Problem 63 - Algebra and Trigonometry

Question

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\(/ z_{1}\) . $$ z_{1}=2 \sqrt{3}-2 i, \quad z_{2}=-1+i $$

Step-by-Step Solution

Answer

The polar forms are \(z_1 = 4\,\text{cis}(-\pi/6)\) and \(z_2 = \sqrt{2}\,\text{cis}(3\pi/4)\). The product is \(4\sqrt{2}\,\text{cis}(\pi/12)\), the quotient is \(2\sqrt{2}\,\text{cis}(-11\pi/12)\), and the reciprocal is \(\frac{1}{4}\,\text{cis}(\pi/6)\).

Step 1: Convert Rectangular to Polar Form for \(z_1\)

To convert \(z_1 = 2\sqrt{3} - 2i\) to polar form, calculate the magnitude: \[ |z_1| = \sqrt{(2\sqrt{3})^2 + (-2)^2} = \sqrt{12 + 4} = \sqrt{16} = 4. \]Find the argument (angle \(\theta\)) using \( \tan^{-1}\left(\frac{-2}{2\sqrt{3}}\right) = \tan^{-1}\left(-\frac{1}{\sqrt{3}}\right)\). This corresponds to \(-\pi/6\) in the fourth quadrant. Therefore, the polar form is:\[ z_1 = 4\left(\cos\left(-\frac{\pi}{6}\right) + i\sin\left(-\frac{\pi}{6}\right)\right). \]

Step 2: Convert Rectangular to Polar Form for \(z_2\)

To convert \(z_2 = -1 + i\) to polar form, calculate the magnitude: \[ |z_2| = \sqrt{(-1)^2 + 1^2} = \sqrt{2}. \]Find the argument (angle \(\theta\)) using \( \tan^{-1}\left(\frac{1}{-1}\right) = \tan^{-1}(-1)\). This corresponds to \(\frac{3\pi}{4}\) since \(z_2\) is in the second quadrant. Therefore, the polar form is:\[ z_2 = \sqrt{2}\left(\cos\left(\frac{3\pi}{4}\right) + i\sin\left(\frac{3\pi}{4}\right)\right). \]

Step 3: Find Product \(z_1z_2\) in Polar Form

Use the formula \(z_1z_2 = |z_1||z_2|\left(\cos(\theta_1 + \theta_2) + i\sin(\theta_1 + \theta_2)\right)\).Calculate:\[ |z_1||z_2| = 4\cdot \sqrt{2} = 4\sqrt{2}, \]\[ \theta_1 + \theta_2 = -\frac{\pi}{6} + \frac{3\pi}{4} = \frac{\pi}{12}. \]Thus, the product is:\[ z_1z_2 = 4\sqrt{2}(\cos(\frac{\pi}{12}) + i\sin(\frac{\pi}{12})). \]

Step 4: Find Quotient \(\frac{z_1}{z_2}\) in Polar Form

Use the formula \(\frac{z_1}{z_2} = \frac{|z_1|}{|z_2|}\left(\cos(\theta_1 - \theta_2) + i\sin(\theta_1 - \theta_2)\right)\).Calculate:\[ \frac{|z_1|}{|z_2|} = \frac{4}{\sqrt{2}} = 2\sqrt{2}, \]\[ \theta_1 - \theta_2 = -\frac{\pi}{6} - \frac{3\pi}{4} = -\frac{11\pi}{12}. \]Thus, the quotient is:\[ \frac{z_1}{z_2} = 2\sqrt{2}(\cos(-\frac{11\pi}{12}) + i\sin(-\frac{11\pi}{12})). \]

Step 5: Find Quotient \(\frac{1}{z_1}\) in Polar Form

Use the formula \(\frac{1}{z_1} = \frac{1}{|z_1|}\left(\cos(-\theta_1) + i\sin(-\theta_1)\right)\).Calculate:\[ \frac{1}{|z_1|} = \frac{1}{4}, \]\[ -\theta_1 = \frac{\pi}{6}. \]Thus, the reciprocal is:\[ \frac{1}{z_1} = \frac{1}{4}(\cos(\frac{\pi}{6}) + i\sin(\frac{\pi}{6})). \]

Key Concepts

Rectangular to Polar Conversion

When dealing with complex numbers, they can be represented in two primary forms: rectangular and polar. The rectangular form of a complex number is noted as \( a + bi \), where \( a \) is the real part and \( b \) is the imaginary part. To convert this into its polar form, we need to determine two key elements: the magnitude and the angle (argument) of the complex number.

First, the magnitude is calculated using the formula: \(|z| = \sqrt{a^2 + b^2}\). This represents the length of the vector from the origin to the point \((a, b)\) in the complex plane. Next, to find the angle \(\theta\), we use the formula \(\theta = \tan^{-1}\left(\frac{b}{a}\right)\). This angle indicates the direction of the vector. However, note that the angle’s exact value might need to be adjusted based on which quadrant \((a, b)\) is located in the complex plane.

The polar form is expressed as \( |z| (\cos(\theta) + i \sin(\theta)) \). This format gives a compact representation using just two parts: magnitude and angle.

Complex Number Multiplication

Multiplying complex numbers in polar form is elegantly straightforward. It's all about the scales and rotation of the numbers. Given two complex numbers \( z_1 = r_1 (\cos(\theta_1) + i\sin(\theta_1)) \) and \( z_2 = r_2 (\cos(\theta_2) + i\sin(\theta_2)) \), their product is found by just multiplying their magnitudes and adding their angles.

The product’s magnitude is simply obtained by \( r_1 \times r_2 \). As per the angles, they add up: \( \theta_1 + \theta_2 \). Therefore, the product \( z_1 z_2 \) in polar form becomes \( r_1r_2 (\cos(\theta_1 + \theta_2) + i\sin(\theta_1 + \theta_2)) \).

When viewed geometrically, this kind of multiplication corresponds to scaling the vector’s length by the product of magnitudes and rotating it by the sum of angles. This makes polar multiplication particularly intuitive yet effective when working with complex numbers.

Complex Number Division

Dividing complex numbers in polar form is much like reversing the multiplication process. For any two complex numbers \( z_1 = r_1(\cos(\theta_1) + i\sin(\theta_1)) \) and \( z_2 = r_2(\cos(\theta_2) + i\sin(\theta_2)) \), their quotient is calculated by dividing their magnitudes and subtracting their angles.

The resulting magnitude is \( \frac{r_1}{r_2} \), and the angle becomes \( \theta_1 - \theta_2 \). Thus, the quotient \( \frac{z_1}{z_2} \) can be expressed as \( \frac{r_1}{r_2} (\cos(\theta_1 - \theta_2) + i\sin(\theta_1 - \theta_2)) \).

This operation is like shrinking or expanding the length of the vector based on the division of magnitudes, and the directional change is determined by the difference between the angles. It's an efficient way of simplifying division, making the operation clear and concise in polar form.

Complex Number Reciprocal

Finding the reciprocal of a complex number is a crucial operation, especially for solving equations involving division by complex numbers. In polar form, the reciprocal of a complex number \( z = r (\cos(\theta) + i\sin(\theta)) \) is determined by inversely scaling the magnitude and reflecting the angle.

The reciprocal’s magnitude is simply \( \frac{1}{r} \), indicating the inverse effect on the vector’s length. Meanwhile, the angle is negated, resulting in \( -\theta \). This gives the reciprocal \( \frac{1}{z} = \frac{1}{r} (\cos(-\theta) + i\sin(-\theta)) \).

Conceptually, this transformation modifies the vector to point in an opposite direction, while adjusting the scale based on the reciprocal of the original magnitude. Handling reciprocals in polar form thus becomes an intuitive and straightforward task.