Problem 49 - Algebra and Trigonometry

Question

In Exercises \(45-52,\) find the quotient \(\frac{z_{1}}{z_{2}}\) of the complex numbers. Leave answers in polar form. In Exercises \(49-50,\) express the argument as an angle between \(0^{\circ}\) and \(360^{\circ}\). $$ \begin{array}{l} {z_{1}=\cos 80^{\circ}+i \sin 80^{\circ}} \\ {z_{2}=\cos 200^{\circ}+i \sin 200^{\circ}} \end{array} $$

Step-by-Step Solution

Answer

\( \frac{z_{1}}{z_{2}} = \cos 240^{\circ} + i \sin 240^{\circ} \)

Step 1: Express the complex numbers in polar form

In polar form, complex numbers are expressed as \( r(\cos\theta + i \sin\theta) \), where \( r \) is the magnitude and \( \theta \) is the argument. Since the magnitude is 1 for both \( z_{1} \) and \( z_{2} \), they can be simplified as follows: \( z_{1} = \cos 80^{\circ} + i \sin 80^{\circ} = e^{i80^{\circ}} \) and \( z_{2} = \cos 200^{\circ} + i \sin 200^{\circ} = e^{i200^{\circ}} \)

Step 2: Calculate the quotient

The quotient \( \frac{z_{1}}{z_{2}} \) can be calculated using the property: \( \frac{e^{ia}}{e^{ib}} = e^{i(a-b)} \). So, \( \frac{z_{1}}{z_{2}} = e^{i(80^{\circ}-200^{\circ})} = e^{-i120^{\circ}} \)

Step 3: Express the argument as an angle

To get the argument to lie in the range \([0^{\circ},360^{\circ}]\), we can add \(360^{\circ}\) to the argument in step 2, which gives us \( e^{i240^{\circ}} = \cos 240^{\circ} + i \sin 240^{\circ} \).

Key Concepts

Polar Form

Complex numbers can be represented in two forms: the rectangular form and the polar form. The polar form emphasizes the angle and magnitude of a complex number and is written as \( r(\cos\theta + i \sin\theta) \). In this form, \( r \) is the magnitude (distance from the origin in the complex plane) and \( \theta \) is the argument (angle from the positive real axis).

For example, in the exercise, we started with \( z_{1} = \cos 80^{\circ} + i \sin 80^{\circ} \) and \( z_{2} = \cos 200^{\circ} + i \sin 200^{\circ} \). These are both already in polar form with a magnitude of 1.
By simplifying, we can also express them using the exponential function: \( z_{1} = e^{i80^{\circ}} \) and \( z_{2} = e^{i200^{\circ}} \). This form is particularly useful for calculations involving multiplication and division.

Argument of a Complex Number

The argument of a complex number is the angle from the positive real axis to the line representing the complex number in the complex plane. It is denoted as \( \theta \) and is a crucial part of the polar form.

In degrees, arguments are typically expressed as angles between \( 0^{\circ} \) and \( 360^{\circ} \). In the given problem, the arguments for \( z_1 \) and \( z_2 \) were \( 80^{\circ} \) and \( 200^{\circ} \) respectively.

• For \( z_1 \), the argument \( \theta \) is \( 80^{\circ} \).
• For \( z_2 \), the argument \( \theta \) is \( 200^{\circ} \).

Understanding these angles helps in performing operations on complex numbers like finding their quotient.

Complex Number Division

Dividing complex numbers can be simplified using the polar form. When we divide one complex number by another, the magnitudes are divided, and the arguments are subtracted. This process beautifully showcases why the polar or exponential form is very effective.

Using the property \( \frac{e^{i\alpha}}{e^{i\beta}} = e^{i(\alpha - \beta)} \), we find the quotient by subtracting the argument\( s \). For the problem given:

• Calculate \( 80^{\circ} - 200^{\circ} = -120^{\circ} \).
• We express this solution as \( e^{-i120^{\circ}} \).
• Since \( -120^{\circ} \) is not in the standard range, add \( 360^{\circ} \) to get \( 240^{\circ} \).

Thus the quotient \( \frac{z_1}{z_2} \) in polar form is \( e^{i240^{\circ}} \), or \( \cos 240^{\circ} + i \sin 240^{\circ} \). This method is efficient and avoids the need for complex arithmetic in rectangular form.