Problem 66 - 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\(/ z_{1}\) . $$ z_{1}=4 \sqrt{3}-4 i, \quad z_{2}=8 i $$

Step-by-Step Solution

Answer

\( z_1 = 8(\cos(-\frac{\pi}{6}) + i\sin(-\frac{\pi}{6})) \), \( z_2 = 8 (\cos(\frac{\pi}{2}) + i\sin(\frac{\pi}{2})) \); \( z_1 z_2 = 32 + 32i\sqrt{3} \); \( \frac{z_1}{z_2} = -\frac{1}{2} - i\frac{\sqrt{3}}{2} \); \( \frac{1}{z_1} = \frac{\sqrt{3}}{16} + i\frac{1}{16} \).

Step 1: Convert to Polar Form - \( z_1 \)

First, determine the modulus of \( z_1 = 4\sqrt{3} - 4i \). The modulus is given by \( r_1 = \sqrt{(4\sqrt{3})^2 + (-4)^2} = \sqrt{48 + 16} = \sqrt{64} = 8 \). Next, calculate the argument \( \theta_1 \) using \( \tan\theta_1 = \frac{-4}{4\sqrt{3}} = -\frac{1}{\sqrt{3}} \), which means \( \theta_1 = -\frac{\pi}{6} \). Thus, \( z_1 = 8(\cos(-\frac{\pi}{6}) + i\sin(-\frac{\pi}{6})) \).

Step 2: Convert to Polar Form - \( z_2 \)

For \( z_2 = 8i \), calculate the modulus \( r_2 = \sqrt{0^2 + 8^2} = 8 \). The argument \( \theta_2 \) is \( \frac{\pi}{2} \) since it lies on the positive imaginary axis. Therefore, \( z_2 = 8 (\cos(\frac{\pi}{2}) + i\sin(\frac{\pi}{2})) \).

Step 3: Product of \( z_1 \) and \( z_2 \)

To find \( z_1 z_2 \), multiply the moduli and add the arguments: \( r_1 \cdot r_2 = 8 \times 8 = 64 \) and \( \theta_1 + \theta_2 = -\frac{\pi}{6} + \frac{\pi}{2} = \frac{\pi}{3} \). Thus, \( z_1 z_2 = 64 (\cos(\frac{\pi}{3}) + i\sin(\frac{\pi}{3})) = 64(\frac{1}{2} + i\frac{\sqrt{3}}{2}) = 32 + 32i\sqrt{3} \).

Step 4: Quotient \( \frac{z_1}{z_2} \)

To find \( \frac{z_1}{z_2} \), divide the moduli and subtract the arguments: \( \frac{r_1}{r_2} = \frac{8}{8} = 1 \) and \( \theta_1 - \theta_2 = -\frac{\pi}{6} - \frac{\pi}{2} = -\frac{\pi}{6} - \frac{3\pi}{6} = -\frac{4\pi}{6} = -\frac{2\pi}{3} \). Therefore, \( \frac{z_1}{z_2} = 1 (\cos(-\frac{2\pi}{3}) + i\sin(-\frac{2\pi}{3})) = -\frac{1}{2} - i\frac{\sqrt{3}}{2} \).

Step 5: Quotient \( \frac{1}{z_1} \)

To find \( \frac{1}{z_1} \), take the reciprocal of the modulus and change the sign of the argument: \( r_1^{-1} = \frac{1}{8} \) and \( -\theta_1 = \frac{\pi}{6} \). Hence, \( \frac{1}{z_1} = \frac{1}{8} (\cos(\frac{\pi}{6}) + i\sin(\frac{\pi}{6})) = \frac{1}{8}(\frac{\sqrt{3}}{2} + i\frac{1}{2}) = \frac{\sqrt{3}}{16} + i\frac{1}{16} \).

Key Concepts

Complex Numbers

Complex numbers are numbers that include both a real part and an imaginary part. They are written in the form \( z = a + bi \), where \( a \) is the real part and \( b \) is the imaginary part. The imaginary unit \( i \) is defined as \( \sqrt{-1} \), which leads to the useful property \( i^2 = -1 \). A complex number can be visualized as a point or a vector in a two-dimensional plane called the complex plane, where the horizontal axis represents the real part and the vertical axis represents the imaginary part.

• The real part tells how far the number is along the horizontal axis.
• The imaginary part tells how far the number is along the vertical axis.

Understanding complex numbers is important, as they provide a natural extension of real numbers and are used extensively in various fields including engineering, physics, and applied mathematics. They allow solutions to equations like \( x^2 + 1 = 0 \), which have no real number solutions.

Modulus and Argument

The modulus and the argument are key components in expressing complex numbers in polar form. The modulus of a complex number \( z = a + bi \) is its distance from the origin in the complex plane. It is calculated using the formula \( |z| = \sqrt{a^2 + b^2} \). The modulus represents the magnitude of the complex number, similar to the length of a vector.The argument of a complex number is the angle \( \theta \) between the positive real axis and the line connecting the origin to the point \( (a, b) \). It is typically measured in radians. The argument can be calculated using trigonometric functions such as the tangent: \( \tan \theta = \frac{b}{a} \). To find the argument, you need to be mindful of the quadrant in which the complex number lies, as this affects the angle returned by the inverse tangent function.With the modulus \( r \) and the argument \( \theta \), a complex number \( z \) can be expressed in polar form as \( z = r(\cos \theta + i\sin \theta) \). This polar representation is particularly useful for multiplication and division, as we'll see next.

Multiplication and Division of Complex Numbers

Multiplying and dividing complex numbers becomes straightforward when they are in polar form. For two complex numbers \( z_1 = r_1(\cos \theta_1 + i\sin \theta_1) \) and \( z_2 = r_2(\cos \theta_2 + i\sin \theta_2) \):

• Multiplication: Adjust the modulus by multiplying them, \( r_1 \times r_2 \), and add the arguments, \( \theta_1 + \theta_2 \). The product \( z_1 z_2 \) is \( r_1r_2(\cos(\theta_1 + \theta_2) + i\sin(\theta_1 + \theta_2)) \).
• Division: Adjust the modulus by dividing them, \( \frac{r_1}{r_2} \), and subtract the arguments, \( \theta_1 - \theta_2 \). The quotient \( \frac{z_1}{z_2} \) becomes \( \frac{r_1}{r_2}(\cos(\theta_1 - \theta_2) + i\sin(\theta_1 - \theta_2)) \).

Thus, polar form simplifies otherwise potentially complex computations involving multiplication and division by reducing them to basic arithmetic operations that involve only the magnitudes and angles.

Reciprocal of a Complex Number

Finding the reciprocal of a complex number is a useful operation and is easily performed using polar form. The reciprocal of a complex number \( z = r(\cos \theta + i\sin \theta) \) is another complex number such that when multiplied together, they yield 1. In the context of polar coordinates:

• Take the reciprocal of the modulus, \( \frac{1}{r} \).
• Negate the argument, \( -\theta \).

Thus, the reciprocal is given by \( \frac{1}{z} = \frac{1}{r}(\cos(-\theta) + i\sin(-\theta)) \). This reciprocal maintains the magnitude of the original number adjusted to be its inverse, while reflecting the argument about the real axis. Such a reflection is equivalent to rotating the angle of the complex number in the opposite direction. Understanding the reciprocal of a complex number assists in completing division operations involving complex numbers by simplifying the division to a multiplication.