Problem 47 - College Trigonometry

Question

Use \(z=-\frac{3 \sqrt{3}}{2}+\frac{3}{2} i\) and \(w=3 \sqrt{2}-3 i \sqrt{2}\) to compute the quantity. Express your answers in polar form using the principal argument. $$ z^{3} w^{2} $$

Step-by-Step Solution

Answer

The value of \( z^3 w^2 \) is 972 in polar form \( 972 ( \cos(0) + i \sin(0) ) \).

Step 1: Convert z to Polar Form

First, we find the modulus of the complex number \( z = -\frac{3\sqrt{3}}{2} + \frac{3}{2} i \). The modulus is calculated as \( |z| = \sqrt{ \left( -\frac{3\sqrt{3}}{2} \right)^2 + \left( \frac{3}{2} \right)^2 } \). Simplifying, \( |z| = 3 \).Next, determine the argument: \( \text{arg}(z) = \tan^{-1} \left( \frac{\frac{3}{2}}{-\frac{3\sqrt{3}}{2}} \right) = \tan^{-1} \left( -\frac{1}{\sqrt{3}} \right) = \tan^{-1}(-\frac{\sqrt{3}}{3}) \), which is equal to \( -\frac{\pi}{6} \) since \( z \) is in the second quadrant.Thus, \( z = 3 \left( \cos \left( -\frac{\pi}{6} \right) + i \sin \left( -\frac{\pi}{6} \right) \right) \).

Step 2: Convert w to Polar Form

Find the modulus of \( w = 3\sqrt{2} - 3i\sqrt{2} \). The modulus is \( |w| = \sqrt{ (3\sqrt{2})^2 + (-3\sqrt{2})^2 } = \sqrt{18 + 18} = 6 \).Next, determine the argument: \( \text{arg}(w) = \tan^{-1} \left( \frac{-3\sqrt{2}}{3\sqrt{2}} \right) = \tan^{-1}(-1) = -\frac{\pi}{4} \) since \( w \) is in the fourth quadrant.Hence, \( w = 6 \left( \cos \left( -\frac{\pi}{4} \right) + i \sin \left( -\frac{\pi}{4} \right) \right) \).

Step 3: Expression for z^3

Using De Moivre's Theorem, calculate \( z^3 \). Given \( z = 3 ( \cos(-\frac{\pi}{6}) + i \sin(-\frac{\pi}{6}) ) \), we have:\[ z^3 = 3^3 \left( \cos\left(3 \times -\frac{\pi}{6}\right) + i \sin\left(3 \times -\frac{\pi}{6}\right) \right) \]This simplifies to \( z^3 = 27 \left( \cos(-\frac{\pi}{2}) + i \sin(-\frac{\pi}{2}) \right) = 27 (0 - i) = -27i \).

Step 4: Expression for w^2

Using De Moivre's Theorem, calculate \( w^2 \). Given \( w = 6 ( \cos(-\frac{\pi}{4}) + i \sin(-\frac{\pi}{4}) ) \), we have:\[ w^2 = 6^2 \left( \cos\left(2 \times -\frac{\pi}{4}\right) + i \sin\left(2 \times -\frac{\pi}{4}\right) \right) \]This simplifies to \( w^2 = 36 \left( \cos(-\frac{\pi}{2}) + i \sin(-\frac{\pi}{2}) \right) = 36 (0 - i) = -36i \).

Step 5: Calculate z^3 w^2 in Polar Form

Combine the results from steps 3 and 4. We have:\[ z^3 w^2 = (-27i) (-36i) = 972(-i^2) = 972 \]Since the product is a positive real number, expressed in polar form: \( 972 ( \cos(0) + i \sin(0) ) \). Its principal argument is 0.

Key Concepts

Complex Number Modulus

The modulus of a complex number provides us with a measure of its size or length. For a complex number expressed as \( z = a + bi \), the modulus is calculated using the formula \( |z| = \sqrt{a^2 + b^2} \). It gives you the distance from the origin in the complex plane.

In our example, for \( z = -\frac{3\sqrt{3}}{2} + \frac{3}{2}i \), the modulus is found using:

• Squaring the real part: \( \left( -\frac{3\sqrt{3}}{2} \right)^2 \)
• Squaring the imaginary part: \( \left( \frac{3}{2} \right)^2 \)
• Adding these values: \( \frac{27}{4} + \frac{9}{4} = 9 \)
• Taking the square root: \( |z| = 3 \)

This modulus represents the magnitude of \( z \) on the complex plane, which is crucial for converting complex numbers into polar form. Similarly, for the complex number \( w = 3\sqrt{2} - 3i\sqrt{2} \), the modulus \( |w| \) is \( 6 \).

Knowing the modulus is the first step in expressing a complex number in polar form.

De Moivre's Theorem

De Moivre's Theorem allows us to easily compute powers of complex numbers when they are in polar form. This theorem states that if a complex number is expressed as \( z = r(\cos \theta + i\sin \theta) \), then \( z^n = r^n (\cos(n\theta) + i \sin(n\theta)) \).

Let's look at how we used this theorem in our example:

• Given \( z^3 \), with \( z = 3(\cos( -\frac{\pi}{6}) + i\sin( -\frac{\pi}{6}) ) \), the theorem is applied to yield: \( z^3 = 27( \cos(-\frac{3\pi}{6}) + i \sin(-\frac{3\pi}{6}) ) \)
• The resulting expression simplifies to \( 27(\cos(-\frac{\pi}{2}) + i \sin(-\frac{\pi}{2}) ) = -27i \)
• For \( w^2 \), with \( w = 6(\cos(-\frac{\pi}{4}) + i\sin(-\frac{\pi}{4})) \), the theorem gives: \( w^2 = 36( \cos(-\frac{\pi}{2}) + i \sin(-\frac{\pi}{2}) ) = -36i \)

By applying De Moivre's Theorem, you can simplify complex exponentiation to a matter of multiplying and rotating angles effectively.

Principal Argument

In the polar form of complex numbers, the argument represents the angle a complex number makes with the positive real axis. The principal argument is the unique angle, usually denoted by \( \theta \), located between \(-\pi\) and \(\pi\). It ensures consistency in expressing the angle.

For example, for \( z = -\frac{3\sqrt{3}}{2} + \frac{3}{2} i \), the argument is calculated via the inverse tangent function:

• Calculate the tangent: \( \tan^{-1}\left(\frac{3/2}{-3\sqrt{3}/2}\right) = \tan^{-1}(-\frac{1}{\sqrt{3}}) = \tan^{-1}(-\frac{\sqrt{3}}{3}) \)
• This obtains \( -\frac{\pi}{6} \) as the angle lies in the second quadrant

Similarly, the argument of \( w = 3\sqrt{2} - 3i\sqrt{2} \) is \(-\frac{\pi}{4}\), positioned in the fourth quadrant.

Finding the principal argument is crucial for accurately expressing complex numbers in a consistent polar representation.

Complex Number Arithmetic

Arithmetic operations on complex numbers can be convenient in different forms. In Cartesian form, you add or subtract the real and imaginary parts separately. However, multiplication and division often become simpler in polar form.

We can observe this in the exercise:

• When multiplying \( z^3 \) and \( w^2 \), both already converted to polar form using De Moivre's Theorem, you find their product as \( (-27i) \times (-36i) \).
• This results in \( 972(-i^2) = 972 \), exploiting the fact that \( i^2 = -1 \).
• In terms of polar coordinates, the result \( 972(\cos(0) + i\sin(0)) \), emphasizes that the principal argument is 0.

Converting computations to polar form using modulus and argument leverages trigonometric identities to simplify complex arithmetic operations.