Problem 60 - Algebra and Trigonometry

Question

\(57-64=\) Write \(z_{1}\) and \(z_{2}\) in polar form, and then find the product \(z_{1} z_{2}\) and the quotients \(z_{1} / z_{2}\) and 1\(/ z_{1}\) . $$ z_{1}=-\sqrt{2} i, \quad z_{2}=-3-3 \sqrt{3} i $$

Step-by-Step Solution

Answer

The product is \( z_1 z_2 = 6\sqrt{2} \text{cis}(\frac{5\pi}{6}) \), the quotient is \( \frac{z_1}{z_2} = \frac{\sqrt{2}}{6} \text{cis}(-\frac{11\pi}{6}) \), and the inverse is \( \frac{1}{z_1} = \frac{1}{\sqrt{2}} \text{cis}(\frac{\pi}{2}) \).

Step 1: Convert \( z_1 \) to polar form

The complex number \( z_1 = -\sqrt{2} i \) is purely imaginary. Its modulus is the absolute value of the imaginary part: \( |z_1| = \sqrt{2} \). As \( z_1 \) lies on the negative imaginary axis, its argument is \( \text{arg}(z_1) = -\frac{\pi}{2} \). Hence, the polar form is: \( z_1 = \sqrt{2} \text{cis}(-\pi/2) \).

Step 2: Convert \( z_2 \) to polar form

For \( z_2 = -3 - 3\sqrt{3}i \), calculate the modulus: \( |z_2| = \sqrt{(-3)^2 + (-3\sqrt{3})^2} = \sqrt{9 + 27} = 6 \). To find the argument, use \( \tan^{-1} \left( \frac{-3\sqrt{3}}{-3} \right) = \tan^{-1}(\sqrt{3}) = \frac{4\pi}{3} \). Thus, the polar form is: \( z_2 = 6 \text{cis}(\frac{4\pi}{3}) \).

Step 3: Calculate the product \( z_1 z_2 \)

The product of two complex numbers in polar form is the product of their moduli and the sum of their arguments: \( |z_1 z_2| = |z_1||z_2| = \sqrt{2} \times 6 = 6\sqrt{2} \) and \( \text{arg}(z_1 z_2) = -\frac{\pi}{2} + \frac{4\pi}{3} = \frac{5\pi}{6} \). Therefore, \( z_1 z_2 = 6\sqrt{2}\, \text{cis}\left( \frac{5\pi}{6} \right) \).

Step 4: Calculate the quotient \( \frac{z_1}{z_2} \)

Divide the moduli and subtract the arguments for the quotient: \( \left| \frac{z_1}{z_2} \right| = \frac{\sqrt{2}}{6} \) and \( \text{arg} \left( \frac{z_1}{z_2} \right) = -\frac{\pi}{2} - \frac{4\pi}{3} = -\frac{11\pi}{6} \). Thus, \( \frac{z_1}{z_2} = \frac{\sqrt{2}}{6} \text{cis}\left( -\frac{11\pi}{6} \right) \).

Step 5: Calculate the inverse \( \frac{1}{z_1} \)

To find the inverse of \( z_1 \), invert the modulus and negate the argument: \( \left| \frac{1}{z_1} \right| = \frac{1}{\sqrt{2}} \) and \( \text{arg} \left( \frac{1}{z_1} \right) = \frac{\pi}{2} \). Therefore, \( \frac{1}{z_1} = \frac{1}{\sqrt{2}} \text{cis}\left( \frac{\pi}{2} \right) \).

Key Concepts

Polar Form

Complex numbers can be expressed in several ways, and one of the most useful forms is the polar form. This representation utilizes an angle and a distance to express the complex number on the complex plane. The polar form of a complex number is given by:

• Modulus (Distance from origin)
• Argument (Angle from the positive x-axis)

The general formula for polar form is \( r \text{cis} \theta \), where \( r \) is the modulus and \( \theta \) is the argument.
The term \( \text{cis} \theta \) is a shorthand for \( \cos \theta + i \sin \theta \). This form not only makes calculations simpler but also provides geometric insight into complex numbers.

Modulus

The modulus of a complex number is akin to the concept of magnitude or length in geometry. It indicates how far the complex number is from the origin on the complex plane. For a complex number \( z = a + bi \), the modulus \( |z| \) is calculated using the formula \( \sqrt{a^2 + b^2} \).
This measurement is always a non-negative value. For example, for \( z_2 = -3 - 3\sqrt{3}i \), the modulus is calculated as:

• \( |z_2| = \sqrt{(-3)^2 + (-3\sqrt{3})^2} = \sqrt{9 + 27} = 6 \)

The modulus tells us that \( z_2 \) is six units away from the origin.

Argument

In the polar form of complex numbers, the argument refers to the angle formed by the complex number with the positive real axis (x-axis) on the complex plane. This angle, \( \theta \), can be found by applying the arctangent function: \( \theta = \tan^{-1}\left(\frac{b}{a}\right) \). It's important to note the quadrant in which your complex number lies, as this affects the sign and value of \( \theta \).
The argument can be positive or negative and is often expressed in radians. For example, the argument of \( z_2 = -3 - 3\sqrt{3}i \) is \( \frac{4\pi}{3} \), indicating the angle measured counterclockwise from the positive real axis.

Quotient of Complex Numbers

When dividing complex numbers, using their polar form simplifies the process greatly. The quotient involves dividing the moduli and subtracting the arguments. If you have two complex numbers \( z_1 = r_1 \text{cis} \theta_1 \) and \( z_2 = r_2 \text{cis} \theta_2 \), the division is:

• \( \left| \frac{z_1}{z_2} \right| = \frac{r_1}{r_2} \)
• \( \text{arg} \left( \frac{z_1}{z_2} \right) = \theta_1 - \theta_2 \)

For example, dividing \( z_1 \) by \( z_2 \) results in a modulus of \( \frac{\sqrt{2}}{6} \) and an argument of \( -\frac{11\pi}{6} \). This approach makes handling division much more systematic and geometric.

Product of Complex Numbers

The product of two complex numbers in polar form involves multiplying their moduli and adding their arguments. Given two numbers in polar form, \( z_1 = r_1 \text{cis} \theta_1 \) and \( z_2 = r_2 \text{cis} \theta_2 \), their product is:

• Modulus: \( |z_1 z_2| = r_1 \times r_2 \)
• Argument: \( \text{arg}(z_1 z_2) = \theta_1 + \theta_2 \)

This rule simplifies multiplying complex numbers significantly since you can simply work with their modulus and arguments instead of their rectangular components. As an example, the product of \( z_1 \) and \( z_2 \) is \( 6\sqrt{2} \text{cis} \frac{5\pi}{6} \), revealing both the resulting magnitude and direction.