Problem 58 - Algebra and Trigonometry

Question

Find the product \(z_{1} z_{2}\) and the quotient \(z_{1} / z_{2}\) . Express your answer in polar form. $$ \begin{array}{l}{z_{1}=\sqrt{2}\left(\cos 75^{\circ}+i \sin 75^{\circ}\right)} \\\ {z_{2}=3 \sqrt{2}\left(\cos 60^{\circ}+i \sin 60^{\circ}\right)}\end{array} $$

Step-by-Step Solution

Answer

Product: \(6(\cos 135^\circ + i \sin 135^\circ)\), Quotient: \(\frac{1}{3}(\cos 15^\circ + i \sin 15^\circ)\)."

Step 1: Convert to Polar Form

Given the complex numbers in the form \(z = r(\cos \theta + i \sin \theta)\), they are already in polar form.\For \(z_1 = \sqrt{2}(\cos 75^\circ + i \sin 75^\circ)\), we have \(r_1 = \sqrt{2}\) and \(\theta_1 = 75^\circ\).\For \(z_2 = 3\sqrt{2}(\cos 60^\circ + i \sin 60^\circ)\), we have \(r_2 = 3\sqrt{2}\) and \(\theta_2 = 60^\circ\).

Step 2: Find the Product

To find the product \(z_1 z_2\), multiply the moduli and add the angles: \[|z_1 z_2| = |z_1||z_2| = \sqrt{2} \times 3\sqrt{2} = 6\]\[\arg(z_1 z_2) = \theta_1 + \theta_2 = 75^\circ + 60^\circ = 135^\circ\]Thus, \(z_1 z_2 = 6(\cos 135^\circ + i \sin 135^\circ)\).

Step 3: Find the Quotient

To find the quotient \(\frac{z_1}{z_2}\), divide the moduli and subtract the angles: \[\left|\frac{z_1}{z_2}\right| = \frac{|z_1|}{|z_2|} = \frac{\sqrt{2}}{3\sqrt{2}} = \frac{1}{3}\]\[\arg\left(\frac{z_1}{z_2}\right) = \theta_1 - \theta_2 = 75^\circ - 60^\circ = 15^\circ\]Hence, \(\frac{z_1}{z_2} = \frac{1}{3}(\cos 15^\circ + i \sin 15^\circ)\).

Key Concepts

Polar Form

Complex numbers can be expressed in different ways, and one of the most insightful representations is the polar form. In polar form, a complex number is written in terms of a modulus and an angle. The modulus, often denoted as \( r \), indicates the distance of the complex number from the origin on the complex plane. The angle, known as the argument, \( \theta \), represents the counterclockwise rotation from the positive x-axis to the line representing the complex number.

The expression for a complex number in polar form is \( z = r(\cos \theta + i \sin \theta) \). This representation is particularly useful for multiplication and division of complex numbers, as it simplifies the arithmetic operations to operations on real numbers.

For example, consider a complex number given in polar form as \( z_1 = \sqrt{2}(\cos 75^{\circ} + i \sin 75^{\circ}) \). Here, \( r_1 = \sqrt{2} \) is the modulus, and \( \theta_1 = 75^{\circ} \) is the argument.

Complex Multiplication

Multiplying complex numbers in polar form is quite straightforward. It involves two main steps: multiplying their moduli and adding their angles. This simplicity is one of the reasons why the polar form is highly favored in complex arithmetic.

To multiply two complex numbers \( z_1 \) and \( z_2 \) given by \( z_1 = r_1(\cos \theta_1 + i \sin \theta_1) \) and \( z_2 = r_2(\cos \theta_2 + i \sin \theta_2) \), follow these steps:

• Find the product of the moduli: \( |z_1 z_2| = r_1 \times r_2 \).
• Add the angles: \( \arg(z_1 z_2) = \theta_1 + \theta_2 \).

For instance, to multiply \( z_1 = \sqrt{2}(\cos 75^{\circ} + i \sin 75^{\circ}) \) and \( z_2 = 3\sqrt{2}(\cos 60^{\circ} + i \sin 60^{\circ}) \):

• The product of the moduli is \( \sqrt{2} \times 3\sqrt{2} = 6 \).
• The sum of the angles is \( 75^{\circ} + 60^{\circ} = 135^{\circ} \).

Therefore, the product is \( z_1 z_2 = 6(\cos 135^{\circ} + i \sin 135^{\circ}) \).

Complex Division

Just as multiplication is simplified in polar form, so is division. Dividing complex numbers in polar form involves dividing the moduli and subtracting the angles. This makes the entire operation intuitive and less prone to error.

To divide one complex number \( z_1 \) by another \( z_2 \), where \( z_1 = r_1(\cos \theta_1 + i \sin \theta_1) \) and \( z_2 = r_2(\cos \theta_2 + i \sin \theta_2) \), follow these steps:

• Divide the moduli: \( \left| \frac{z_1}{z_2} \right| = \frac{r_1}{r_2} \).
• Subtract the angles: \( \arg\left(\frac{z_1}{z_2}\right) = \theta_1 - \theta_2 \).

Taking an example, if \( z_1 = \sqrt{2}(\cos 75^{\circ} + i \sin 75^{\circ}) \) and \( z_2 = 3\sqrt{2}(\cos 60^{\circ} + i \sin 60^{\circ}) \):

• The division of the moduli is \( \frac{\sqrt{2}}{3\sqrt{2}} = \frac{1}{3} \).
• The difference in angles is \( 75^{\circ} - 60^{\circ} = 15^{\circ} \).

Thus, the quotient is \( \frac{z_1}{z_2} = \frac{1}{3}(\cos 15^{\circ} + i \sin 15^{\circ}) \). This method removes the complexity involved with the multiplication of two binomials, making calculations both easier and faster.