Evaluating Complex Number Expressions Using Polar Form and De Moivre's Theorem

We are asked to evaluate the expression \({\left[ {\frac{{\sin \frac{{\rm{\pi }}}{6} + {\rm{i}}\left( {1 - \cos \frac{{\rm{\pi }}}{6}} \right)}}{{\sin \frac{{\rm{\pi }}}{6} - {\rm{i}}\left( {1 - \cos \frac{{\rm{\pi }}}{6}} \right)}}} \right]^3}\), where \({\rm{i}} = \sqrt { - 1}\).

First, let's find the values of the trigonometric functions at \( \frac{\pi}{6} \):

• \( \sin \frac{\pi}{6} = \frac{1}{2} \)
• \( \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2} \)

Substitute these values into the expression inside the bracket:

The numerator is \( \sin \frac{{\rm{\pi }}}{6} + {\rm{i}}\left( {1 - \cos \frac{{\rm{\pi }}}{6}} \right) = \frac{1}{2} + {\rm{i}}\left( {1 - \frac{\sqrt{3}}{2}} \right) = \frac{1}{2} + {\rm{i}}\frac{2 - \sqrt{3}}{2} \).

The denominator is \( \sin \frac{{\rm{\pi }}}{6} - {\rm{i}}\left( {1 - \cos \frac{{\rm{\pi }}}{6}} \right) = \frac{1}{2} - {\rm{i}}\left( {1 - \frac{\sqrt{3}}{2}} \right) = \frac{1}{2} - {\rm{i}}\frac{2 - \sqrt{3}}{2} \).

Let \( z = \frac{1}{2} + {\rm{i}}\frac{2 - \sqrt{3}}{2} \). Notice that the denominator is the complex conjugate of the numerator, \( \bar{z} \).

The expression inside the bracket is of the form \( \frac{z}{\bar{z}} \).

We can simplify this ratio using the polar form of complex numbers. If \( z = r e^{i\theta} \), then \( \bar{z} = r e^{-i\theta} \), and \( \frac{z}{\bar{z}} = \frac{r e^{i\theta}}{r e^{-i\theta}} = e^{i(2\theta)} = \cos(2\theta) + i \sin(2\theta) \). Here, \(\theta\) is the argument of \(z\).

Let's find the argument \(\theta\) for \( z = \frac{1}{2} + {\rm{i}}\frac{2 - \sqrt{3}}{2} \). The real part is \( A = \frac{1}{2} \) and the imaginary part is \( B = \frac{2 - \sqrt{3}}{2} \). Since both parts are positive, the complex number lies in the first quadrant.

The argument \(\theta\) is given by \( \tan \theta = \frac{B}{A} \):

\( \tan \theta = \frac{\frac{2 - \sqrt{3}}{2}}{\frac{1}{2}} = 2 - \sqrt{3} \).

We know that \( \tan \frac{\pi}{12} = 2 - \sqrt{3} \). Therefore, the argument of \(z\) is \( \theta = \frac{\pi}{12} \).

The argument of the ratio \( \frac{z}{\bar{z}} \) is \( 2\theta = 2 \times \frac{\pi}{12} = \frac{\pi}{6} \). The modulus of \( \frac{z}{\bar{z}} \) is \( \frac{|z|}{|\bar{z}|} = \frac{|z|}{|z|} = 1 \).

So, the expression inside the bracket is equal to \( \cos \frac{\pi}{6} + i \sin \frac{\pi}{6} \).

Now, we need to raise this expression to the power of 3. We can use De Moivre's Theorem, which states that \( (\cos \theta + i \sin \theta)^n = \cos(n\theta) + i \sin(n\theta) \).

Applying De Moivre's Theorem with \( \theta = \frac{\pi}{6} \) and \( n = 3 \):

\( {\left[ {\cos \frac{{\rm{\pi }}}{6} + {\rm{i}}\sin \frac{{\rm{\pi }}}{6}} \right]^3} = \cos \left( 3 \times \frac{{\rm{\pi }}}{6} \right) + {\rm{i}}\sin \left( 3 \times \frac{{\rm{\pi }}}{6} \right) \)

\( = \cos \frac{{\rm{\pi }}}{2} + {\rm{i}}\sin \frac{{\rm{\pi }}}{2} \)

Now, evaluate the trigonometric functions at \( \frac{\pi}{2} \):

• \( \cos \frac{\pi}{2} = 0 \)
• \( \sin \frac{\pi}{2} = 1 \)

So, the expression simplifies to \( 0 + {\rm{i}}(1) = {\rm{i}} \).

Step-by-Step Calculation

1. Identify the expression inside the bracket as a ratio of a complex number and its conjugate. Let \( z = \sin \frac{{\rm{\pi }}}{6} + {\rm{i}}\left( {1 - \cos \frac{{\rm{\pi }}}{6}} \right) \).
2. Calculate the values of \( \sin \frac{\pi}{6} \) and \( \cos \frac{\pi}{6} \): \( \sin \frac{\pi}{6} = \frac{1}{2} \), \( \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2} \).
3. Substitute these values into \(z\): \( z = \frac{1}{2} + {\rm{i}}\left( 1 - \frac{\sqrt{3}}{2} \right) = \frac{1}{2} + {\rm{i}}\frac{2 - \sqrt{3}}{2} \).
4. Recognize that the denominator is \( \bar{z} = \frac{1}{2} - {\rm{i}}\frac{2 - \sqrt{3}}{2} \).
5. Find the argument \(\theta\) of \(z\). \( \tan \theta = \frac{\text{Im}(z)}{\text{Re}(z)} = \frac{\frac{2 - \sqrt{3}}{2}}{\frac{1}{2}} = 2 - \sqrt{3} \).
6. Identify that \( \tan \frac{\pi}{12} = 2 - \sqrt{3} \), so \( \theta = \frac{\pi}{12} \).
7. The ratio \( \frac{z}{\bar{z}} \) has an argument of \( 2\theta = 2 \times \frac{\pi}{12} = \frac{\pi}{6} \) and a modulus of 1.
8. Express the ratio in polar form: \( \frac{z}{\bar{z}} = \cos \frac{\pi}{6} + i \sin \frac{\pi}{6} \).
9. Raise this polar form to the power of 3 using De Moivre's Theorem: \( {\left[ {\cos \frac{{\rm{\pi }}}{6} + {\rm{i}}\sin \frac{{\rm{\pi }}}{6}} \right]^3} = \cos \left( 3 \times \frac{{\rm{\pi }}}{6} \right) + {\rm{i}}\sin \left( 3 \times \frac{{\rm{\pi }}}{6} \right) \).
10. Simplify the angle: \( 3 \times \frac{\pi}{6} = \frac{\pi}{2} \).
11. Evaluate \( \cos \frac{\pi}{2} + i \sin \frac{\pi}{2} \). \( \cos \frac{\pi}{2} = 0 \), \( \sin \frac{\pi}{2} = 1 \).
12. The final result is \( 0 + i(1) = i \).

Revision Table: Key Concepts

Additional Information on Complex Numbers and Polar Form

Complex numbers are a fundamental concept in mathematics, widely used in fields like engineering, physics, and signal processing. The rectangular form \( a + bi \) is intuitive for addition and subtraction, while the polar form \( r(\cos \theta + i \sin \theta) \) or \( re^{i\theta} \) is particularly useful for multiplication, division, and exponentiation, thanks to De Moivre's Theorem and Euler's formula \( e^{i\theta} = \cos \theta + i \sin \theta \).

The ratio of a complex number and its conjugate, \( \frac{z}{\bar{z}} \), always lies on the unit circle in the complex plane. If \( z = r e^{i\theta} \), then \( \frac{z}{\bar{z}} = e^{i2\theta} \), which has a modulus of 1 and an argument of \( 2\theta \). This property significantly simplifies expressions of the form \( \left( \frac{z}{\bar{z}} \right)^n \), reducing the problem to finding the argument of \(z\) and applying De Moivre's Theorem.

Finding the argument \(\theta\) of \( z = a + bi \) involves calculating \( \tan \theta = \frac{b}{a} \). One must be careful to determine the correct quadrant for \(\theta\) based on the signs of \(a\) and \(b\) to get the principal argument, which is typically in the range \( (-\pi, \pi] \) or \( [0, 2\pi) \). In this problem, the real and imaginary parts of the numerator \( \frac{1}{2} + i\frac{2 - \sqrt{3}}{2} \) are both positive (\(2 - \sqrt{3} \approx 2 - 1.732 = 0.268 > 0\)), so the argument is in the first quadrant, and \( \arctan \left( \frac{2 - \sqrt{3}}{1} \right) = \frac{\pi}{12} \) is the correct principal argument.