Problem 71 - Algebra and Trigonometry

Question

So far, we have worked only with polynomials that have real coefficients. These exercises involve polynomials with real and imaginary coefficients. (a) Show that 2\(i\) and \(1-i\) are both solutions of the equation $$x^{2}-(1+i) x+(2+2 i)=0$$ but that their complex conjugates \(-2 i\) and \(1+i\) are not. (b) Explain why the result of part (a) does not violate the Conjugate Zeros Theorem.

Step-by-Step Solution

Answer

The roots \(2i\) and \(1-i\) satisfy the equation, while their conjugates do not due to non-real coefficients not adhering to the Conjugate Zeros Theorem.

Step 1: Identify the given equation

The equation provided is a quadratic equation: \(x^2 - (1+i)x + (2+2i) = 0\). We will test if the given solutions satisfy this equation.

Step 2: Verify if \(x = 2i\) is a solution

Substitute \(x = 2i\) into the equation: \((2i)^2 - (1+i)(2i) + (2+2i) = 0\). Simplifying gives \(-4 - 2i + 2 - 2i + 2 + 2i = 0\), which simplifies further to \(-4 + 4 = 0\). Since both sides equal zero, \(x = 2i\) is indeed a solution.

Step 3: Verify if \(x = 1-i\) is a solution

Substitute \(x = 1-i\) into the equation: \((1-i)^2 - (1+i)(1-i) + (2+2i) = 0\). Simplifying gives \((1-2i+i^2) - (1 + i - i - i^2) + 2 + 2i = 0\), which leads to \(1 - 2i - 1 + 1 + 2 + 2i = 0\). Since both sides equal zero, \(x = 1-i\) is a solution.

Step 4: Check if the conjugate \(x = -2i\) is a solution

Substitute \(x = -2i\) into the equation: \((-2i)^2 - (1+i)(-2i) + (2+2i) = 0\). Simplifying gives \(4 + 2i - 2 + 2i + 2 + 2i = 10i eq 0\). The left side does not equal zero, so \(x = -2i\) is not a solution.

Step 5: Check if the conjugate \(x = 1+i\) is a solution

Substitute \(x = 1+i\) into the equation: \((1+i)^2 - (1+i)(1+i) + (2+2i) = 0\). Upon simplifying, both sides of the equation do not equal and hence, \(x = 1+i\) is not a solution.

Step 6: Explain why Conjugate Zeros Theorem is not violated

The Conjugate Zeros Theorem states that if a polynomial has all real coefficients, then non-real complex roots must occur in conjugate pairs. However, the given polynomial \(x^2 - (1+i)x + (2+2i)\) does not have all real coefficients, which is why it is not subject to this theorem. Thus, the theorem is not violated by the solutions.

Key Concepts

Polynomials with Complex Coefficients

When dealing with polynomials, we often encounter those with coefficients that are strictly real numbers. However, sometimes these coefficients can also be complex numbers.
A polynomial with complex coefficients is one where some or all of the coefficients in the expression are complex numbers. For instance, in the quadratic equation from the exercise:

• Given equation: \(x^2 - (1+i)x + (2+2i) = 0\)
• The coefficients \(-(1+i)\) and \((2+2i)\) are complex.

These coefficients include both a real part and an imaginary part (e.g., \(1+i\) has a real part 1 and an imaginary part 1).
This aspect changes the behavior of the solutions of the polynomial. Unlike real-coefficient polynomials which have solutions in the real domain, those with complex coefficients can have solutions that are uniquely complex.

Complex Conjugates

Complex conjugates are a fundamental concept in complex numbers. If you have a complex number, such as \(z = a + bi\), its conjugate is \(\overline{z} = a - bi\).

• Original: \(z = 1+i\)
• Conjugate: \(\overline{z} = 1-i\)

Conjugates are special because they have unique properties. When a number is multiplied by its conjugate, the result is a real number. This is because the imaginary parts cancel each other out in the process:

• \((a+bi)(a-bi) = a^2 - b^2i^2 = a^2 + b^2\), since \(i^2 = -1\).

Understanding conjugates helps particularly in simplifying expressions and solving equations involving complex numbers. This is significant in computational scenarios like finding module or norm of a complex number and ensuring stability in mathematical processes.

Conjugate Zeros Theorem

The Conjugate Zeros Theorem is a fascinating principle in polynomial algebra. It asserts that if you have a polynomial with real coefficients, any non-real complex roots must appear in conjugate pairs. However, this theorem does not apply to polynomials with complex coefficients.
In the exercise, the polynomial \(x^2 - (1+i)x + (2+2i)\) does not have purely real coefficients; hence, the Conjugate Zeros Theorem is not applicable. This explains why solutions \(2i\) and \(1-i\) do not necessarily require their conjugates to be solutions as well.

• Real coefficients require complex roots to be conjugate pairs.
• Complex coefficients do not impose this requirement.

This subtlety is crucial while dealing with complex polynomials and must be understood to manage the roots effectively.