If ` [ (sqrt(3) // 2 + (1 //2)i)/(sqrt(3) // 2 - (1//2)i)] ^(120) = a + ib ` then

Options

A. ` a = cos 20^(@) , b = sin 20^(@)`
B. ` a = - cos 20^(@), b = - sin 20^(@)`
C. ` a = cos 20^(@), b = - sin 20^(@)`
D. ` a = 1, b = 0 `

Answer

To solve the problem, we need to simplify the expression \( \left( \frac{\sqrt{3}/2 + (1/2)i}{\sqrt{3}/2 - (1/2)i} \right)^{120} \) and express it in the form \( a + ib \).

### Step-by-Step Solution:

1. **Identify the components of the complex number**:
We have:
\[ z = \frac{\sqrt{3}/2 + (1/2)i}{\sqrt{3}/2 - (1/2)i} \]

2. **Convert to polar form**:
To convert the complex number to polar form, we can express it in terms of its modulus and argument. We can rewrite the numerator and denominator:
- Numerator: \( \sqrt{3}/2 + (1/2)i \)
- Denominator: \( \sqrt{3}/2 - (1/2)i \)

The modulus of both the numerator and denominator is:
\[ r = \sqrt{\left(\frac{\sqrt{3}}{2}\right)^2 + \left(\frac{1}{2}\right)^2} = \sqrt{\frac{3}{4} + \frac{1}{4}} = \sqrt{1} = 1 \]

3. **Calculate the argument**:
- For the numerator:
\[ \theta_1 = \tan^{-1}\left(\frac{1/2}{\sqrt{3}/2}\right) = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6} \]
- For the denominator:
\[ \theta_2 = \tan^{-1}\left(\frac{-1/2}{\sqrt{3}/2}\right) = \tan^{-1}\left(-\frac{1}{\sqrt{3}}\right) = -\frac{\pi}{6} \]

4. **Find the argument of the quotient**:
The argument of \( z \) is:
\[ \theta = \theta_1 - \theta_2 = \frac{\pi}{6} - \left(-\frac{\pi}{6}\right) = \frac{\pi}{6} + \frac{\pi}{6} = \frac{\pi}{3} \]

5. **Express \( z \) in polar form**:
Since the modulus is 1, we can express \( z \) as:
\[ z = e^{i \frac{\pi}{3}} \]

6. **Raise to the power of 120**:
Now we raise \( z \) to the power of 120:
\[ z^{120} = \left(e^{i \frac{\pi}{3}}\right)^{120} = e^{i \frac{120\pi}{3}} = e^{i 40\pi} \]

7. **Simplify \( e^{i 40\pi} \)**:
Since \( e^{i 40\pi} \) corresponds to a full rotation (as \( e^{i 2\pi} = 1 \)), we have:
\[ e^{i 40\pi} = \cos(40\pi) + i \sin(40\pi) = 1 + 0i \]

8. **Identify \( a \) and \( b \)**:
Comparing with \( a + ib \), we find:
\[ a = 1, \quad b = 0 \]

### Final Answer:
Thus, the values of \( a \) and \( b \) are:
\[ a = 1, \quad b = 0 \]