Problem 25 - Precalculus

Question

Find the result of each expression using De Moivre's theorem. Write the answer in rectangular form. $$(1-\sqrt{3} i)^{4}$$

Step-by-Step Solution

Answer

The result in rectangular form is \(-8 - 8\sqrt{3}i\).

Step 1: Determine the modulus of the complex number

The complex number given is \(1 - \sqrt{3}i\). To find the modulus, use the formula \(|z| = \sqrt{a^2 + b^2}\), where \(a = 1\) and \(b = -\sqrt{3}\). So, \(|1 - \sqrt{3}i| = \sqrt{1^2 + (-\sqrt{3})^2} = \sqrt{1 + 3} = 2\).

Step 2: Find the argument of the complex number

The argument \(\theta\) is determined using the tangent function: \(\tan \theta = \frac{b}{a} = \frac{-\sqrt{3}}{1} = -\sqrt{3}\). This corresponds to an angle of \(\theta = -\frac{\pi}{3}\) (since the complex number is in the fourth quadrant).

Step 3: Express the complex number in polar form

Now that we have the modulus and argument, we can express the number in polar form: \(z = r(\cos \theta + i \sin \theta)\), which is \(2(\cos(-\frac{\pi}{3}) + i \sin(-\frac{\pi}{3})\).

Step 4: Apply De Moivre's theorem

To raise the complex number to the 4th power, we use De Moivre's theorem: \(z^n = r^n (\cos(n\theta) + i \sin(n\theta))\). Here, \(n = 4\), \(r = 2\), and \(\theta = -\frac{\pi}{3}\). Thus, \((1-\sqrt{3}i)^4 = 2^4 \left(\cos\left(4\cdot(-\frac{\pi}{3})\right) + i \sin\left(4\cdot(-\frac{\pi}{3})\right)\right)\).

Step 5: Simplify using trigonometric identities

Calculate \(2^4 = 16\). Next, compute the angle: \(4\cdot(-\frac{\pi}{3}) = -\frac{4\pi}{3}\). Using periodicity of cosine and sine, \(\cos(-\frac{4\pi}{3}) = -\frac{1}{2}\) and \(\sin(-\frac{4\pi}{3}) = -\frac{\sqrt{3}}{2}\). Substitute back: \(16\left(-\frac{1}{2} + i\left(-\frac{\sqrt{3}}{2}\right)\right)\).

Step 6: Convert back to rectangular form

Distribute the 16 to get: \(-8 - 8\sqrt{3}i\). Thus, the rectangular form of \((1-\sqrt{3}i)^4\) is \(-8 - 8\sqrt{3}i\).

Key Concepts

Modulus of a Complex Number

The modulus of a complex number, often denoted as \(|z|\), is the length of the vector representing the complex number in the complex plane. If a complex number is written as \(a + bi\), where \(a\) is the real part and \(b\) is the imaginary part, its modulus can be calculated using the formula:

• \[|z| = \sqrt{a^2 + b^2}\]

This formula gives the Euclidean distance from the origin (0,0) to the point \(a, b\) on the complex plane. For instance, in the problem \(1 - \sqrt{3}i\), the real part \(a\) is 1 and the imaginary part \(b\) is \(-\sqrt{3}\), which makes the modulus equal to 2. Just like the length of a hypotenuse in a right-angle triangle, the modulus is useful for transitioning between different forms of complex numbers.

Polar Form of Complex Numbers

The polar form of complex numbers provides a way to represent complex numbers using magnitude (modulus) and angle (argument). Transforming a complex number from its rectangular form \(a + bi\) to polar form involves:

• Finding the modulus \(r\) using \(r = \sqrt{a^2 + b^2}\).
• Determining the argument \(\theta\) using \(|\tan \theta| = |\frac{b}{a}|\).

The polar form is then expressed as \(z = r(\cos \theta + i\sin \theta)\), also noted as \(re^{i\theta}\) using Euler's formula. In the exercise, converting \(1 - \sqrt{3}i\) to polar form gives you \(2(\cos(-\frac{\pi}{3}) + i\sin(-\frac{\pi}{3}))\), showing how geometry and trigonometry converge to represent complex numbers concisely.

Rectangular Form of Complex Numbers

The rectangular form of a complex number is expressed as \(a + bi\), where \(a\) and \(b\) are real numbers. This form is useful for addition and subtraction of complex numbers because it aligns with the Cartesian coordinate system where \(a\) is the x-coordinate and \(b\) is the y-coordinate. In many applications, including in our exercise, the polar form is often converted back to rectangular form for interpretation or solution. Using De Moivre's theorem, the expression \(16(-\frac{1}{2} + i(-\frac{\sqrt{3}}{2}))\) was simplified back to its rectangular form \(-8 - 8\sqrt{3}i\), a clear way to present the real and imaginary parts after complex calculations.

Trigonometric Identities

Trigonometric identities play a key role in converting and simplifying expressions within complex number calculations. In polar form, the trigonometric functions \(\cos\) and \(\sin\) describe the angle and amplitude in parts like \(\cos(\theta) + i\sin(\theta)\). During calculations such as those performed by De Moivre's theorem, understanding identities like:

• \(\cos(2\pi + \theta) = \cos(\theta)\)
• \(\sin(2\pi + \theta) = \sin(\theta)\)

These help in recognizing periodicity and simplifying angles. In our example, simplifying \(\cos(-\frac{4\pi}{3})\) and \(\sin(-\frac{4\pi}{3})\) required using the periodicity properties of trigonometric functions to ensure the correct positioning on the unit circle. These identities support the transition between angles and help simplify expressions back to the synonymous rectangular forms.