Problem 59 - 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\(/ z_{1}\) . $$ z_{1}=2 \sqrt{3}-2 i, \quad z_{2}=-1+i $$

Step-by-Step Solution

Answer

\(z_1 = 4(\cos(-\frac{\pi}{6}) + i \sin(-\frac{\pi}{6}))\), \(z_2 = \sqrt{2}(\cos(\frac{3\pi}{4}) + i \sin(\frac{3\pi}{4}))\), \(z_1z_2 = 4\sqrt{2}(\cos(\frac{5\pi}{12}) + i\sin(\frac{5\pi}{12}))\), \(\frac{z_1}{z_2} = 2\sqrt{2}(\cos(-\frac{11\pi}{12}) + i\sin(-\frac{11\pi}{12}))\), \(\frac{1}{z_1} = \frac{1}{4}(\cos(\frac{\pi}{6}) + i\sin(\frac{\pi}{6}))\).

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

The complex number \( z_1 = 2\sqrt{3} - 2i \) must be converted to polar form. First, calculate the modulus \( r_1 \):\[ r_1 = \sqrt{(2\sqrt{3})^2 + (-2)^2} = \sqrt{12 + 4} = \sqrt{16} = 4. \]Next, find the argument \( \theta_1 \) using \( \tan \theta_1 = \frac{-2}{2\sqrt{3}} = -\frac{1}{\sqrt{3}} \). This implies \( \theta_1 = -\frac{\pi}{6} \). Therefore, the polar form of \( z_1 \) is:\[ z_1 = 4 \left( \cos\left(-\frac{\pi}{6}\right) + i \sin\left(-\frac{\pi}{6}\right) \right). \]

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

For the complex number \( z_2 = -1+i \), calculate the modulus \( r_2 \):\[ r_2 = \sqrt{(-1)^2 + (1)^2} = \sqrt{2}. \]Determine the argument \( \theta_2 \) using \( \tan \theta_2 = \frac{1}{-1} = -1 \). The argument is \( \theta_2 = \frac{3\pi}{4} \) (since it is in the second quadrant). Therefore, the polar form of \( z_2 \) 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 the Product \(z_1 z_2\)

To find the product, multiply the moduli and add the arguments:\[ \text{Modulus: } r_1 r_2 = 4 \times \sqrt{2} = 4\sqrt{2}, \]\[ \text{Argument: } \theta_1 + \theta_2 = -\frac{\pi}{6} + \frac{3\pi}{4} = \frac{5\pi}{12}. \]Thus, the product \( z_1 z_2 \) is:\[ z_1 z_2 = 4\sqrt{2} \left( \cos\left(\frac{5\pi}{12}\right) + i \sin\left(\frac{5\pi}{12}\right) \right). \]

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

For the quotient, divide the moduli and subtract the arguments:\[ \text{Modulus: } \frac{r_1}{r_2} = \frac{4}{\sqrt{2}} = 2\sqrt{2}, \]\[ \text{Argument: } \theta_1 - \theta_2 = -\frac{\pi}{6} - \frac{3\pi}{4} = -\frac{11\pi}{12}. \]Therefore, the quotient \( \frac{z_1}{z_2} \) is:\[ \frac{z_1}{z_2} = 2\sqrt{2} \left( \cos\left(-\frac{11\pi}{12}\right) + i \sin\left(-\frac{11\pi}{12}\right) \right). \]

Step 5: Find the Reciprocal \(\frac{1}{z_1}\)

For the reciprocal, take the inverse of the modulus and negate the argument:\[ \text{Modulus: } \frac{1}{r_1} = \frac{1}{4}, \]\[ \text{Argument: } -\theta_1 = \frac{\pi}{6}. \]Thus, the reciprocal \( \frac{1}{z_1} \) is:\[ \frac{1}{z_1} = \frac{1}{4} \left( \cos\left(\frac{\pi}{6}\right) + i \sin\left(\frac{\pi}{6}\right) \right). \]

Key Concepts

Complex Numbers

Complex numbers are numbers that consist of a real part and an imaginary part. They are usually expressed in the form \(a + bi\), where \(a\) is the real part and \(bi\) is the imaginary part with \(i\) being the imaginary unit satisfying \(i^2 = -1\). Complex numbers can be graphed on a plane known as the complex plane, where the horizontal axis represents the real part, and the vertical axis represents the imaginary part.

• Real Part: This is the \(a\) in a complex number, representing its real component.
• Imaginary Part: This is the \(b\) in a complex number, noted as \(bi\), which involves the imaginary unit \(i\).

Understanding complex numbers is essential for converting them into other forms such as polar form, which allows easier manipulation in multiplications and divisions.

Modulus and Argument

The modulus and argument are two critical components that help convert complex numbers from Cartesian form \(a + bi\) to polar form.

• Modulus: It measures the distance from the origin to the point on the complex plane and is calculated using \(r = \sqrt{a^2 + b^2}\).
• Argument: This is the angle formed with the positive real axis, often calculated using the arctangent function, \(\tan\theta = \frac{b}{a}\).

In the original exercise, we computed the modulus \(r\) and argument \(\theta\) for both complex numbers \(z_1\) and \(z_2\), which are essential for expressing them in polar coordinates.

Polar Coordinates

Polar coordinates represent complex numbers in terms of their modulus and argument. In polar form, a complex number \( z \) is expressed as \( r(\cos \theta + i \sin \theta) \). This form is particularly useful in simplifying multiplication and division operations involving complex numbers.

• The modulus \(r\) indicates the radial distance from the origin.
• The argument \(\theta\) specifies the direction or angle in radians.

For example, in the exercise, \(z_1 = 4(\cos(-\pi/6) + i\sin(-\pi/6))\) and \(z_2 = \sqrt{2}(\cos(3\pi/4) + i\sin(3\pi/4))\) are the polar forms of the complex numbers, showing both magnitude and direction.

Multiplication and Division of Complex Numbers

Multiplication and division of complex numbers are significantly simplified when using their polar form.

Multiplication

To multiply two complex numbers in polar form, multiply their moduli and add their arguments:

• Resultant Modulus: \(r_1 \times r_2\)
• Resultant Argument: \(\theta_1 + \theta_2\)

In the exercise, the product of \(z_1\) and \(z_2\) was \(4\sqrt{2}(\cos(5\pi/12) + i\sin(5\pi/12))\).

Division

For division, divide their moduli and subtract their arguments:

• Resultant Modulus: \(\frac{r_1}{r_2}\)
• Resultant Argument: \(\theta_1 - \theta_2\)

The quotient calculated in the example was \(2\sqrt{2}(\cos(-11\pi/12) + i\sin(-11\pi/12))\). This streamlined calculation using polar coordinates saves time and simplifies the process when compared to handling the numbers in Cartesian form.