Problem 24 - Advanced Engineering Mathematics

Question

Use \((8)\) to compute the indicated power. $$ (-\sqrt{2}+\sqrt{6} i)^{4} $$

Step-by-Step Solution

Answer

The power is \(-32 - 32i\sqrt{3}\).

Step 1: Convert to Polar Form

First, express the complex number \(-\sqrt{2} + \sqrt{6} i\) in polar form. To do this, find the modulus \(r\) and the argument \(\theta\). The modulus is\[r = \sqrt{(-\sqrt{2})^2 + (\sqrt{6})^2} = \sqrt{2 + 6} = \sqrt{8} = 2\sqrt{2}\]and the argument is\[\theta = \tan^{-1}\left(\frac{\sqrt{6}}{-\sqrt{2}}\right) = \tan^{-1}(-\sqrt{3}) = -\frac{\pi}{3}.\]

Step 2: Use De Moivre's Theorem

According to De Moivre’s Theorem, for a complex number in polar form \(re^{i\theta}\), the \(n\)-th power is calculated as \[r^n e^{i n \theta}\].So,\[(-\sqrt{2}+\sqrt{6} i)^{4} = (2\sqrt{2})^{4} \left[\cos\left(4 \times -\frac{\pi}{3}\right) + i \sin\left(4 \times -\frac{\pi}{3}\right)\right].\]

Step 3: Calculate the Modulus Raised to the Power

Calculate the modulus \((2\sqrt{2})\) raised to the 4th power:\[(2\sqrt{2})^{4} = (2^4)(\sqrt{2})^4 = 16 \times 4 = 64.\]

Step 4: Compute the Argument Multiplication

Multiply the argument by 4:\[4 \times -\frac{\pi}{3} = -\frac{4\pi}{3}.\]

Step 5: Evaluate Trigonometric Functions

Evaluate the trigonometric functions:\[\cos\left(-\frac{4\pi}{3}\right) = -\frac{1}{2}\]\[\sin\left(-\frac{4\pi}{3}\right) = -\frac{\sqrt{3}}{2}.\]

Step 6: Construct the Complex Result

Using the results from Steps 3, 4, and 5, construct the complex number:\[64 \left(-\frac{1}{2} + i \left(-\frac{\sqrt{3}}{2}\right)\right) = 64 \left(-\frac{1}{2} - i \frac{\sqrt{3}}{2}\right).\]Distribute the 64:\[-32 - 32i\sqrt{3}.\]

Key Concepts

Complex Numbers

Complex numbers are numbers that have both a real part and an imaginary part. These are written in the form \(a + bi\), where \(a\) is the real part, and \(bi\) is the imaginary part, with \(i\) being the imaginary unit, satisfying \(i^2 = -1\). For example, in the complex number \(-\sqrt{2} + \sqrt{6}i\), \(-\sqrt{2}\) is the real part, while \(\sqrt{6}i\) is the imaginary part.
Complex numbers have a broad range of applications in fields such as engineering, physics, and computer science.
They are essential when dealing with equations that do not have real number solutions.

• They extend the idea of dimensional representation, allowing two-dimensional number space.
• Operations like addition, subtraction, multiplication, and division can be performed on complex numbers, turning them into a powerful mathematical tool.

As one becomes comfortable with complex numbers, it becomes easier to explore more advanced topics, such as transforming them into different forms.

Polar Form

Complex numbers can also be expressed in polar form, which is another representation that leverages the modulus and angle concepts.
This form is particularly helpful when multiplying or raising complex numbers to power.
In polar form, a complex number is represented as \(r(\cos \theta + i\sin \theta)\) or as \(re^{i\theta}\), where \(r\) is the modulus and \(\theta\) is the argument (or angle).

• Polar form simplifies complex number calculations, especially under operations like exponentiation.
• De Moivre's Theorem, which states that \((r e^{i\theta})^n = r^n e^{i n\theta}\), is more straightforward when using polar form.

This method is advantageous in both theoretical mathematics and applied sciences, particularly for solving polynomial equations and analyzing periodic functions.

Modulus and Argument

The modulus and argument are crucial components of a complex number, especially when converting to polar form.
The modulus of a complex number \(a + bi\) is its distance from the origin in the complex plane.
It is calculated using the formula \(r = \sqrt{a^2 + b^2}\). For the given complex number \(-\sqrt{2} + \sqrt{6}i\), the modulus is calculated as \(2\sqrt{2}\).
The argument is the angle formed with the real axis, determining the direction of the complex number on the plane. It can be found using the tangent function: \(\theta = \tan^{-1}\left(\frac{b}{a}\right)\). In this case, it yields \(-\frac{\pi}{3}\).

• These two parameters, modulus and argument, allow for easy transformation into polar form, which makes them invaluable in complex number math.
• Accurate calculation of these helps in the flawless application of De Moivre's Theorem, aiding in efficient power computations of complex numbers.

Understanding the concept of modulus and argument lays the groundwork for exploring and mastering polar coordinates and trigonometric functions in complex analysis.