What is `((sqrt(3)+i)/(sqrt(3)-i))^(6)` equal to, where `I = sqrt(-1)` ?

Options

A. 1
B. `1//6`
C. 6
D. 2

Explanation

To solve the expression \(\left(\frac{\sqrt{3}+i}{\sqrt{3}-i}\right)^{6}\), where \(i = \sqrt{-1}\), we will follow these steps:

### Step 1: Rationalize the denominator
We start with the expression:
\[ \frac{\sqrt{3}+i}{\sqrt{3}-i} \]
To rationalize the denominator, we multiply the numerator and the denominator by the conjugate of the denominator:
\[ \frac{(\sqrt{3}+i)(\sqrt{3}+i)}{(\sqrt{3}-i)(\sqrt{3}+i)} \]

### Step 2: Simplify the denominator
Calculating the denominator:
\[ (\sqrt{3}-i)(\sqrt{3}+i) = \sqrt{3}^2 - i^2 = 3 - (-1) = 3 + 1 = 4 \]

### Step 3: Simplify the numerator
Calculating the numerator:
\[ (\sqrt{3}+i)(\sqrt{3}+i) = \sqrt{3}^2 + 2\sqrt{3}i + i^2 = 3 + 2\sqrt{3}i - 1 = 2 + 2\sqrt{3}i \]

### Step 4: Combine the results
Now we can combine the results from the numerator and denominator:
\[ \frac{2 + 2\sqrt{3}i}{4} = \frac{1 + \sqrt{3}i}{2} \]

### Step 5: Express in polar form
Next, we express \(\frac{1 + \sqrt{3}i}{2}\) in polar form. We can identify the real part \(x = \frac{1}{2}\) and the imaginary part \(y = \frac{\sqrt{3}}{2}\).

The modulus \(r\) is given by:
\[ r = \sqrt{x^2 + y^2} = \sqrt{\left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} = \sqrt{\frac{1}{4} + \frac{3}{4}} = \sqrt{1} = 1 \]

The argument \(\theta\) is given by:
\[ \theta = \tan^{-1}\left(\frac{y}{x}\right) = \tan^{-1}\left(\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}\right) = \tan^{-1}(\sqrt{3}) = \frac{\pi}{3} \]

Thus, we can write:
\[ \frac{1 + \sqrt{3}i}{2} = 1 \left( \cos\frac{\pi}{3} + i\sin\frac{\pi}{3} \right) \]

### Step 6: Use Euler's formula
Using Euler's formula, we have:
\[ \frac{1 + \sqrt{3}i}{2} = e^{i\frac{\pi}{3}} \]

### Step 7: Raise to the power of 6
Now we raise this to the power of 6:
\[ \left(e^{i\frac{\pi}{3}}\right)^{6} = e^{i\frac{6\pi}{3}} = e^{i2\pi} \]

### Step 8: Simplify the result
Since \(e^{i2\pi} = 1\), we conclude:
\[ \left(\frac{\sqrt{3}+i}{\sqrt{3}-i}\right)^{6} = 1 \]

### Final Answer
Thus, the final answer is:
\[ \boxed{1} \]