Problem 36 - Algebra 1 Course in Mathematics for the IIT-JEE and Other Engineering Exams

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) 6 (b) 12 (c) 18 (d) 24

Step-by-Step Solution

Answer

The value is (a) 6.

Step 1: Solve the Equation for z

Given that \(z^2 + z + 1 = 0\), we recognize this as a quadratic equation. The solutions for \(z\) can be found using the quadratic formula: \(z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a=1\), \(b=1\), and \(c=1\). After substituting the values and simplifying, we get \(z = \frac{-1 \pm \sqrt{-3}}{2}\), leading to \(z = \frac{-1 \pm i\sqrt{3}}{2}\). These roots are complex numbers, often known as \(\omega\) and \(\omega^2\), where \(\omega\) is the cube root of unity \(\omega = e^{2\pi i / 3}\), satisfying \(\omega^3 = 1\) and \(\omega^2 = \overline{\omega}\).

Step 2: Find z and 1/z Coefficient

Recall that \(z^3 = 1\), so for solutions \(\omega\) or \(\omega^2\), \(\frac{1}{z} = z^2\). For the sequence terms \(\left(z^n + \frac{1}{z^n}\right)^2\), use the property that \(z^{n} = (z^3)^{k} z^m = z^m\) because \(z^3 = 1\).

Step 3: Evaluate Each Expression

Calculate each term: \(z^1 + \frac{1}{z^1} = z + z^2\), simplifying \(z + z^2 = -1\). Hence, \((z + \frac{1}{z})^2 = (-1)^2 = 1\). For higher powers \(z^2, z^3, ..., z^6\), we reuse the symmetry in \(z^3 = 1\) leading to these expressions being equal at modulo 3 due to periodicity. Each results in \((z^n + \frac{1}{z^n})^2 = 1\).

Step 4: Sum the Values

Sum the values from the sequence: \(\left(z^1 + \frac{1}{z^1}\right)^2 + \left(z^2 + \frac{1}{z^2}\right)^2 + \left(z^3 + \frac{1}{z^3}\right)^2 + \ldots + \left(z^6 + \frac{1}{z^6}\right)^2\). Since every term resulted in 1, there are 6 terms, each equal to 1. Sum is \(6 \times 1 = 6\).

Step 5: Verify with Known Values

Cross-check with known expression periodically repeating every three powers as \(z^3 = 1\). Each cycle just distresses over cube rotations and reforming results show correctness for roots of unity over the range.

Key Concepts

Quadratic Equations

Quadratic equations are fundamental in algebra. They typically have the form \(ax^2 + bx + c = 0\). In the specific problem we have, the equation \(z^2 + z + 1 = 0\) fits this structure, though here \(z\) represents a complex number.

The quadratic formula \(z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) is used to solve these equations. It helps us find the values of \(z\) when \(a = 1\), \(b = 1\), and \(c = 1\).

Using this formula, we find that \(z\) resolves to complex roots. The presence of \(\sqrt{-3}\) under the square root gives us an imaginary component, indicating complex solutions. These roots are extremely helpful in exploring deeper algebraic structures like roots of unity.

Roots of Unity

Roots of unity are special complex numbers that, when raised to a specific power, yield 1. Particularly, the cube roots of unity solve the equation \(z^3 = 1\).

In this scenario, the roots of the equation \(z^2 + z + 1 = 0\) relate to cube roots of unity, specifically \(\omega = e^{2\pi i / 3}\). These roots, \(\omega\) and \(\omega^2\), keep the property that \(\omega^3 = 1\) and \(\omega^2 = \overline{\omega}\).

These properties imply cyclical behavior since raising \(\omega\) to powers in patterns of three (like \(\omega^3, \omega^6\)) will bring you back to 1.

• \(\omega^1 = \omega\)
• \(\omega^2 = \omega^2\)
• \(\omega^3 = 1\)

This repetitiveness helps in simplifying larger algebraic expressions, as seen in periodic evaluations in the textbook problem.

Properties of Complex Numbers

Complex numbers consist of a real part and an imaginary part, written as \(z = a + bi\). Here, \(i\) is the imaginary unit with the property \(i^2 = -1\).

The complex roots of our quadratic equation possess both real and imaginary components: \(z = \frac{-1 \pm i\sqrt{3}}{2}\). Using complex roots, computations enhance our understanding of symmetrical properties seen in formulas and expressions like \(z^n + \frac{1}{z^n}\).

When working with expressions like \(z^3\), these properties help reduce complexity as multiplication of imaginary units simplifies following standard rules.

• Multiplication: \((a+bi)(c+di) = ac + adi + bci - bd\)
• Addition: \((a+bi) + (c+di) = (a+c) + (b+d)i\)
• Conjugate: If \(z = a+bi\), then conjugate \(\overline{z} = a-bi\)

Algebraic Expressions

Algebraic expressions involve numbers, variables, and operation signs. These expressions can get quite complex when including higher powers or fractions.

In our example, the expression \(\left(z + \frac{1}{z}\right)^2\) involves both a complex number and its reciprocal. Solving such expressions requires applying formulas, and often utilizing properties of the numbers involved (like roots of unity).

Here's a step-by-step guide on how to evaluate such expressions:

• For specific powers, simplify using principles like \(z^3 = 1\).
• Process cyclic behavior where applicable, taking each periodic solution into account.
• Compute distinct parts separately when applicable.

Ultimately, learning to manage complex algebraic expressions empowers problem-solving capabilities across mathematics.