Problem 64 - 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\(/ z_{1}\) . $$ z_{1}=-\sqrt{2} i, \quad z_{2}=-3-3 \sqrt{3} i $$

Step-by-Step Solution

Answer

Product: \(6\sqrt{2}(\cos \frac{11\pi}{6} + i \sin \frac{11\pi}{6})\), Quotient: \(\frac{\sqrt{2}}{6}(\cos -\frac{\pi}{6} + i \sin -\frac{\pi}{6})\), Reciprocal: \(\frac{i}{\sqrt{2}}\).

Step 1: Understand Polar Form

Polar form of a complex number is given by \( z = r(\cos \theta + i \sin \theta) \), where \( r \) is the modulus and \( \theta \) is the argument of the complex number.

Step 2: Find the Modulus of \(z_1\)

For \( z_1 = -\sqrt{2}i \), the modulus \( r_1 \) is given by \( r_1 = \sqrt{0^2 + (-\sqrt{2})^2} = \sqrt{2} \).

Step 3: Find the Argument of \(z_1\)

Since \( z_1 \) lies on the negative imaginary axis, the argument \( \theta_1 \) is \( \frac{3\pi}{2} \) radians because it is at 270 degrees.

Step 4: Write \(z_1\) in Polar Form

Using the modulus and argument, \( z_1 \) in polar form is \( \sqrt{2}(\cos \frac{3\pi}{2} + i \sin \frac{3\pi}{2}) \).

Step 5: Find the Modulus of \(z_2\)

For \( z_2 = -3 - 3\sqrt{3}i \), the modulus \( r_2 \) is given by \( r_2 = \sqrt{(-3)^2 + (-3\sqrt{3})^2} = \sqrt{9 + 27} = \sqrt{36} = 6 \).

Step 6: Find the Argument of \(z_2\)

\( z_2 \) is in the third quadrant, thus the argument \( \theta_2 \) is \( \pi + \arctan\left(\frac{3\sqrt{3}}{3}\right) = \pi + \frac{\pi}{3} = \frac{4\pi}{3} \).

Step 7: Write \(z_2\) in Polar Form

Using the modulus and argument, \( z_2 \) in polar form is \( 6(\cos \frac{4\pi}{3} + i \sin \frac{4\pi}{3}) \).

Step 8: Calculate Product \(z_1 z_2\)

The product \( z_1 z_2 \) is given by \( r_1r_2(\cos(\theta_1 + \theta_2) + i \sin(\theta_1 + \theta_2)) \). Thus, \( z_1 z_2 = \sqrt{2} \cdot 6 (\cos(\frac{11\pi}{6}) + i \sin(\frac{11\pi}{6})) = 6\sqrt{2} \left(\frac{\sqrt{3}}{2} - i \frac{1}{2}\right) \).

Step 9: Calculate Quotient \(\frac{z_1}{z_2}\)

The quotient \( \frac{z_1}{z_2} \) is \( \frac{r_1}{r_2}(\cos(\theta_1 - \theta_2) + i \sin(\theta_1 - \theta_2)) \). So, the quotient is \( \frac{\sqrt{2}}{6}(\cos(-\frac{\pi}{6}) + i \sin(-\frac{\pi}{6})) = \frac{\sqrt{2}}{6} \left(\frac{\sqrt{3}}{2} - i \frac{1}{2}\right) \).

Step 10: Calculate Reciprocal \(\frac{1}{z_1}\)

The reciprocal \( \frac{1}{z_1} \) is \( \frac{1}{r_1}(\cos(-\theta_1) + i \sin(-\theta_1)) \). Thus, \( \frac{1}{z_1} = \frac{1}{\sqrt{2}}(\cos(-\frac{3\pi}{2}) + i \sin(-\frac{3\pi}{2})) = \frac{1}{\sqrt{2}}(0 + i \cdot 1) = \frac{i}{\sqrt{2}} \).

Key Concepts

Modulus of a Complex Number

The modulus of a complex number, sometimes referred to as the absolute value, is a measure of the "size" or "distance" of the complex number from the origin in the complex plane. It's simply the length of the line segment representing the complex number.
For a complex number expressed as \( z = a + bi \), the modulus is given by the formula:

• \( |z| = \sqrt{a^2 + b^2} \)

Applying this to our problem:- For \( z_1 = -\sqrt{2}i \), the modulus is calculated as \( |z_1| = \sqrt{0^2 + (-\sqrt{2})^2} = \sqrt{2} \).- For \( z_2 = -3 - 3\sqrt{3}i \), the modulus is \( |z_2| = \sqrt{(-3)^2 + (-3\sqrt{3})^2} = 6 \).
These moduli represent the distances of \( z_1 \) and \( z_2 \) from the origin, helping to transition them into polar form.

Argument of a Complex Number

The argument of a complex number is the angle the line representing the complex number makes with the positive real axis on the complex plane. This angle, often denoted \( \theta \), is measured in radians.
For a complex number \( z = a + bi \), you can find the argument through:

• \( \theta = \arctan\left(\frac{b}{a}\right) \)
• Adjust based on the quadrant:
  • Add \( \pi \) if the complex number is in the second or third quadrant.
  • Subtract \( 2\pi \) if the result is greater than \( \pi \).

In our example problem:- For \( z_1 \), the argument is \( \frac{3\pi}{2} \) radians since it lies on the negative imaginary axis.- For \( z_2 \), the calculation is \( \theta_2 = \pi + \frac{\pi}{3} = \frac{4\pi}{3} \), noting it’s in the third quadrant.
Understanding the argument helps express complex numbers in polar form.

Product and Division of Complex Numbers in Polar Form

Once in polar form, multiplying and dividing complex numbers becomes a bit more intuitive.
For the product of two complex numbers in polar form, \( z_1 = r_1 (\cos \theta_1 + i \sin \theta_1) \) and \( z_2 = r_2 (\cos \theta_2 + i \sin \theta_2) \), you use:

• Multiply moduli: \( r_1 \times r_2 \)
• Add arguments: \( \theta_1 + \theta_2 \)
• Result: \( z_1 z_2 = r_1 r_2 (\cos(\theta_1 + \theta_2) + i \sin(\theta_1 + \theta_2)) \)

For division, the process is similar but slightly reversed:

• Divide moduli: \( \frac{r_1}{r_2} \)
• Subtract arguments: \( \theta_1 - \theta_2 \)
• Result: \( \frac{z_1}{z_2} = \frac{r_1}{r_2} (\cos(\theta_1 - \theta_2) + i \sin(\theta_1 - \theta_2)) \)

Using this knowledge in our exercise:- For the product \( z_1 z_2 \), we find the polar form to be \( 6\sqrt{2} \left(\frac{\sqrt{3}}{2} - i \frac{1}{2}\right) \).- For the quotient \( \frac{z_1}{z_2} \), it is \( \frac{\sqrt{2}}{6} \left(\frac{\sqrt{3}}{2} - i \frac{1}{2}\right) \).
These operations demonstrate the power of using polar forms for simplifying complex calculations.