Problem 49 - 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. $$ \frac{w}{z^{2}} $$

Step-by-Step Solution

Answer

\(\frac{w}{z^2} = \frac{2}{3} \text{cis}\left(\frac{\pi}{12}\right)\) in polar form.

Step 1: Understanding the Problem

We need to compute \(\frac{w}{z^2}\) using the given complex numbers \(z = -\frac{3 \sqrt{3}}{2} + \frac{3}{2} i\) and \(w = 3 \sqrt{2} - 3 i \sqrt{2}\). The final answer should be in polar form with the principal argument.

Step 2: Convert Complex Numbers to Polar Form

First, we convert \(z = -\frac{3 \sqrt{3}}{2} + \frac{3}{2} i\) to polar form. The modulus is \(\lvert z \rvert = \sqrt{\left(-\frac{3\sqrt{3}}{2}\right)^2 + \left(\frac{3}{2}\right)^2} = 3\). The argument \(\theta\) can be found using \(\tan(\theta) = \frac{\text{Im}(z)}{\text{Re}(z)} = -\frac{1}{\sqrt{3}}\), giving \(\theta = 5\pi/6\) in the principal range \([-\pi, \pi)\). Thus, \(z = 3 \text{cis}(5\pi/6)\). Similarly for \(w = 3\sqrt{2} - 3i\sqrt{2}\), the modulus is \(\lvert w \rvert = 6\) and the argument is \(\tan^{-1}\left(-1\right) = -\pi/4\). Therefore, \(w = 6 \text{cis}(-\pi/4)\).

Step 3: Calculate \(z^2\)

Using the polar form of \(z\), we find \(z^2 = (3\text{cis}(5\pi/6))^2 = 9\text{cis}(5\pi/6 \times 2) = 9\text{cis}(5\pi/3)\).

Step 4: Perform Division in Polar Form

Now divide \(w\) by \(z^2\): \[ \frac{w}{z^2} = \frac{6\text{cis}(-\pi/4)}{9\text{cis}(5\pi/3)} = \frac{6}{9} \text{cis}(-\pi/4 - 5\pi/3) = \frac{2}{3} \text{cis}(-\pi/4 - 5\pi/3). \]Simplifying the argument: \[-\pi/4 - 5\pi/3 = -\frac{3\pi}{12} - \frac{20\pi}{12} = -\frac{23\pi}{12}.\] Convert to the principal argument by adding \(24\pi/12\) to bring it within \([-\pi, \pi)\):\[-\frac{23\pi}{12} + 2\pi = \frac{\pi}{12}.\]

Step 5: Final Answer in Polar Form

Thus, the division in polar form is \[ \frac{w}{z^2} = \frac{2}{3} \text{cis}\left(\frac{\pi}{12}\right). \]

Key Concepts

Polar Form

Polar form is a way of representing complex numbers, particularly useful when performing multiplication and division. Rather than expressing a complex number in its standard format of \(a + bi\) (where \(a\) is the real part and \(bi\) is the imaginary part), we express it in terms of its modulus (absolute value) and argument (angle). The polar form looks like \(r \text{cis}(\theta)\), which can also be expressed as \(r(\cos(\theta) + i\sin(\theta))\). Here, \(r\) is the modulus and \(\theta\) is the argument.

• The modulus \(r\) is calculated using \(r = \sqrt{a^2 + b^2}\).
• The argument \(\theta\) is found using \(\theta = \tan^{-1}(\frac{b}{a})\).

By converting complex numbers into polar form, multiplying them becomes as simple as multiplying their moduli and adding their arguments. Likewise, division involves dividing the moduli and subtracting the arguments.

Principal Argument

The principal argument is the angle part of a complex number in polar form, specifically chosen to be within the range \([-\pi, \pi)\). This is important for finding a unique representation of the angle, making computations consistent and avoiding ambiguity.

• When the argument calculated results in a value outside this range, adjustments are made by adding or subtracting \(2\pi\) to bring it back within the range.
• The principal argument is generally more intuitive for calculations and is commonly used because any two equivalent angles differ by an integer multiple of \(2\pi\).

For instance, if after performing division, you find an argument like \(-\frac{23\pi}{12}\), you would convert it to the principal argument by adding \(2\pi\) enough times to bring it within \([-\pi, \pi)\). This ensures your final answer is uniquely represented without ambiguity.

Modulus and Argument

The modulus and the argument give a comprehensive description of a complex number in polar form.

• Modulus: This is similar to the distance of a point from the origin in a plane and is given by \(\lvert z \rvert = \sqrt{a^2 + b^2}\) for a complex number \(z = a + bi\). It represents the 'size' or 'magnitude' of the complex number.
• Argument: The argument \(\theta\) represents the direction and is calculated with \(\theta = \tan^{-1}(\frac{b}{a})\). It shows the angle the line (from origin to the complex number in the complex plane) makes with the positive real axis.

When performing operations like multiplication or division on complex numbers, working with their moduli and arguments simplifies the process. Instead of dealing with real and imaginary parts separately, the operation is reduced to manipulating magnitudes and angles, which is often easier to visualize and compute. Adjusting arguments into their principal range further aids in maintaining consistency across computations.