Problem 20 - Advanced Engineering Mathematics

Question

$$\frac{2 \sqrt{2}\left[\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}\right]}{2\left[\cos \frac{2 \pi}{3}+i \sin \frac{2 \pi}{3}\right]}=\sqrt{2}\left[\cos \left(\frac{\pi}{3}-\frac{2 \pi}{3}\right)+i \sin \left(\frac{\pi}{3}-\frac{2 \pi}{3}\right)\right]=\frac{\sqrt{2}}{2}-\frac{\sqrt{6}}{2} i$$

Step-by-Step Solution

Answer

The expression simplifies to \( \frac{\sqrt{2}}{2} - \frac{\sqrt{6}}{2} i \).

Step 1: Simplify the Given Expression

The expression given is \( \frac{2 \sqrt{2} [\cos \frac{\pi}{3} + i \sin \frac{\pi}{3}]}{2 [\cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3}]} \). To simplify this, first, the 2's in the numerator and denominator can be canceled out, leaving \( \frac{\sqrt{2} [\cos \frac{\pi}{3} + i \sin \frac{\pi}{3}]}{[\cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3}]} \).

Step 2: Apply De Moivre's Theorem for Division

To divide two complex numbers in polar form \( z_1 = r_1 (\cos \theta_1 + i \sin \theta_1) \) and \( z_2 = r_2 (\cos \theta_2 + i \sin \theta_2) \), the formula is: \( \frac{r_1}{r_2} [\cos (\theta_1 - \theta_2) + i \sin (\theta_1 - \theta_2)] \). Applying this to our expression: \[ \sqrt{2} \left[ \cos \left( \frac{\pi}{3} - \frac{2\pi}{3} \right) + i \sin \left( \frac{\pi}{3} - \frac{2\pi}{3} \right) \right] \].

Step 3: Compute the Angles

Compute \( \theta_1 - \theta_2 = \frac{\pi}{3} - \frac{2\pi}{3} = -\frac{\pi}{3} \). Substitute this back into the equation: \[ \sqrt{2} \left[ \cos \left( -\frac{\pi}{3} \right) + i \sin \left( -\frac{\pi}{3} \right) \right] \].

Step 4: Evaluate the Trigonometric Functions

Utilize the trigonometric identities: \( \cos(-\theta) = \cos(\theta) \) and \( \sin(-\theta) = -\sin(\theta) \). Thus we have:\( \cos(-\frac{\pi}{3}) = \frac{1}{2} \)\( \sin(-\frac{\pi}{3}) = -\frac{\sqrt{3}}{2} \).

Step 5: Substitute Back and Simplify

Substituting back, we obtain:\[ \sqrt{2} \left( \cos \left( -\frac{\pi}{3} \right) + i \sin \left( -\frac{\pi}{3} \right) \right) = \sqrt{2} \left( \frac{1}{2} - i \frac{\sqrt{3}}{2} \right) \]Simplify to get:\[ \frac{\sqrt{2}}{2} - i \frac{\sqrt{6}}{2} \].

Key Concepts

Trigonometry

The world of trigonometry deals with the relationships between the angles and sides of triangles. It is essential in handling complex numbers because these numbers often involve calculations with angles. Trigonometric functions like cosine and sine become tools to express complex numbers in a more accessible form.
When you encounter expressions like \( \cos \theta + i \sin \theta \), they represent points on the unit circle. This relationship is fundamental to understanding how we rotate or transform complex numbers. Trigonometric identities, such as \( \cos(-\theta) = \cos(\theta) \) and \( \sin(-\theta) = -\sin(\theta) \), help simplify expressions. These properties were pivotal in the original problem, where they helped simplify the angle calculations.

De Moivre's Theorem

This theorem is a powerful tool for raising complex numbers to powers or finding their roots in polar form. It states that for a complex number \( z = r(\cos \theta + i \sin \theta) \), its power can be calculated as \( z^n = r^n(\cos(n\theta) + i \sin(n\theta)) \).
In the context of division, as seen in the step-by-step solution, De Moivre’s Theorem helps us manage two complex numbers. It allows us to express the division of complex numbers using subtraction of their angles, as demonstrated by the formula used: \( \frac{r_1}{r_2} \left[ \cos (\theta_1 - \theta_2) + i \sin (\theta_1 - \theta_2) \right] \). This simplifies both the computation and conceptual understanding, especially when dealing with negative angles or fractions of \( \pi \).

Polar Form of Complex Numbers

The polar form is a way of presenting complex numbers which utilizes both a magnitude \( r \) and an angle \( \theta \). This form is especially useful because it simplifies multiplication and division through easy manipulation of angles and magnitudes, based on trigonometric identities.
For instance, in the given exercise, you saw complex numbers expressed as \( r(\cos \theta + i \sin \theta) \). Here, \( r \) is the modulus representing the distance from the origin to the point, and \( \theta \) is the argument, indicating the angle relative to the positive x-axis. This allows us to elegantly handle operations like division using De Moivre's Theorem, by simply subtracting or adding angles, and dividing or multiplying magnitudes, making complex arithmetic more intuitive.