Problem 20 - Precalculus

Question

Find the quotient \(\frac{z_{1}}{z_{2}}\) and express it in rectangular form. $$z_{1}=22\left[\cos \left(\frac{11 \pi}{18}\right)+i \sin \left(\frac{11 \pi}{18}\right)\right] \text { and } z_{2}=11\left[\cos \left(\frac{5 \pi}{18}\right)+i \sin \left(\frac{5 \pi}{18}\right)\right]$$

Step-by-Step Solution

Answer

The quotient is \(1 + i\sqrt{3}\) in rectangular form.

Step 1: Write given complex numbers in polar form

The complex numbers are given in the form \( z = r(\cos(\theta) + i \sin(\theta)) \). Here, \( z_1 = 22(\cos(\frac{11\pi}{18}) + i \sin(\frac{11\pi}{18})) \) and \( z_2 = 11(\cos(\frac{5\pi}{18}) + i \sin(\frac{5\pi}{18})) \).

Step 2: Apply the division rule for complex numbers in polar form

The quotient of two complex numbers in polar form \( \frac{z_1}{z_2} = \frac{r_1}{r_2}(\cos(\theta_1 - \theta_2) + i \sin(\theta_1 - \theta_2)) \). Substitute \( r_1 = 22 \), \( r_2 = 11 \), \( \theta_1 = \frac{11\pi}{18} \), and \( \theta_2 = \frac{5\pi}{18} \).

Step 3: Calculate the magnitude of the quotient

Calculate \( \frac{r_1}{r_2} = \frac{22}{11} = 2 \).

Step 4: Calculate the angle of the quotient

Calculate the angle difference \( \theta_1 - \theta_2 = \frac{11\pi}{18} - \frac{5\pi}{18} = \frac{6\pi}{18} = \frac{\pi}{3} \).

Step 5: Express the quotient in polar form

Substitute the calculated magnitude and angle into the polar form formula: \( z = 2( \cos(\frac{\pi}{3}) + i \sin(\frac{\pi}{3})) \).

Step 6: Convert polar form to rectangular form

Use the exact values for \( \cos(\frac{\pi}{3}) = \frac{1}{2} \) and \( \sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2} \) to get \( z = 2( \frac{1}{2} + i \frac{\sqrt{3}}{2} ) = 1 + i \sqrt{3} \).

Key Concepts

Polar form

Complex numbers can be represented in polar form as \( z = r(\cos(\theta) + i \sin(\theta)) \). This format might seem complex at first but when broken down, is actually quite simple.
The polar form expresses a complex number using two parameters:

• Magnitude \( r \): the distance of the complex number from the origin.
• Angle \( \theta \): the direction from the positive x-axis, measured in radians.

In this representation:- \( r \) is the modulus, calculated as \( \sqrt{x^2 + y^2} \) for a complex number \( z = x + yi \).- \( \theta \) is the argument, derived from \( \tan^{-1}(\frac{y}{x}) \).
For our given complex numbers, \( z_1 = 22(\cos(\frac{11\pi}{18}) + i \sin(\frac{11\pi}{18})) \) and \( z_2 = 11(\cos(\frac{5\pi}{18}) + i \sin(\frac{5\pi}{18})) \), both are conveniently presented in polar form which simplifies the operation tasks we will perform on them.

Rectangular form

Now, let's focus on transforming complex numbers from polar to rectangular form. The rectangular or Cartesian form is expressed as \( z = x + yi \), where:

• \( x \) is the real part
• \( y \) is the imaginary part

The conversion from polar to rectangular relies on basic trigonometry:\[ x = r \cos(\theta)\]\[ y = r \sin(\theta)\]For example, we found \( z = 2( \cos(\frac{\pi}{3}) + i \sin(\frac{\pi}{3})) \). Using the set angles \( \cos(\frac{\pi}{3}) = \frac{1}{2} \) and \( \sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2} \), we can easily translate the polar form to:\[ z = 2 \left( \frac{1}{2} + i \frac{\sqrt{3}}{2} \right) = 1 + i\sqrt{3}\]Expressing the number in rectangular form helps in performing arithmetic operations like addition and subtraction.

Complex number division

Dividing complex numbers might seem daunting, but it's straightforward with the polar form. The division of two complex numbers in polar form is achieved by dividing their magnitudes and subtracting their angles:
\[ \frac{z_1}{z_2} = \frac{r_1}{r_2} \left( \cos(\theta_1 - \theta_2) + i \sin(\theta_1 - \theta_2) \right)\]Here's the breakdown:

• Divide the magnitudes: \( \frac{r_1}{r_2} \). Example: \( \frac{22}{11} = 2 \).
• Subtract the angles: \( \theta_1 - \theta_2 \). Example: \( \frac{11\pi}{18} - \frac{5\pi}{18} = \frac{6\pi}{18} = \frac{\pi}{3} \).

This yields the result in polar form, making it simple to convert back to rectangular form if needed. The end product here becomes easy to apply in various mathematical problems or when performing complex analysis.
This method simplifies calculations as there's no direct need to handle the expansions and conjugate multiplications you'd face when using rectangular form directly.