Problem 204 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(z^{2}+z+1=0\), where \(z\) is a complex number, then the value of \(\left(z+\frac{1}{z}\right)^{2}+\left(z^{2}+\frac{1}{z^{2}}\right)^{2}+\left(z^{3}+\frac{1}{z^{3}}\right)^{2}+\ldots+\left(z^{6}+\frac{1}{z^{6}}\right)^{2}\) is (A) 18 (B) 54 (C) 6 (D) 12

Step-by-Step Solution

Answer

The answer is 12; option (D).

Step 1: Solve for the Roots

First, solve the quadratic equation \(z^2 + z + 1 = 0\). The roots can be found using the quadratic formula:\[ z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]where \(a = 1\), \(b = 1\), and \(c = 1\). Plug these into the formula:\[ z = \frac{-1 \pm \sqrt{1 - 4}}{2} = \frac{-1 \pm i\sqrt{3}}{2} \].These roots are primitive cube roots of unity, often denoted as \(\omega\) and \(\omega^2\), where \( \omega = \frac{-1 + i\sqrt{3}}{2} \), \( \omega^2 = \frac{-1 - i\sqrt{3}}{2} \), and \( \omega^3 = 1 \).

Step 2: Simplify Powers of z

Using the fact that \( \omega^3 = 1 \), determine the subsequent powers of \( \omega \):- \( \omega^4 = \omega \)- \( \omega^5 = \omega^2 \)- \( \omega^6 = 1 \).Now, determine \( z^n + \frac{1}{z^n} \) using these powers. For example:- For \( n = 1 \), \( \omega + \frac{1}{\omega} = -1 \).- For \( n = 2 \), \( \omega^2 + \frac{1}{\omega^2} = -1 \).- For \( n = 3 \), \( \omega^3 + \frac{1}{\omega^3} = 2 \).Continue this pattern for \(n = 4, 5, 6\).

Step 3: Calculate Each Square in Series

Compute \( \left(z^n + \frac{1}{z^n}\right)^2 \) for each \(n = 1, 2, 3, 4, 5, 6\):- \( \left(\omega + \frac{1}{\omega}\right)^2 = (-1)^2 = 1 \).- \( \left(\omega^2 + \frac{1}{\omega^2}\right)^2 = (-1)^2 = 1 \).- \( \left(\omega^3 + \frac{1}{\omega^3}\right)^2 = (2)^2 = 4 \).- \( \left(\omega^4 + \frac{1}{\omega^4}\right)^2 = (-1)^2 = 1 \).- \( \left(\omega^5 + \frac{1}{\omega^5}\right)^2 = (-1)^2 = 1 \).- \( \left(\omega^6 + \frac{1}{\omega^6}\right)^2 = (2)^2 = 4 \).

Step 4: Sum the Squares

Add all squared terms from Step 3:\(1 + 1 + 4 + 1 + 1 + 4 = 12\).

Step 5: Conclusion

The total sum of the series \( \left(z + \frac{1}{z}\right)^{2} + \left(z^{2} + \frac{1}{z^{2}}\right)^{2} + \dots + \left(z^{6} + \frac{1}{z^{6}}\right)^{2} \) is 12. Hence, the answer is (D) 12.

Key Concepts

Quadratic Equations

Quadratic equations form the basis for solving problems involving complex numbers. A standard quadratic equation is expressed as \( ax^2 + bx + c = 0 \). To find the solutions or 'roots' of the equation, we can use the quadratic formula:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

For the given problem, \( z^2 + z + 1 = 0 \), the coefficients are \( a = 1 \), \( b = 1 \), and \( c = 1 \). Plugging these values into the quadratic formula helps us find the roots. The roots are complex numbers, \( \frac{-1 \pm i\sqrt{3}}{2} \), which are not real and show the beauty of complex roots in quadratic equations. Often, complex roots appear in conjugate pairs, the case here being classic for a quadratic equation with no real solutions.
Understanding complex solutions is crucial. In this case, the roots \( \omega \) and \( \omega^2 \) lead to an elegant property in complex numbers known as the Roots of Unity, an important concept in deeper mathematical studies.

Roots of Unity

The Roots of Unity are specific solutions to the equation \( x^n = 1 \) in the complex number system. Particularly, for cube roots of unity as encountered here, the solutions to \( z^2 + z + 1 = 0 \) give us primitive cube roots. They are called 'primitive' because they generate all other cube roots through their powers. The roots found, \( \omega = \frac{-1 + i\sqrt{3}}{2} \) and \( \omega^2 = \frac{-1 - i\sqrt{3}}{2} \), satisfy \( \omega^3 = 1 \).
These cube roots have fascinating properties; they are located equidistantly on the unit circle in the complex plane. In terms of practical calculation:

• \( \omega^3 = 1 \), resulting in simple powers like \( \omega^4 = \omega \), \( \omega^5 = \omega^2 \).
• These properties simplify many complex calculations, as seen when determining the sums and dimensions of series involving higher powers of \( z \).

This insight is powerful because it allows us to simplify complex expressions easily. Understanding these roots simplifies calculations in fields extending beyond pure mathematics, like engineering and signal processing.

Power of Complex Numbers

Handling powers of complex numbers becomes intuitive when you take advantage of the properties of roots of unity. When tasked with solving \( z^n + \frac{1}{z^n} \) for various powers of \( n \), knowing the relationship \( \omega^3 = 1 \) is critical. This means that the powers cycle, and every third power returns to one. Such cycling properties make it straightforward to compute expressions involving complex powers.
For example:

• For \( n = 1 \), \( z + \frac{1}{z} = \omega + \frac{1}{\omega} = -1 \).
• For \( n = 3 \), \( z^3 = 1 \) and thus \( z^3 + \frac{1}{z^3} = 2 \).

These computational tactics reduce complex operations to simple arithmetic. With these powers, expressions like \( \left(z^n + \frac{1}{z^n}\right)^2 \) become easily manageable, allowing for quick summation in sequences. Understanding how to compute these powers not only aids in this specific problem set but also develops deeper insights into complex number theory.