Problem 100 - Precalculus Mathematics for Calculus

Question

Complex Coefficients and the Quadratic Formula The quadratic formula works whether the coefficients of the equation are real or complex. Solve these equations using the quadratic formula and, if necessary, De Moivre's Theorem. (a) \(z^{2}+(1+i) z+i=0\) (b) \(z^{2}-i z+1=0\) (c) \(z^{2}-(2-i) z-\frac{1}{4} i=0\)

Step-by-Step Solution

Answer

(a) Two complex roots, (b) Two complex roots, (c) Two complex roots.

Step 1: Identify the coefficients for equation (a)

For the equation \(z^{2} + (1+i)z + i = 0\), identify the coefficients: \(a = 1\), \(b = 1+i\), and \(c = i\).

Step 2: Apply the quadratic formula to equation (a)

Use the quadratic formula: \(z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Substitute the values of \(a\), \(b\), and \(c\): \(z = \frac{-(1+i) \pm \sqrt{(1+i)^2 - 4 \times 1 \times i}}{2}\).

Step 3: Simplify the discriminant for equation (a)

Calculate \((1+i)^2 = 1 + 2i + i^2 = 1 + 2i - 1 = 2i\). Now the discriminant is \(2i - 4i = -2i\).

Step 4: Solve the discriminant using De Moivre’s Theorem for equation (a)

Convert \(-2i\) to polar form: magnitude is \(2\), angle is \(-\frac{\pi}{2}\). So, \(-2i = 2\text{cis}\left(-\frac{\pi}{2}\right)\). Use De Moivre's: \(\sqrt{-2i}= \sqrt{2}\text{cis}\left(-\frac{\pi}{4}\right)\).

Step 5: Calculate the roots for equation (a)

Now the solutions are \(z = \frac{-(1+i) \pm \sqrt{2}\text{cis}\left(-\frac{\pi}{4}\right)}{2}\). Simplify to find two possible roots: \(z_1\) and \(z_2\).

Step 6: Identify the coefficients for equation (b)

For the equation \(z^{2} - iz + 1 = 0\), identify the coefficients: \(a = 1\), \(b = -i\), and \(c = 1\).

Step 7: Apply the quadratic formula to equation (b)

Use the quadratic formula \(z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Substitute: \(z = \frac{i \pm \sqrt{(-i)^2-4 \times 1 \times 1}}{2}\).

Step 8: Simplify the discriminant for equation (b)

Calculate \((-i)^2 = -1\). The discriminant becomes \(-1 - 4 = -5\).

Step 9: Solve the discriminant using De Moivre’s Theorem for equation (b)

Convert \(-5\) to polar form: magnitude is \(\sqrt{5}\), angle is \(\pi\). Thus: \(\sqrt{-5} = \sqrt{5}\text{cis}\left(\frac{\pi}{2}\right)\).

Step 10: Calculate the roots for equation (b)

Now the solutions are \(z = \frac{i \pm \sqrt{5}i}{2}\). Simplify to find \(z_1\) and \(z_2\).

Step 11: Identify the coefficients for equation (c)

For the equation \(z^{2} - (2-i)z - \frac{1}{4}i = 0\), identify \(a = 1\), \(b = -(2-i)\), \(c = -\frac{1}{4}i\).

Step 12: Apply the quadratic formula to equation (c)

Use the formula \(z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Substitute: \(z = \frac{2-i \pm \sqrt{(2-i)^2 + i}}{2}\).

Step 13: Simplify the discriminant for equation (c)

Calculate \((2-i)^2 = 4 - 4i + i^2 = 3 - 4i\). The discriminant becomes \(3 - 4i + i\).

Step 14: Solve the discriminant using De Moivre’s Theorem for equation (c)

Combine: the discriminant is \(3 - 4i\). Convert to polar form for square root \(\sqrt{3-4i}\).

Step 15: Calculate the roots for equation (c)

Solve \(z = \frac{2-i \pm \sqrt{r}\text{cis}\left(\theta/2\right)}{2}\) for magnitude \(r\) and angle \(\theta\). Simplify to find \(z_1\) and \(z_2\).

Key Concepts

Quadratic Formula

The quadratic formula is a powerful tool that helps solve quadratic equations of the form \(az^2 + bz + c = 0\). The beauty of this formula is that it remains effective even when the coefficients \(a\), \(b\), and \(c\) are complex numbers. The formula is given by:
\[z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
To use the quadratic formula, we first identify the coefficients from the equation. For instance, in \(z^2 + (1+i)z + i = 0\), \(a = 1\), \(b = 1+i\), and \(c = i\). After identifying these coefficients, substitute them into the formula. The challenge often lies in simplifying the discriminant \(b^2 - 4ac\). This discriminant can sometimes be a complex number itself, which needs further simplification using other techniques, such as converting to polar form and applying De Moivre's Theorem.

The quadratic formula is not limited to just providing root values; it also offers insight into the nature of the roots based on the discriminant. A positive discriminant indicates two distinct real or complex roots, zero signifies a double root, and a negative discriminant implies complex conjugate roots.

De Moivre's Theorem

De Moivre's Theorem is crucial for simplifying powers and roots of complex numbers. It states that for a complex number in polar form \(r \text{cis} \theta\) (where "cis" stands for \(\cos \theta + i\sin \theta\)), the \(n\)-th power is:
\[r^n \text{cis} (n\theta)\]
This theorem becomes very useful, particularly when working with the square roots of complex numbers derived from the discriminant in the quadratic formula. For instance, to find \(\sqrt{-2i}\), first convert \(-2i\) to polar form with magnitude 2 and angle \(-\frac{\pi}{2}\). Hence, \(-2i = 2 \text{cis} (-\frac{\pi}{2})\). Using De Moivre's Theorem for the square root, compute:
\[\sqrt{2} \text{cis}\left(-\frac{\pi}{4}\right)\]

By simplifying complex expressions in polar form, De Moivre's Theorem enables us to provide solutions that are otherwise cumbersome to handle in standard cartesian form. This approach is particularly helpful for quadratic equations with complex coefficients, allowing us to find roots more efficiently.

Polar Form

The polar form of a complex number expresses it in terms of its magnitude and angle relative to the positive x-axis. Any complex number \(z = a + bi\) can be represented in polar form as \(r \text{cis} \theta\), where \(r = \sqrt{a^2 + b^2}\) (magnitude), and \(\theta = \tan^{-1}(\frac{b}{a})\) (angle).

One of the advantages of using polar form is the simplification of operations like multiplication, division, and finding powers or roots. In polar form, multiplication and division are more straightforward: multiply/divide the magnitudes and add/subtract the angles. For example, to solve \(z^2 = -2i\) converted into polar form, start with identifying the magnitude \(r\) and then compute using De Moivre’s Theorem.

• Polar form is handy when combining angles and magnitudes, providing a comprehensive view of complex numbers.
• It also aids in visualizing complex numbers on the Argand plane, making it easier to understand operations like rotation and dilation.

For many complex calculations, converting complex numbers to polar form is a strategic move that dramatically simplifies the process. This form is particularly useful when solving equations involving powers and roots of complex numbers.