Problem 71 - Precalculus Mathematics for Calculus

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

(a) The roots 2i and 1-i satisfy the equation. (b) The Conjugate Zeros Theorem is not violated as the coefficients are not all real.

Step 1: Use the Quadratic Formula

To find the roots of the equation \(x^2 - (1+i)x + (2+2i) = 0\), we use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]For this equation, \(a = 1\), \(b = -(1+i)\), and \(c = (2+2i)\). Substitute these values into the formula.

Step 2: Calculate the Discriminant

The discriminant \(b^2 - 4ac\) is the key to finding the roots. Calculate:\[(1+i)^2 - 4(2+2i) = (1 + 2i - 1) - (8 + 8i) = -7i \]This result is a complex number, indicating non-real roots.

Step 3: Calculate the Roots

Substitute the discriminant into the quadratic formula:\[x = \frac{1+i \pm \sqrt{-7i}}{2} \]We need to find \(\sqrt{-7i}\). Approximate it as \( \sqrt{7/2} + i\sqrt{7/2} \) through complex root techniques. Solve each solution to verify \(2i\) and \(1-i\) satisfy the equation.

Step 4: Verify Roots 2i and 1-i

Plug \(x = 2i\) into the polynomial:\[(2i)^2 - (1+i)(2i) + (2+2i) = -4 - 2 + 2i + 2 + 2i = 0 \]Similarly, plugging \(x = 1-i\):\[(1-i)^2 - (1+i)(1-i) + (2+2i) = 1 - 2i + 1 - (1 + 1) + 2 + 2i = 0 \]Both values satisfy the equation.

Step 5: Verify Complex Conjugates -2i and 1+i

Test \(x = -2i\):\[ (-2i)^2 - (1+i)(-2i) + (2 + 2i) = -4 + 2i - 2i + 2 + 2i eq 0 \]Test \(x = 1+i\):\[ (1+i)^2 - (1+i)(1+i) + (2 + 2i) = 0 \] shows no simplifying to 0.Neither satisfies the equation, reconfirming them as non-roots.

Step 6: Explain Why the Conjugate Zeros Theorem Does Not Apply

The Conjugate Zeros Theorem holds when polynomial coefficients are real. Here, coefficients appear as complex. The theorem remains unviolated, given the complex nature of the coefficients and solutions in the polynomial.

Key Concepts

quadratic formula

The quadratic formula is a fundamental tool for solving quadratic equations. A quadratic equation is typically in the form \( ax^2 + bx + c = 0 \). The quadratic formula helps to find the roots of such equations by providing a straightforward method: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]In the given problem, we have a quadratic equation with complex coefficients: \( a = 1 \), \( b = -(1+i) \), and \( c = (2+2i) \). To apply the quadratic formula, identify these coefficients and use them in the formula. This requires calculation of the discriminant, \( b^2 - 4ac \). The discriminant is crucial because it dictates the nature of the roots:

• If it is positive, the roots are real and distinct.
• If zero, they are real and repeated.
• If negative, the roots are complex.

In our case, the discriminant is negative, indicating complex roots. Recognizing this factor is essential for solving equations with non-real roots, especially when they involve complex coefficients.

polynomials with complex coefficients

When dealing with polynomials that have complex coefficients, it is important to understand that their roots can behave differently compared to polynomials with real coefficients only. Complex coefficients mean that either the real, imaginary, or both parts of the coefficients are non-zero and imaginary. For the equation \( x^2 - (1+i)x + (2+2i) = 0 \), the coefficients involve imaginary numbers, specifically \( b = -(1+i) \) and \( c = (2+2i) \). This changes the approach we take with the quadratic formula, as we must also consider complex arithmetic. When solving, you must be comfortable with calculations that involve:

• Adding and subtracting complex numbers.
• Multiplying complex numbers, especially when expanding expressions like \( (1+i)^2 \).
• Simplifying square roots of complex numbers, which may require converting them into a more manageable form, like using polar coordinates or other techniques.

A crucial skill when pondering complex coefficients is recognizing that they potentially lead to non-real, complex roots, as seen in the problem where \( 2i \) and \( 1-i \) are solutions.

Conjugate Zeros Theorem

The Conjugate Zeros Theorem is a principle that applies to polynomials with real coefficients. It states that if a polynomial has real coefficients and a complex root, then its complex conjugate must also be a root. However, in this exercise, the polynomial includes complex coefficients.Because the coefficients are not entirely real, the theorem does not apply. Instead, each complex root does not necessarily mandate the presence of its conjugate as another root. In simpler terms, the theorem needs:

• Real coefficients in the polynomial.
• If \( a + bi \) (where \( b eq 0 \)) is a root, then \( a - bi \) must also be a root.

However, if any coefficient has an imaginary part, as in our current equation, using complex coefficients, the theorem does not enforce the presence of the conjugate root. In the exercise, this allowed solutions like \( 2i \) and \( 1-i \) without needing their conjugates to also solve the polynomial.