Problem 60 - 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. $$z_{1}=\frac{4}{5}\left(\cos 25^{\circ}+i \sin 25^{\circ}\right)$$ $$z_{2}=\frac{1}{3}\left(\cos 155^{\circ}+i \sin 155^{\circ}\right)$$

Step-by-Step Solution

Answer

The product is \(-\frac{4}{15}\), and the quotient is \(\frac{12}{5}(\cos 230^{\circ} + i \sin 230^{\circ})\).

Step 1: Understanding Polar Form of Complex Numbers

Each complex number is given in the form of \( r (\cos \theta + i \sin \theta) \). This is the polar form of a complex number, where \( r \) is the modulus and \( \theta \) is the argument.

Step 2: Calculate the Product of Two Complex Numbers

To find the product \( z_1 z_2 \) of two complex numbers in polar form, multiply their moduli and add their arguments: \[ z_1 z_2 = \left( \frac{4}{5} \times \frac{1}{3} \right) \left( \cos(25^{\circ} + 155^{\circ}) + i \sin(25^{\circ} + 155^{\circ}) \right) \]Calculate \( \frac{4}{5} \times \frac{1}{3} = \frac{4}{15} \) and \( 25^{\circ} + 155^{\circ} = 180^{\circ} \). So, \( z_1 z_2 = \frac{4}{15}(\cos 180^{\circ} + i \sin 180^{\circ}) \).

Step 3: Express Product in Polar Form

From the calculations, we translate \( \cos 180^{\circ} = -1 \) and \( \sin 180^{\circ} = 0 \). So, the expression becomes: \[ z_1 z_2 = \frac{4}{15}(-1 + 0i) \]Therefore, the product \( z_1 z_2 \) in polar form is \( -\frac{4}{15} + i0 \).

Step 4: Calculate the Quotient of Two Complex Numbers

To find the quotient \( \frac{z_1}{z_2} \) of two complex numbers, divide their moduli and subtract their arguments: \[ \frac{z_1}{z_2} = \left( \frac{\frac{4}{5}}{\frac{1}{3}} \right) \left( \cos(25^{\circ} - 155^{\circ}) + i \sin(25^{\circ} - 155^{\circ}) \right) \]Calculate \( \frac{4}{5} \div \frac{1}{3} = \frac{12}{5} \) and \( 25^{\circ} - 155^{\circ} = -130^{\circ} \).

Step 5: Express Quotient in Polar Form

Since angles in polar form must be between \(0^{\circ}\) and \(360^{\circ}\), convert \(-130^{\circ}\) to a positive angle by adding \(360^{\circ}\). Thus, \(-130^{\circ} + 360^{\circ} = 230^{\circ}\). Therefore, \( \frac{z_1}{z_2} = \frac{12}{5}(\cos 230^{\circ} + i \sin 230^{\circ}) \).

Key Concepts

Product of Complex Numbers

In mathematics, the **product of complex numbers in polar form** is straightforward when you understand the core principles. Imagine two complex numbers represented by their magnitude (or modulus) and direction (or argument) in the plane. In polar form, these numbers are given as follows:

• Complex number 1: \( z_1 = r_1 (\cos \theta_1 + i \sin \theta_1) \)
• Complex number 2: \( z_2 = r_2 (\cos \theta_2 + i \sin \theta_2) \)

To find their product \( z_1 z_2 \), multiply the moduli \( r_1 \times r_2 \) and add the arguments \( \theta_1 + \theta_2 \). This is due to the famous trigonometric identities and simplifications in complex numbers. Applying this to our values, \( r_1 = \frac{4}{5} \) and \( r_2 = \frac{1}{3} \), so their product is \( \frac{4}{5} \times \frac{1}{3} = \frac{4}{15} \). The arguments \( 25^\circ + 155^\circ \) sum up to \( 180^\circ \).

Thus, the product in polar form becomes \( \frac{4}{15} \left( \cos 180^\circ + i \sin 180^\circ \right) \). Remember that a key feature of angles on the unit circle is consistency, as angle measurements repeat every \( 360^\circ \). At \( 180^\circ \), \( \cos(180^\circ) = -1 \) and \( \sin(180^\circ) = 0 \), simplifying the outcome to \( -\frac{4}{15} + i0 \).

Quotient of Complex Numbers

Dividing complex numbers in polar form might seem more daunting than multiplication, but it follows a similar, accessible pathway. When faced with the task of division, focus on these steps:
To express the division \( \frac{z_1}{z_2} \), divide their moduli, \( \frac{r_1}{r_2} \), and subtract the arguments, \( \theta_1 - \theta_2 \). This process transforms the **quotient of complex numbers in polar form**. In our case, the moduli division becomes \( \frac{\left(\frac{4}{5}\right)}{\left(\frac{1}{3}\right)} = \frac{12}{5} \). The difference in angles, \( 25^\circ - 155^\circ \), results in \( -130^\circ \).
It is crucial to ensure the angle remains within a standard range for polar representation, typically between \( 0^\circ \) and \( 360^\circ \). Hence, adjust \( -130^\circ \) by adding \( 360^\circ \), giving us the positive angle \( 230^\circ \). The result is \( \frac{12}{5} \left( \cos 230^\circ + i \sin 230^\circ \right) \). This reflects how division translates into geometric interpretation: decreasing spread and contrasting angle difference of the vectors on the plane.

Modulus and Argument in Polar Form

Understanding the **modulus and argument in polar form** unlocks the door to mastering complex numbers. Each complex number can be viewed as a point on the coordinate plane, but when seen as \( r \) (distance) and \( \theta \) (direction), its behavior becomes intriguing and meaningful.

• The modulus \( r \) symbolizes the "length" of the complex number from the origin. It indicates how far out from the center the number is located on the plane.
• The argument \( \theta \) represents the angle from the positive real axis to the line connecting the origin and the complex number, measured in degrees or radians.

This polar representation makes calculations such as multiplication and division easy, as it focuses on altering magnitude through simple multiplication or division, while direction is managed by adding or subtracting angles. Polar form not only simplifies operations but also provides insight into the rotational and scaling effects of complex numbers.
In our exercise, \( z_1 \) and \( z_2 \) are given with both a modulus (\( \frac{4}{5} \) and \( \frac{1}{3} \), respectively) and an argument (\( 25^\circ \) and \( 155^\circ \), respectively). These parameters define each complex number's position and orientation in space, essential for performing operations and understanding their outcomes effectively.