omega^2 \), where \( \omega = e^{i 2\pi/3} = \frac{-1 + i\sqrt{3}}{2} \) and \( \omega^2 = e^{i 4\pi/3} = \frac{-1 - i\sqrt{3}}{2} \).
• Also, \( 1 + \omega + \omega^2 = 0 \) and \( \omega^3 = 1 \).

Alternatively, we can use the quadratic formula for \( z^2 + z + 1 = 0 \):
\( z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

Here, \( a=1, b=1, c=1 \).
\( z = \frac{-1 \pm \sqrt{(1)^2 - 4(1)(1)}}{2(1)} \)
\( z = \frac{-1 \pm \sqrt{1 - 4}}{2} \)
\( z = \frac{-1 \pm \sqrt{-3}}{2} \)
\( z = \frac{-1 \pm i\sqrt{3}}{2} \)

These two roots are indeed \( \omega \) and \( \omega^2 \).

So, the roots of \( z^2 + z + 1 = 0 \) are \( \omega \) and \( \omega^2 \).

Summary of the Roots of Equation-I

Combining the roots from both factors, the roots of the equation \( z^3 + 2z^2 + 2z + 1 = 0 \) are \( -1, \omega, \) and \( \omega^2 \).

Let's compare this with the given options:

• Option 1: 1, \( \omega \), \( \omega^2 \) (Incorrect, 1 is not a root)
• Option 2: -1, \( \omega \), \( \omega^2 \) (Correct)
• Option 3: 1, -\( \omega \), \( \omega^2 \) (Incorrect, 1 is not a root, and the complex roots are \( \omega \) and \( \omega^2 \))
• Option 4: -1, -\( \omega \), -\( \omega^2 \) (Incorrect, the complex roots are \( \omega \) and \( \omega^2 \))

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

Revision Table: Key Concepts for Equation Roots

Additional Information: Properties of Complex Cube Roots

The complex cube roots of unity, \( \omega \) and \( \omega^2 \), have several important properties that are frequently used in problems involving complex numbers:

• \( 1 + \omega + \omega^2 = 0 \)
• \( \omega^3 = 1 \)
• \( \omega = e^{i 2\pi/3} = \cos(2\pi/3) + i\sin(2\pi/3) = -\frac{1}{2} + i\frac{\sqrt{3}}{2} \)
• \( \omega^2 = e^{i 4\pi/3} = \cos(4\pi/3) + i\sin(4\pi/3) = -\frac{1}{2} - i\frac{\sqrt{3}}{2} \) (which is also \( \bar{\omega} \) or \( \frac{1}{\omega} \))
• Powers of \( \omega \) cycle with a period of 3: \( \omega^4 = \omega, \omega^5 = \omega^2, \omega^6 = \omega^3 = 1 \), and so on. For any integer n, \( \omega^n = \omega^{n \pmod 3} \).

Understanding these properties is crucial for solving problems involving \( \omega \), such as the second equation (Equation-II: \( z^{1985} + z^{100} + 1 = 0 \)) mentioned in the problem context, though solving Equation-II is not required for answering the current question about the roots of Equation-I.