Problem 70 - Algebra and Trigonometry

Question

Find the indicated power using De Moivre's Theorem. $$ (1-\sqrt{3} i)^{5} $$

Step-by-Step Solution

Answer

The result is \(16 + 16\sqrt{3}i\).

Step 1: Express the Complex Number in Polar Form

The complex number is given as \(1 - \sqrt{3}i\). To convert it to polar form, we find the modulus \(r\) and the argument \(\theta\). The modulus is \(r = \sqrt{1^2 + (-\sqrt{3})^2} = \sqrt{4} = 2\). The argument \(\theta\) can be determined using \(\tan \theta = \frac{-\sqrt{3}}{1}\), which gives \(\theta = \frac{5\pi}{3}\) (since the point is in the fourth quadrant). Thus, the polar form is \(2\left(\cos \frac{5\pi}{3} + i \sin \frac{5\pi}{3}\right)\).

Step 2: Apply De Moivre's Theorem

De Moivre's Theorem states that \((r(\cos(\theta) + i\sin(\theta)))^n = r^n (\cos(n\theta) + i\sin(n\theta))\). Here, \(r=2\), \(\theta=\frac{5\pi}{3}\), and \(n=5\). Applying the theorem: \[ (2^5) \left( \cos \left(5 \times \frac{5\pi}{3}\right) + i \sin \left(5 \times \frac{5\pi}{3}\right) \right) \].

Step 3: Simplify the Expression

First calculate \(2^5 = 32\). Next, simplify \(5 \times \frac{5\pi}{3} = \frac{25\pi}{3}\). Since \(\cos\) and \(\sin\) functions are periodic with a period of \(2\pi\), reduce \(\frac{25\pi}{3}\) by subtracting \(8\pi\) (which is \(2\pi \times 4\)): \(\frac{25\pi}{3} \text{ mod } 2\pi = \frac{\pi}{3}\). Thus, the expression becomes \[32 \left( \cos \frac{\pi}{3} + i \sin \frac{\pi}{3} \right)\].

Step 4: Evaluate the Trigonometric Functions

Now substitute the values for \(\cos\) and \(\sin\), knowing \(\cos \frac{\pi}{3} = \frac{1}{2}\) and \(\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}\). So the expression becomes: \[ 32 \left( \frac{1}{2} + i \frac{\sqrt{3}}{2} \right) = 32 \cdot \frac{1}{2} + 32 \cdot i \frac{\sqrt{3}}{2} \].

Step 5: Compute the Final Result

Perform the multiplication: \[ 32 \cdot \frac{1}{2} = 16 \] and \[ 32 \cdot \frac{\sqrt{3}}{2} = 16\sqrt{3} \]. Combine these results to obtain the complex number: \( 16 + 16\sqrt{3}i \).

Key Concepts

Complex Numbers

Complex numbers are an extension of the real numbers, allowing for the inclusion of an imaginary unit, denoted as \(i\), where \(i^2 = -1\). They can be represented as \(a + bi\), where \(a\) is the real part and \(b\) is the imaginary part. These numbers provide a comprehensive way to solve equations that do not have solutions in just the real numbers. For instance, the equation \(x^2 + 1 = 0\) has no real solution, but it does have complex solutions \(i\) and \(-i\).
In our exercise, \(1 - \sqrt{3}i\) is a complex number. Here, the real part is 1, and the imaginary part is \(-\sqrt{3}\). Understanding complex numbers is foundational for comprehending operations such as addition, subtraction, and multiplication within the complex plane.

Polar Coordinates

The polar coordinate system represents a complex number in terms of a modulus and an angle, known as \[ r(\cos \theta + i\sin \theta) \].Here, \(r\) is the distance from the origin in the complex plane, and \(\theta\) is the angle from the positive real axis. This representation is incredibly useful in simplifying the processes like raising a complex number to a power.
To convert the given complex number \(1 - \sqrt{3}i\) to polar form, we calculate its modulus \[ r = \sqrt{1^2 + (-\sqrt{3})^2} = 2\] and its argument \[ \theta = \tan^{-1} \left(\frac{-\sqrt{3}}{1}\right) = \frac{5\pi}{3} \].The calculation considers the quadrant where the complex number resides.
With this, the polar form becomes \(2(\cos \frac{5\pi}{3} + i \sin \frac{5\pi}{3})\).

Trigonometric Functions

Trigonometric functions like sine and cosine are crucial when dealing with angles in polar coordinates. They help in translating the angle into the respective positions on the unit circle. For example, at the angle \(\frac{\pi}{3}\), cosine gives \(\frac{1}{2}\) while sine provides \(\frac{\sqrt{3}}{2}\).
In the polar form \(2(\cos \frac{5\pi}{3} + i \sin \frac{5\pi}{3})\), these trigonometric functions guide us in finding the real and imaginary components of the resultant complex number. When multiplying or applying De Moivre's Theorem, these periodic functions make it easier to handle angles larger than \(2\pi\) by understanding symmetry and periodicity on the unit circle.

Algebraic Manipulation

Algebraic manipulation allows us to derive a simpler form of equations or expressions through techniques like distribution, factoring, or reduction. In the context of complex numbers and De Moivre's Theorem, it helps simplify the expression after raising to a power.
Starting from the polar form expression \(32(\cos \frac{\pi}{3} + i\sin \frac{\pi}{3})\), multiplying through by 32, we conduct the following steps:

• Calculate \(32 \times \frac{1}{2} = 16\)
• Calculate \(32 \times \frac{\sqrt{3}}{2} = 16\sqrt{3}\)

Through algebraic manipulations, combining these results gives us the final form of the complex number: \(16 + 16\sqrt{3}i\), illustrating both the power of algebraic manipulation and trigonometric understanding in handling complex numbers.