Problem 26 - Advanced Engineering Mathematics

Question

$$(\sqrt{3})^{6}\left[\cos \frac{12 \pi}{9}+i \sin \frac{12 \pi}{9}\right]=-\frac{27}{2}-\frac{27 \sqrt{3}}{2} i$$

Step-by-Step Solution

Answer

Yes, the given expression is correct in its Cartesian form.

Step 1: Interpreting the Expression

The given expression \((\sqrt{3})^{6}\left[\cos \frac{12 \pi}{9}+i \sin \frac{12 \pi}{9}\right]\)is written in polar form, where \((\sqrt{3})^{6}\) is the magnitude and the argument is given by \(\frac{12 \pi}{9}\). We need to simplify this expression.

Step 2: Simplifying the Magnitude

Calculate \((\sqrt{3})^{6}\):\[(\sqrt{3})^{6} = (3^{1/2})^6 = 3^{6/2} = 3^3 = 27.\]The magnitude of the polar form expression is \(27\).

Step 3: Simplifying the Argument

Simplify the argument \(\frac{12\pi}{9}\):\[\frac{12\pi}{9} = \frac{4\pi}{3}.\]Now the expression becomes \[27 \left[\cos \frac{4\pi}{3} + i \sin \frac{4\pi}{3}\right].\]

Step 4: Convert to Cartesian Form

Evaluate \(\cos \frac{4\pi}{3}\) and \(\sin \frac{4\pi}{3}\): - \(\cos \frac{4\pi}{3} = -\frac{1}{2}\)- \(\sin \frac{4\pi}{3} = -\frac{\sqrt{3}}{2}\)Thus, \[27 \left[-\frac{1}{2} + i \left(-\frac{\sqrt{3}}{2}\right)\right] = 27 \left(-\frac{1}{2} - i \frac{\sqrt{3}}{2}\right).\]

Step 5: Calculate the Cartesian Result

Multiply the magnitude by the Cartesian components:\[27 \left(-\frac{1}{2}\right) = -\frac{27}{2},\]\[27 \left(-\frac{\sqrt{3}}{2}\right) = -\frac{27 \sqrt{3}}{2},\]Thus, the Cartesian form is\[-\frac{27}{2} - \frac{27 \sqrt{3}}{2} i.\]

Step 6: Verify the Result

Compare the calculated result with the given expression \[-\frac{27}{2} - \frac{27 \sqrt{3}}{2} i.\]Since both expressions match, the calculation is correct.

Key Concepts

Polar Form

Polar form is a way to represent complex numbers using the magnitude and the angle they make with the positive real axis, known as the argument. Instead of writing a complex number as a sum of a real part and an imaginary part, polar form expresses it as a product of its magnitude and a trigonometric expression. This is particularly useful for multiplication and division of complex numbers.

In polar form, a complex number is typically represented as:

• Magnitude: The size or length of the vector, often noted as \( r \).
• Argument: The angle \( \theta \), measured in radians, between the positive real axis and the vector.

The expression \( r(\cos \theta + i\sin \theta) \) fully describes a complex number in polar form. The problem given uses this form initially with the magnitude \((\sqrt{3})^6\) and an argument \(\frac{12\pi}{9}\). Calculating these simplifies the problem considerably.

Cartesian Form

The Cartesian form of a complex number expresses it in terms of its real and imaginary parts, labeled as \( a+bi \), where \( a \) is the real part and \( b \) is the imaginary part. Cartesian form is often more straightforward for addition and subtraction operations.

To convert a complex number from polar to Cartesian form, you compute the cosine and sine of the argument and multiply these values by the magnitude:

• The real part \( a = r \cos \theta \)
• The imaginary part \( b = r \sin \theta \)

In the example, using the magnitude \(27\) and the simplified angle \(\frac{4\pi}{3}\), we find the Cartesian form to be \(-\frac{27}{2} - \frac{27 \sqrt{3}}{2} i\). These values are derived from evaluating the trigonometric functions for the argument \(\theta\).

Trigonometric Functions

Trigonometric functions play a crucial role in converting between polar and Cartesian forms. They are used to determine the real and imaginary components of a complex number from its polar representation.

When you have an angle \(\theta\), the cosine and sine functions help locate the point on the complex plane:

• \( \cos \theta \) gives you the horizontal component (how far along the real axis).
• \( \sin \theta \) provides the vertical component (how far along the imaginary axis).

In this problem, calculating \( \cos \frac{4\pi}{3} \) and \( \sin \frac{4\pi}{3} \) gives us the values used to express the complex number in Cartesian form. That's because these trigonometric values directly inform the real and imaginary parts of the expression.

Magnitude and Argument

The magnitude and argument are fundamental in understanding complex numbers' polar representation. The magnitude, sometimes called the modulus, is the distance from the origin to the point in the complex plane, and the argument, or phase, is the angle that vector forms with the positive x-axis.

The magnitude \( r \) is calculated as:

• It is simply \(|z|\), computed as \( \sqrt{a^2 + b^2} \) if you start from Cartesian coordinates.
• For this exercise, \((\sqrt{3})^6\) simplified the task with a magnitude \(27\).

The argument \( \theta \) involves trigonometry. It directs the angle, describing where the complex number lies in relation to the real axis. This helps in converting back to Cartesian form as demonstrated, where \( \frac{4\pi}{3} \) was reshaped from the original \( \frac{12\pi}{9} \). Understanding these allows for efficient movement between the representations.