Problem 26 - Advanced Engineering Mathematics

Question

In Problems 21-26, use (8) to compute the indicated power. $$ \left[\sqrt{3}\left(\cos \frac{2 \pi}{9}+i \sin \frac{2 \pi}{9}\right)\right]^{6} $$

Step-by-Step Solution

Answer

The simplified expression is \(-\frac{27}{2} - i \frac{27\sqrt{3}}{2}\).

Step 1: Understand the Expression

The given expression is of the form \([r(\cos \theta + i \sin \theta)]^n\), where \(r = \sqrt{3}\), \(\theta = \frac{2\pi}{9}\), and \(n = 6\). We will use the formula for powers of complex numbers in polar form: \[[r(\cos \theta + i \sin \theta)]^n = r^n [\cos(n\theta) + i \sin(n\theta)]\]

Step 2: Calculate \(r^n\)

Calculate \(r^n\) using \(r = \sqrt{3}\) and \(n = 6\):\[(\sqrt{3})^6 = (3^{1/2})^6 = 3^{3} = 27\]

Step 3: Calculate \(n\theta\)

Calculate \(n\theta\) using \( \theta = \frac{2\pi}{9} \) and \( n = 6 \):\[n\theta = 6 \cdot \frac{2\pi}{9} = \frac{12\pi}{9} = \frac{4\pi}{3}\]

Step 4: Express Complex Number in Standard Form

Use the results from the previous steps to express the complex number:\[\text{Final expression} = 27 [\cos(\frac{4\pi}{3}) + i \sin(\frac{4\pi}{3})]\]Knowing that \(\cos(\frac{4\pi}{3}) = -\frac{1}{2}\) and \(\sin(\frac{4\pi}{3}) = -\frac{\sqrt{3}}{2}\), we have:\[= 27 \left(-\frac{1}{2} + i (-\frac{\sqrt{3}}{2})\right)\]Simplify:\[= 27 (-\frac{1}{2} - i \frac{\sqrt{3}}{2}) = -\frac{27}{2} - i \frac{27\sqrt{3}}{2}\]

Step 5: Conclusion

The expression simplifies to:\[-\frac{27}{2} - i \frac{27\sqrt{3}}{2}\]

Key Concepts

Polar Form

Complex numbers are typically written in the form \(a + bi\), where \(a\) is the real part and \(b\) is the imaginary part. However, it can be easier to work with complex numbers when they are expressed in polar form. This form uses a magnitude \(r\) (or modulus) and an angle \(\theta\) (or argument). The polar form of a complex number \(z\) is written as:

\[ z = r(\cos \theta + i \sin \theta) \]
Here’s why polar form is useful:

• Magnitude: The magnitude \(r\) gives you the length of the vector representing the complex number.
• Angle: The angle \(\theta\) tells you the direction of the vector.

To convert a complex number from its standard form \(a + bi\) to polar form, you need:

• Magnitude: \(r = \sqrt{a^2 + b^2}\)
• Angle: \(\theta = \text{atan2}(b, a)\)

The polar form simplifies operations like multiplication and exponentiation, making it particularly useful in problems such as raising a complex number to a power.

De Moivre's Theorem

De Moivre's Theorem is a powerful tool for working with complex numbers, especially when they are in polar form. This theorem simplifies the process of raising a complex number to a power. De Moivre's Theorem states that:

\[ (r(\cos \theta + i \sin \theta))^n = r^n (\cos(n\theta) + i \sin(n\theta)) \]
This theorem provides a straightforward way to calculate powers of complex numbers:

• Apply the Power: Raise the modulus \(r^n\).
• Multiply the Angle: Multiply the angle \(\theta\) by the exponent \(n\).

For example, if you wish to compute \((\sqrt{3}(\cos \frac{2\pi}{9} + i \sin \frac{2\pi}{9}))^6\), De Moivre's theorem tells us:

• Compute \(\sqrt{3}^6\) to find the new modulus, which equals 27.
• Calculate \(6 \times \frac{2\pi}{9}\) to find the new angle, resulting in \(\frac{4\pi}{3}\).

Thus, the theorem allows you to express complex powers simply, a task that would otherwise involve tedious calculations if done purely in rectangular form.

Complex Number Powers

Finding the power of a complex number is much more manageable when using the polar form and De Moivre's theorem. When dealing with complex number powers:

• Initial Form: Start by converting the complex number to its polar representation \(r(\cos \theta + i \sin \theta)\).
• Application of De Moivre's Theorem: Use the theorem to compute the power efficiently.
• Simplification: Finally, simplify the expression to reach the standard form, if needed.

In our exercise, we were tasked to compute the power \(\left[\sqrt{3}(\cos \frac{2\pi}{9}+i \sin \frac{2\pi}{9})\right]^{6}\):

• Step 1: Convert the angle and raise the modulus \(\sqrt{3}\) to the power of 6, resulting in 27.
• Step 2: Multiply the angle \(\frac{2\pi}{9}\) by 6, giving us \(\frac{4\pi}{3}\).
• Step 3: Use trigonometric identities to compute \(\cos(\frac{4\pi}{3}) = -\frac{1}{2}\) and \(\sin(\frac{4\pi}{3}) = -\frac{\sqrt{3}}{2}\).
• Step 4: Simplify the polar form into its standard complex form: \(-\frac{27}{2} - i \frac{27\sqrt{3}}{2}\).

This systematic approach allows for quick, clean computation of complex number powers without much room for error.