Problem 57 - 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. $$z_{1}=4\left(\cos 120^{\circ}+i \sin 120^{\circ}\right)$$ $$z_{2}=2\left(\cos 30^{\circ}+i \sin 30^{\circ}\right)$$

Step-by-Step Solution

Answer

\(z_{1}z_{2} = 8(\cos 150^{\circ} + i \sin 150^{\circ})\) and \(\frac{z_{1}}{z_{2}} = 2(\cos 90^{\circ} + i \sin 90^{\circ})\).

Step 1: Identify the Polar Forms

Given complex numbers are already in polar form: \(z_{1}=4\left(\cos 120^{\circ}+i \sin 120^{\circ}\right)\)and \(z_{2}=2\left(\cos 30^{\circ}+i \sin 30^{\circ}\right)\). The modulus and the argument for each complex number are: - For \(z_1\), modulus is 4 and argument is \(120^{\circ}\).- For \(z_2\), modulus is 2 and argument is \(30^{\circ}\).

Step 2: Product of Two Complex Numbers in Polar Form

To find the product \(z_{1}z_{2}\), multiply their moduli and add their arguments. \( |z_{1}z_{2}| = |z_{1}||z_{2}| = 4 \times 2 = 8 \)\( \text{Argument of } z_{1}z_{2} = 120^{\circ} + 30^{\circ} = 150^{\circ} \)Thus, \( z_{1}z_{2} = 8 \left(\cos 150^{\circ} + i \sin 150^{\circ}\right) \)

Step 3: Quotient of Two Complex Numbers in Polar Form

To find the quotient \(\frac{z_{1}}{z_{2}}\), divide their moduli and subtract their arguments. \( \left| \frac{z_{1}}{z_{2}} \right| = \frac{|z_{1}|}{|z_{2}|} = \frac{4}{2} = 2 \)\( \text{Argument of } \frac{z_{1}}{z_{2}} = 120^{\circ} - 30^{\circ} = 90^{\circ} \)Thus, \( \frac{z_{1}}{z_{2}} = 2 \left(\cos 90^{\circ} + i \sin 90^{\circ}\right) \)

Key Concepts

Understanding Complex Numbers

Complex numbers are fascinating mathematical entities that have a real and an imaginary part. They are usually expressed as \( a + bi \), where \( a \) is the real part and \( b \) is the imaginary part with the imaginary unit \( i \) satisfying \( i^2 = -1 \). However, when dealing with them in a rotational context such as physics or engineering, the polar form is incredibly useful.

The polar form of a complex number expresses the number in terms of its modulus (absolute value) and argument (angle). It is represented as \( r(\cos \theta + i \sin \theta) \) or \( r\text{cis}\theta \), where \( r \) is the modulus and \( \theta \) is the argument. Converting to polar form involves calculating the modulus as \( \sqrt{a^2 + b^2} \) and the argument \( \theta \), which is the angle formed with the positive real axis, using \( \tan^{-1}(b/a) \).

Polar form is especially advantageous when multiplying or dividing complex numbers, as operations become a mere manipulation of modulus and argument rather than dealing with individual real and imaginary parts.

Multiplying Complex Numbers in Polar Form

Multiplying complex numbers in their polar form is much simpler than when they are in rectangular form. Let's call the two complex numbers \( z_1 \) and \( z_2 \), expressed in polar form as \( r_1\text{cis}\theta_1 \) and \( r_2\text{cis}\theta_2 \), respectively.

The product \( z_1z_2 \) is found by multiplying their moduli and adding their arguments. This is more straightforward:

• Multiply the moduli: \( |z_1z_2| = |z_1||z_2| \)
• Add the arguments: \( \theta_1 + \theta_2 \)

So, the product is given by \( |z_1||z_2|\text{cis}(\theta_1 + \theta_2) \).

In the exercise, given complex numbers \( z_1 = 4\text{cis}120^{\circ} \) and \( z_2 = 2\text{cis}30^{\circ} \), the modulus of the product is \( 8 \) and the argument is \( 150^{\circ} \), resulting in \( z_1z_2 = 8(\cos 150^{\circ} + i \sin 150^{\circ}) \). The ease of this multiplication stems from polar form, showcasing an elegant solution to potentially cumbersome arithmetic.

Dividing Complex Numbers in Polar Form

When dividing complex numbers in polar form, the simplicity is akin to multiplication. Taking two complex numbers \( z_1 \) and \( z_2 \) in the forms \( r_1\text{cis}\theta_1 \) and \( r_2\text{cis}\theta_2 \), the quotient \( \frac{z_1}{z_2} \) is achieved by dividing their moduli and subtracting their arguments.

The steps are:

• Divide the moduli: \( \left| \frac{z_1}{z_2} \right| = \frac{|z_1|}{|z_2|} \)
• Subtract the arguments: \( \theta_1 - \theta_2 \)

Thus, \( \frac{z_1}{z_2} \) is expressed as \( \frac{|z_1|}{|z_2|}\text{cis}(\theta_1 - \theta_2) \).

In the given exercise, \( z_1 = 4\text{cis}120^{\circ} \) and \( z_2 = 2\text{cis}30^{\circ} \), the modulus of the quotient is \( 2 \) and the argument is \( 90^{\circ} \). Therefore, \( \frac{z_1}{z_2} = 2(\cos 90^{\circ} + i \sin 90^{\circ}) \). This method highlights how polar form conveniently simplifies otherwise complex operations.