Problem 66 - Algebra and Trigonometry

Question

\(65-76=\) Find the indicated power using DeMoivre's Theorem. $$ (1-\sqrt{3} i)^{5} $$

Step-by-Step Solution

Answer

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

Step 1: Convert to Polar Form

To use DeMoivre's Theorem, we first convert the complex number to polar form. The complex number is given as \(z = 1 - \sqrt{3}i\). Its modulus is found using the formula \(|z| = \sqrt{a^2 + b^2}\), where \(a = 1\) and \(b = -\sqrt{3}\). This gives \(|z| = \sqrt{1 + 3} = 2\). The argument \(\theta\) is given by \(\theta = \tan^{-1}\left(\frac{b}{a}\right) = \tan^{-1}\left(-\sqrt{3}\right) = -\frac{\pi}{3}\). Hence, \(z\) in polar form is \(2\left(\cos\left(-\frac{\pi}{3}\right) + i\sin\left(-\frac{\pi}{3}\right)\right)\).

Step 2: Apply DeMoivre's Theorem

DeMoivre's Theorem states that for a complex number \(r(\cos \theta + i\sin \theta)\), its \(n^{th}\) power is \(r^n(\cos(n\theta) + i\sin(n\theta))\). Here \(r = 2\), \(\theta = -\frac{\pi}{3}\), and \(n = 5\). Thus, \((1-\sqrt{3}i)^5 = 2^5\left(\cos\left(5 \times -\frac{\pi}{3}\right) + i\sin\left(5 \times -\frac{\pi}{3}\right)\right)\).

Step 3: Calculate the Power of the Modulus

Calculate \(r^n\), where \(r = 2\) and \(n = 5\). So, \(r^n = 32\).

Step 4: Compute the New Argument

Compute the scaled argument: \(5\theta = 5 \times -\frac{\pi}{3} = -\frac{5\pi}{3}\). Since this angle is outside the principal range of \([-\pi, \pi]\), we adjust it to an equivalent angle by adding \(2\pi\), resulting in \(-\frac{5\pi}{3} + 2\pi = \frac{\pi}{3}\).

Step 5: Express the Final Result

Substitute the new argument back into the expression: \((1-\sqrt{3}i)^5 = 32\left(\cos\left(\frac{\pi}{3}\right) + i\sin\left(\frac{\pi}{3}\right)\right)\). As \(\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}\) and \(\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}\), the final result is \(32\left(\frac{1}{2} + i\frac{\sqrt{3}}{2}\right)\). This simplifies to \(16 + 16\sqrt{3}i\).

Key Concepts

Complex Numbers

Complex numbers are numbers in the form of \(a + bi\), where \(a\) and \(b\) are real numbers, and \(i\) is the imaginary unit with the property that \(i^2 = -1\). This means complex numbers are composed of two parts: a real part \(a\) and an imaginary part \(b\).
Complex numbers are helpful because they allow us to solve equations that wouldn't have solutions within the realm of real numbers alone.

• The real part \(a\) contributes to the 'horizontal' component on the complex plane.
• The imaginary part \(b\) incorporates the 'vertical' component, represented as \(bi\).

A complex number like \(1 - \sqrt{3}i\) indicates a real part of 1 and an imaginary part of \(-\sqrt{3}\). Understanding this format is the foundation for more advanced operations like converting these numbers to polar form or calculating powers using DeMoivre's Theorem.

Polar Form

Polar form is a way of expressing complex numbers using a modulus and an angle. Instead of the rectangular form \(a + bi\), we use \(r(\cos \theta + i\sin \theta)\), where \(r\) is the modulus, and \(\theta\) is the argument or angle.
By converting complex numbers to polar form, we can easily perform multiplication, division, and find powers of complex numbers.

• The modulus \(r\) is the distance from the origin to the point \((a, b)\) in the complex plane.
• The argument \(\theta\) is the angle formed with the positive x-axis.

For the complex number \(1 - \sqrt{3}i\), the polar form is found by first calculating the modulus which is \(\sqrt{1^2 + (-\sqrt{3})^2} = 2\), and the argument \(\theta = \tan^{-1}\left(-\sqrt{3}\right) = -\frac{\pi}{3}\). This transforms the number into the polar form \(2(\cos(-\frac{\pi}{3}) + i\sin(-\frac{\pi}{3}))\).

Modulus and Argument

The modulus and argument are crucial components of a complex number in polar form. They provide a different perspective that is often more suitable for multiplication and exponentiation.

• The modulus \(|z|\) is calculated as \(\sqrt{a^2 + b^2}\). For our example, \(a = 1\) and \(b = -\sqrt{3}\), resulting in a modulus of 2.
• The argument \(\theta\) is computed using \(\tan^{-1}\left(\frac{b}{a}\right)\). In this case, \(\theta = \tan^{-1}(-\sqrt{3}) = -\frac{\pi}{3}\).

These components not only allow the conversion into polar form but also streamline the computations in DeMoivre's Theorem, making it possible to determine powers of complex numbers with ease.

Complex Number Powers

DeMoivre's Theorem provides a straightforward method to compute powers of complex numbers when in polar form. It states that for any complex number expressed as \(r(\cos \theta + i\sin \theta)\), the \(n^{th}\) power is formulated as \(r^n(\cos(n\theta) + i\sin(n\theta))\).
This theorem helps to bridge the gap between the complex and real number systems, making computations involving powers far simpler than before.

• In this exercise, \(r = 2\), \(\theta = -\frac{\pi}{3}\), and \(n = 5\).
• The modulus power would be \(2^5 = 32\).
• The argument scales by multiplying \(\theta\) by \(n\) to get \(-\frac{5\pi}{3}\).

After adjusting the argument within the typical range, the new angle becomes \(\frac{\pi}{3}\). Therefore the complex number power results in \(32(\cos(\frac{\pi}{3}) + i\sin(\frac{\pi}{3}))\), which simplifies to \(16 + 16\sqrt{3}i\). This showcases the power and simplicity that DeMoivre's Theorem offers when dealing with complex numbers.