Problem 57 - 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\(/ z_{1}\) . $$ z_{1}=\sqrt{3}+i, \quad z_{2}=1+\sqrt{3} i $$

Step-by-Step Solution

Answer

\(z_1 z_2 = 4i\), \(\frac{z_1}{z_2} = \frac{\sqrt{3}}{2} - \frac{1}{2}i\), and \(\frac{1}{z_1} = \frac{\sqrt{3}}{4} - \frac{1}{4}i\).

Step 1: Convert to Polar Form

To convert a complex number \( z = x + yi \) to polar form, use \( r = \sqrt{x^2 + y^2} \) to find the magnitude and \( \theta = \tan^{-1}\left(\frac{y}{x}\right) \) to find the angle. For \( z_1 = \sqrt{3} + i \), it is \( r_1 = \sqrt{(\sqrt{3})^2 + 1^2} = 2 \) and \( \theta_1 = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6} \). Thus, \( z_1 = 2(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6}) \).For \( z_2 = 1 + \sqrt{3}i \), \( r_2 = \sqrt{1^2 + (\sqrt{3})^2} = 2 \) and \( \theta_2 = \tan^{-1}\left(\frac{\sqrt{3}}{1}\right) = \frac{\pi}{3} \). Thus, \( z_2 = 2(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}) \).

Step 2: Find the Product

The product of two complex numbers in polar form \( z_1 = r_1(\cos\theta_1 + i\sin\theta_1) \) and \( z_2 = r_2(\cos\theta_2 + i\sin\theta_2) \) is \( z_1 z_2 = r_1 r_2 [\cos(\theta_1 + \theta_2) + i\sin(\theta_1 + \theta_2)] \).For \( z_1 z_2 \), \( (2)(2) = 4 \) and \( \theta_1 + \theta_2 = \frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi}{2} \). Thus, \( z_1 z_2 = 4(\cos\frac{\pi}{2} + i\sin\frac{\pi}{2}) = 4i \).

Step 3: Calculate the Quotient \( z_1 / z_2 \)

The division \( \frac{z_1}{z_2} \) in polar form is given by \( \frac{r_1}{r_2} [\cos(\theta_1 - \theta_2) + i\sin(\theta_1 - \theta_2)] \).\( \frac{2}{2} = 1 \) and \( \theta_1 - \theta_2 = \frac{\pi}{6} - \frac{\pi}{3} = -\frac{\pi}{6} \). Thus, \( \frac{z_1}{z_2} = \cos(-\frac{\pi}{6}) + i\sin(-\frac{\pi}{6}) = \sqrt{3}/2 - i/2 \).

Step 4: Calculate the Quotient \( 1 / z_1 \)

The equivalent polar form of \( 1/z_1 \) is \( \frac{1}{2}(\cos(-\theta_1) + i\sin(-\theta_1)) \), where \( \theta_1 = \frac{\pi}{6} \).So, \( \frac{1}{z_1} = \frac{1}{2}(\cos(-\frac{\pi}{6}) + i\sin(-\frac{\pi}{6})) = \frac{1}{2}(\frac{\sqrt{3}}{2} - \frac{1}{2}i) = \frac{\sqrt{3}}{4} - \frac{1}{4}i \).

Key Concepts

Complex Number Operations

Complex number operations form the basis for much of the work done with complex numbers, which are numbers of the form \( z = x + yi \), where \( x \) and \( y \) are real numbers and \( i \) is the imaginary unit. Key complex number operations include addition, subtraction, multiplication, and division. Addition and subtraction with complex numbers are relatively straightforward: simply add or subtract the real parts and the imaginary parts separately. However, when it comes to multiplication and division, especially in polar form, the operations become more intriguing and require using angles and magnitudes. This is because complex numbers can be more easily calculated in polar form when dealing with multiplicative operations.

Product of Complex Numbers

When multiplying complex numbers, especially in polar form, it becomes quite efficient. In polar form, a complex number \( z \) can be expressed as \( r(\cos \theta + i \sin \theta) \), where \( r \) is the magnitude and \( \theta \) is the angle. The product of two complex numbers \( z_1 = r_1(\cos \theta_1 + i \sin \theta_1) \) and \( z_2 = r_2(\cos \theta_2 + i \sin \theta_2) \) is given by multiplying their magnitudes (\( r_1 r_2 \)) and adding their angles \( (\theta_1 + \theta_2) \). This results in a new complex number \( z_1 z_2 = r_1 r_2 [\cos(\theta_1 + \theta_2) + i \sin(\theta_1 + \theta_2)] \). This method simplifies the multiplication process and is particularly useful in complex number calculations, as shown in the example where two complex numbers were multiplied to get \( 4i \).

Quotient of Complex Numbers

Dividing complex numbers, notably in polar form, involves dividing their magnitudes and subtracting their angles. If we have 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 \( \frac{z_1}{z_2} \) is derived by calculating \( \frac{r_1}{r_2} \) for the magnitudes and subtracting the angles \( \theta_1 - \theta_2 \). Thus, the quotient becomes \( \frac{z_1}{z_2} = \frac{r_1}{r_2} [\cos(\theta_1 - \theta_2) + i \sin(\theta_1 - \theta_2)] \). This procedure allows for a streamlined calculation of the division, as shown in the example where dividing two complex numbers results in a particular complex number outcome.

Convert to Polar Form

Converting a complex number from rectangular (or Cartesian) form \( x + yi \) to polar form \( r(\cos \theta + i \sin \theta) \) involves finding the magnitude \( r \) and the angle \( \theta \). To find \( r \), use the formula \( r = \sqrt{x^2 + y^2} \). The angle \( \theta \) is determined using the arctangent function \( \theta = \tan^{-1}\left(\frac{y}{x}\right) \). For example, for \( z_1 = \sqrt{3} + i \), the magnitude is 2, and the angle is \( \frac{\pi}{6} \), giving a polar form of \( 2(\cos \frac{\pi}{6} + i \sin \frac{\pi}{6}) \). Similarly, for \( z_2 = 1 + \sqrt{3}i \), the magnitude is 2, and the angle is \( \frac{\pi}{3} \). Converting to polar form can greatly simplify the handlings of complex numbers, mainly when multiplying or dividing them.