Problem 55 - 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. $$z_{1}=3\left(\cos \frac{\pi}{6}+i \sin \frac{\pi}{6}\right), \quad z_{2}=5\left(\cos \frac{4 \pi}{3}+i \sin \frac{4 \pi}{3}\right)$$

Step-by-Step Solution

Answer

The product is \(15(\cos \frac{3\pi}{2} + i \sin \frac{3\pi}{2})\), and the quotient is \(\frac{3}{5}(\cos \frac{5\pi}{6} + i \sin \frac{5\pi}{6})\).

Step 1: Identify Moduli and Arguments

For the given complex numbers in polar form:\[z_1 = 3\left(\cos \frac{\pi}{6} + i \sin \frac{\pi}{6}\right)\] and \[z_2 = 5\left(\cos \frac{4\pi}{3} + i \sin \frac{4\pi}{3}\right)\],the moduli are \( r_1 = 3 \) and \( r_2 = 5 \). The arguments are \( \theta_1 = \frac{\pi}{6} \) and \( \theta_2 = \frac{4\pi}{3} \).

Step 2: Calculate the Product

To find the product \( z_1 z_2 \), multiply the moduli and add the arguments:\[r_1 r_2 = 3 \times 5 = 15\]\[\theta_1 + \theta_2 = \frac{\pi}{6} + \frac{4\pi}{3} = \frac{\pi}{6} + \frac{8\pi}{6} = \frac{9\pi}{6} = \frac{3\pi}{2}\]Hence, \( z_1 z_2 = 15 \left(\cos \frac{3\pi}{2} + i \sin \frac{3\pi}{2}\right) \).

Step 3: Calculate the Quotient

To find the quotient \( \frac{z_1}{z_2} \), divide the moduli and subtract the arguments:\[\frac{r_1}{r_2} = \frac{3}{5} \]\[\theta_1 - \theta_2 = \frac{\pi}{6} - \frac{4\pi}{3} = \frac{\pi}{6} - \frac{8\pi}{6} = -\frac{7\pi}{6}\]Thus, \( \frac{z_1}{z_2} = \frac{3}{5} \left(\cos \left(-\frac{7\pi}{6}\right) + i \sin \left(-\frac{7\pi}{6}\right)\right) \). Note: This angle can be adjusted to positive angle by adding \(2\pi\), resulting in \(\cos \left(\frac{5\pi}{6}\right) + i \sin \left(\frac{5\pi}{6}\right)\).

Key Concepts

Modulus and Argument

The modulus and argument are key components when discussing complex numbers in their polar form. A complex number, in polar form, is represented as \( z = r(\cos \theta + i \sin \theta) \), where \( r \) is the modulus and \( \theta \) is the argument.
The modulus \( r \) of a complex number is the distance from the origin to the point represented by the complex number in the Argand plane. This tells us how 'large' or 'far out' the number is.
For the given numbers, the moduli are \( r_1 = 3 \) and \( r_2 = 5 \). This means that each point is 3 units and 5 units away from the origin respectively.

The argument \( \theta \) is the angle made with the positive x-axis. It tells you the direction of the complex number in the plane. For \( z_1 \), the argument is \( \frac{\pi}{6} \), while for \( z_2 \), it is \( \frac{4\pi}{3} \). These angles position the complex numbers relative to the horizontal axis.

Multiplication of Complex Numbers

When multiplying complex numbers in polar form, the process is quite straightforward compared to the rectangular form. The key steps involve multiplying their moduli and adding their arguments.

• To find the product of two complex numbers \( z_1 = r_1 (\cos \theta_1 + i \sin \theta_1) \) and \( z_2 = r_2 (\cos \theta_2 + i \sin \theta_2) \), start by multiplying the moduli: \( r_1 r_2 \).
• Next, add the arguments: \( \theta_1 + \theta_2 \).

Applying these steps, with \( r_1 = 3 \) and \( r_2 = 5 \), their product is \( 15 \). Adding the arguments \( \frac{\pi}{6} + \frac{4\pi}{3} = \frac{3\pi}{2} \), the result ends up being \( z_1 z_2 = 15 \left( \cos \frac{3\pi}{2} + i \sin \frac{3\pi}{2} \right) \).
This reflects a new complex number located at three-quarters of a complete circle, or 270 degrees.

Division of Complex Numbers

Dividing complex numbers in polar form involves dividing their moduli and subtracting their arguments. This is another instance where polar form simplifies complex calculations.

• First, divide the moduli: \( \frac{r_1}{r_2} \).
• Then, subtract the arguments: \( \theta_1 - \theta_2 \).

For the given problem, the division of moduli \( \frac{3}{5} \) provides the modulus of the quotient. Subtracting the arguments results in \( \frac{\pi}{6} - \frac{4\pi}{3} = -\frac{7\pi}{6} \).
Since angles are typically expressed in terms of non-negative measures, we can add \( 2\pi \) to bring it to \( \frac{5\pi}{6} \), without changing its orientation. Therefore, the quotient becomes \( \frac{3}{5} \left( \cos \frac{5\pi}{6} + i \sin \frac{5\pi}{6} \right) \).
This results in a new complex number that sits nearly opposite in direction to the initial reference along the circle's diameter.