Problem 56 - Precalculus Mathematics for Calculus

Question

Find the product \(z_{1} z_{2}\) and the quotient \(z_{1} / z_{2}\). Express your answer in polar form. $$z_{1}=7\left(\cos \frac{9 \pi}{8}+i \sin \frac{9 \pi}{8}\right), \quad z_{2}=2\left(\cos \frac{\pi}{8}+i \sin \frac{\pi}{8}\right)$$

Step-by-Step Solution

Answer

The product is \(14e^{i\frac{5\pi}{4}}\) and the quotient is \(\frac{7}{2}e^{i\pi}\).

Step 1: Understand Polar Form

Both \(z_1\) and \(z_2\) are written in polar form \(z = r(\cos \theta + i \sin \theta)\), which is also represented as \(z = re^{i\theta}\). This means \(z_1 = 7e^{i\frac{9\pi}{8}}\) and \(z_2 = 2e^{i\frac{\pi}{8}}\).

Step 2: Multiply the Complex Numbers

To find \(z_1 z_2\), multiply the magnitudes and add the arguments: \(r_1 \times r_2\) and \(\theta_1 + \theta_2\). Thus, \(z_1 z_2 = 7 \times 2 \cdot e^{i(\frac{9\pi}{8} + \frac{\pi}{8})}\).

Step 3: Simplify the Product

Compute the product of magnitudes: \(7 \times 2 = 14\). Then, calculate the sum of angles: \(\frac{9\pi}{8} + \frac{\pi}{8} = \frac{10\pi}{8} = \frac{5\pi}{4}\). Thus, \(z_1 z_2 = 14e^{i\frac{5\pi}{4}}\).

Step 4: Divide the Complex Numbers

To find \(\frac{z_1}{z_2}\), divide the magnitudes and subtract the arguments: \(\frac{r_1}{r_2}\) and \(\theta_1 - \theta_2\). Thus, \(\frac{z_1}{z_2} = \frac{7}{2} e^{i(\frac{9\pi}{8} - \frac{\pi}{8})}\).

Step 5: Simplify the Quotient

Compute the division of magnitudes: \(\frac{7}{2}\). Then, calculate the difference of angles: \(\frac{9\pi}{8} - \frac{\pi}{8} = \frac{8\pi}{8} = \pi\). Thus, \(\frac{z_1}{z_2} = \frac{7}{2}e^{i\pi}\).

Key Concepts

Polar Form

The polar form of a complex number provides a way of expressing it using a magnitude and an angle. It is especially useful when performing multiplications and divisions. The polar form is written as \(z = r(\cos \theta + i \sin \theta)\), where \(r\) is the magnitude, and \(\theta\) is the angle (or argument) with the real axis. This can also be expressed using Euler's formula as \(z = re^{i\theta}\).

This representation makes use of trigonometric functions to locate the number on the complex plane. The angle \(\theta\) is measured in radians from the positive real axis, providing a unique position for each complex number.

For example, in the exercise, \(z_1 = 7\left(\cos \frac{9\pi}{8} + i \sin \frac{9\pi}{8}\right)\) and \(z_2 = 2\left(\cos \frac{\pi}{8} + i \sin \frac{\pi}{8}\right)\) are given in polar form. Here, the magnitudes are 7 and 2, and the angles are \(\frac{9\pi}{8}\) and \(\frac{\pi}{8}\), respectively. Being in polar form makes it easier to multiply and divide these complex numbers, as will be demonstrated in the next sections.

Complex Multiplication

Multiplying complex numbers in polar form involves straightforward operations on the magnitudes and angles. Given two complex numbers \(z_1 = r_1e^{i\theta_1}\) and \(z_2 = r_2e^{i\theta_2}\), the product \(z_1 z_2\) is calculated by:

• Multiplying the magnitudes: \(r_1 \times r_2\)
• Adding the angles: \(\theta_1 + \theta_2\)

This results in \(z_1 z_2 = (r_1 \times r_2)e^{i(\theta_1 + \theta_2)}\).

In the context of the exercise, for \(z_1 = 7e^{i\frac{9\pi}{8}}\) and \(z_2 = 2e^{i\frac{\pi}{8}}\), their product is \(7 \times 2 = 14\) for the magnitude and the angles add up to \(\frac{9\pi}{8} + \frac{\pi}{8} = \frac{10\pi}{8} = \frac{5\pi}{4}\). Thus, the product is \(z_1 z_2 = 14e^{i\frac{5\pi}{4}}\). Using polar form for multiplication simplifies the computations, as there's no need to expand the expressions with trigonometric identities.

Complex Division

Dividing complex numbers in polar form, much like their multiplication, enjoys a significant simplification compared to the typical Cartesian form. For two complex numbers \(z_1 = r_1e^{i\theta_1}\) and \(z_2 = r_2e^{i\theta_2}\), the quotient \(\frac{z_1}{z_2}\) is determined by:

• Dividing the magnitudes: \(\frac{r_1}{r_2}\)
• Subtracting the angles: \(\theta_1 - \theta_2\)

This results in \(\frac{z_1}{z_2} = \left(\frac{r_1}{r_2}\right)e^{i(\theta_1 - \theta_2)}\).

In the provided problem, with \(z_1 = 7e^{i\frac{9\pi}{8}}\) and \(z_2 = 2e^{i\frac{\pi}{8}}\), the division yields a magnitude of \(\frac{7}{2}\) and the angles' difference \(\frac{9\pi}{8} - \frac{\pi}{8} = \frac{8\pi}{8} = \pi\). Thus, the quotient is \(\frac{z_1}{z_2} = \frac{7}{2}e^{i\pi}\). Using polar form here makes the division straightforward, avoiding more complex arithmetic or algebraic manipulation in the conventional approach.