What is `((sqrt3+i)/(sqrt3-i))^(6)` equal to ?

Options

A. `-1`
B. 0
C. 1
D. 2

Explanation

To solve the problem \(\left(\frac{\sqrt{3} + i}{\sqrt{3} - i}\right)^{6}\), 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 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:
\[ \frac{2 + 2\sqrt{3}i}{4} = \frac{1 + \sqrt{3}i}{2} \]

### Step 5: Write in exponential form
Next, we convert \(\frac{1 + \sqrt{3}i}{2}\) into exponential form. We find the modulus and argument:
- The modulus is:
\[ \left| \frac{1 + \sqrt{3}i}{2} \right| = \frac{1}{2} \sqrt{1^2 + (\sqrt{3})^2} = \frac{1}{2} \sqrt{1 + 3} = \frac{1}{2} \cdot 2 = 1 \]
- The argument (angle) \(\theta\) is given by:
\[ \tan^{-1}\left(\frac{\sqrt{3}}{1}\right) = \frac{\pi}{3} \]

Thus, we can express it as:
\[ \frac{1 + \sqrt{3}i}{2} = e^{i\frac{\pi}{3}} \]

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

### Step 7: Evaluate \(e^{i2\pi}\)
Using Euler's formula:
\[ e^{i2\pi} = \cos(2\pi) + i\sin(2\pi) = 1 + 0i = 1 \]

### Final Answer
Thus, the final answer is:
\[ \left(\frac{\sqrt{3} + i}{\sqrt{3} - i}\right)^{6} = 1 \]