Problem 204 - A Complete Resource Book in Mathematics for JEE Main 2017

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 \([\mathbf{2 0 0 6}]\) (A) 18 (B) 54 (C) 6 (D) 12

Step-by-Step Solution

Answer

(A) 18

Step 1: Identify Roots of Equation

Given the equation \( z^2 + z + 1 = 0 \), use the quadratic formula: \( z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here \( a = 1 \), \( b = 1 \), and \( c = 1 \). Substitute these values into the formula.

Step 2: Solve Quadratic Formula

Calculate \( b^2 - 4ac = 1 - 4 = -3 \). The formula becomes \( z = \frac{-1 \pm \sqrt{-3}}{2} = \frac{-1 \pm i\sqrt{3}}{2} \). Thus, the roots are \( z = \frac{-1 + i\sqrt{3}}{2} \) and \( z = \frac{-1 - i\sqrt{3}}{2} \).

Step 3: Use Properties of Roots

For \( z^2 + z + 1 = 0 \), dividing by \( z \) gives \( z + 1 + \frac{1}{z} = 0 \). This implies \( z + \frac{1}{z} = -1 \).

Step 4: Calculate Each Term in Series

We need to find \( \left(z^k + \frac{1}{z^k}\right)^2 \) for \( k = 1 \) to \( 6 \). Using \( z^3 = 1 \) because \( z \) satisfies \( z^2 + z + 1 = 0 \), calculate \( z + \frac{1}{z} = -1 \), \( z^2 + \frac{1}{z^2} = -1 \), and recollect the periodicity: values repeat every three powers.

Step 5: Consider Periodicity and Series

Combine the terms based on periodicity: \( \left(z + \frac{1}{z}\right)^2 = 1 \), \( \left(z^2 + \frac{1}{z^2}\right)^2 = 1 \), \( \left(z^3 + \frac{1}{z^3}\right)^2 = 1 \), the series extends to six terms covering two periods, thus multiplying by 2.

Step 6: Sum the Series

Sum the identical terms (there are two sets of 3 identical values): \( 3(1^2 + 1^2 + 1^2) = 3 \times 3 = 9 \). Multiply by 2 for both periods: \( 9 \times 2 = 18 \).

Key Concepts

Quadratic Equations

Quadratic equations are mathematical expressions that include a variable, typically denoted as \(x\), raised to the power of two. The standard form of a quadratic equation is \( ax^2 + bx + c = 0 \), where \(a\), \(b\), and \(c\) are constants and \(a eq 0\). These equations are fundamental in algebra as they model various physical phenomena like projectile motion and optics.
To solve quadratic equations, we often use techniques such as factoring, completing the square, or the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
In our exercise, the equation is \( z^2 + z + 1 = 0 \), with a complex number \(z\). By applying the quadratic formula we find the roots, even if they are complex numbers, meaning they include an imaginary unit \(i\). Understanding this concept lets us explore solutions even in cases where real number solutions might not exist.

Roots of Equations

Roots of equations refer to the values of the variable that satisfy the equation, meaning they make the equation equal to zero. For quadratic equations, these roots can be real or complex. Complex roots arise particularly when the discriminant (the expression \(b^2 - 4ac\) within the quadratic formula) is negative.
In our problem, using \(z^2 + z + 1 = 0\), we find the discriminant to be \(-3\). This negative value indicates complex roots. Solving the quadratic formula gives us \( z = \frac{-1 + i\sqrt{3}}{2} \) and \( z = \frac{-1 - i\sqrt{3}}{2} \).
These represent our complex roots. Roots such as these connect to the symmetrical nature of polynomial functions, which often holds interesting properties discussed in deeper algebra and calculus.

Periodicity of Complex Numbers

Periodicity in complex numbers refers to the repeating pattern they exhibit when raised to successive powers. This is a specific property tied to the concept of their roots of unity. For instance, certain equations, like our \( z^2 + z + 1 = 0 \), illustrate how the powers of a complex root begin cycling after a few steps.
Here, observe that \( z^3 = 1 \), meaning every third power of \(z\) returns to 1, which establishes a periodicity of 3.
This periodicity helps simplify expressions involving complex numbers: one does not need to compute every power anew; recognizing repeating cycles can streamline calculations. In our exercise, knowing \(z^3 = 1\) means that for powers beyond this, the pattern repeats, aiding in series summations.

Series Summation

Series summation is the process of adding a sequence of numbers, and it's a powerful tool that helps simplify complex operations. In the context of complex numbers, series summation can involve powers of complex roots.
For our problem, we consider the sum of squares: \((z + \frac{1}{z})^2, (z^2 + \frac{1}{z^2})^2, \ldots, (z^6 + \frac{1}{z^6})^2\).
Given the nature of periodicity (\(z^3 = 1\)), these expressions cycle every three terms:

• \((z + \frac{1}{z})^2 = 1\)
• \((z^2 + \frac{1}{z^2})^2 = 1\)
• \((z^3 + \frac{1}{z^3})^2 = 1\)

Because of this cycle, the sequence repeats in another period from \(z^4\) to \(z^6\), each giving the same contribution as their respective counterparts. Therefore, calculating the series simply involves recognizing the repeating cycle and summing effectively across two periods, leading to our final result of 18.