Problem 67 - Algebra and Trigonometry

Question

\(66-68=\) 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

(a) \(2i\) and \(1-i\) are roots; \(-2i\) and \(1+i\) are not. (b) The Conjugate Zeros Theorem doesn't apply due to imaginary coefficients.

Step 1: Verify if \(2i\) is a root

Substitute \(x = 2i\) into the equation \(x^2 - (1+i)x + (2+2i) = 0\). Calculate \((2i)^2 = -4\) and \((1+i)(2i) = 2i + 2i^2 = -2 + 2i\). Now substitute these into the equation:\[ -4 - (-2 + 2i) + (2 + 2i) = 0 \]Simplifying, \(-4 + 2 - 2i + 2 + 2i = 0\), which is true, indicating \(2i\) is a root.

Step 2: Verify if \(1-i\) is a root

Substitute \(x = 1-i\) into the equation \(x^2 - (1+i)x + (2+2i) = 0\). Calculate \((1-i)^2 = 1 - 2i + i^2 = 1 - 2i - 1 = -2i\). The term \((1+i)(1-i) = 1-i+i-1 = 0\).Now substitute into the equation:\[ -2i - 0 + (2 + 2i) = 0 \]Simplifying gives \(-2i + 2 + 2i = 0\), which is true, indicating \(1-i\) is a root.

Step 3: Verify if \(-2i\) is not a root

Substitute \(x = -2i\) into the equation \(x^2 - (1+i)x + (2+2i) = 0\). Calculate \((-2i)^2 = -4\) and \((1+i)(-2i) = -2i - 2i^2 = 2 - 2i\).Now substitute these into the equation:\[ -4 - (2 - 2i) + (2 + 2i) = -4 - 2 + 2i + 2 + 2i \]Simplifying gives \(-6 + 4i eq 0\), indicating \(-2i\) is not a root.

Step 4: Verify if \(1+i\) is not a root

Substitute \(x = 1+i\) into the equation \(x^2 - (1+i)x + (2+2i) = 0\). Calculate \((1+i)^2 = 1 + 2i + i^2 = 1 + 2i - 1 = 2i\). The term \((1+i)(1+i) = 1+2i+i^2 = 1+2i-1 = 2i\).Now substitute these into the equation:\[ 2i - 2i + (2 + 2i) = 2 + 2i eq 0 \]Thus, \(1+i\) is not a root.

Step 5: Explanation of the Conjugate Zeros Theorem

The Conjugate Zeros Theorem applies to polynomials with real coefficients, where complex roots must occur in conjugate pairs. The polynomial here has both real and imaginary coefficients, so the theorem doesn't apply, allowing non-conjugate complex roots.

Key Concepts

Complex Roots

Complex roots are solutions to polynomial equations that involve imaginary numbers. Imaginary numbers are based on the imaginary unit, denoted as \(i\), where \(i^2 = -1\). A complex number generally takes the form \(a + bi\), where \(a\) and \(b\) are real numbers.
When solving equations, like the one in our exercise \(x^2 - (1+i)x + (2+2i) = 0\), we substitute complex numbers as possible solutions. For example, substituting \(2i\) as \(x\) in the equation formed a true statement, indicating \(2i\) is a complex root of the polynomial.
Understanding complex roots involves verifying whether substituting these numbers into the polynomial results in an equation that holds true (i.e., it equals zero). This verification is crucial in determining the roots for equations with both real and imaginary components.

Conjugate Zeros Theorem

The Conjugate Zeros Theorem states that if a polynomial has real coefficients, then its non-real complex roots must appear in conjugate pairs. A conjugate pair of a complex number \(a + bi\) is \(a - bi\). They ensure that any imaginary components cancel out, keeping the polynomial's coefficients real.
However, the Conjugate Zeros Theorem applies specifically to polynomials with real coefficients only. Our exercise here includes imaginary coefficients, hence complex roots do not necessarily occur in conjugate pairs.

• If a complex number \(2i\) is a root, one might expect \(-2i\) to also be a root based on this theorem, but it is not the case here as the polynomial includes imaginary coefficients.
• In situations where coefficients are not all real, the requirement for conjugate pairs is relaxed, allowing solutions that would otherwise contradict the theorem.

Imaginary Coefficients

Imaginary coefficients add complexity to solving polynomial equations, as they deviate from typical cases that only involve real numbers. These coefficients include terms with the imaginary unit \(i\).
The presence of such coefficients requires us to use complex arithmetic while verifying solutions. In the exercise, the polynomial \(x^2 - (1+i)x + (2+2i) = 0\) contains imaginary coefficients such as \((1+i)\) and \((2+2i)\), demanding careful attention while calculating roots.

• This means, during substitution for solution verification, each imaginary term must be dealt with by applying complex number multiplication and addition rules.
• This complexity is why regular rules like the Conjugate Zeros Theorem don't apply, as the problem space extends beyond real-only adjustments.

Understanding these concepts helps in handling any polynomial, whether it contains imaginary coefficients or not.