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

Answer

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

### Step 1: Simplify the fraction
We start by multiplying the numerator and the denominator by the conjugate of the denominator:

\[ \frac{\sqrt{3} + i}{\sqrt{3} - i} \cdot \frac{\sqrt{3} + i}{\sqrt{3} + i} = \frac{(\sqrt{3} + i)^2}{(\sqrt{3})^2 - (i)^2} \]

### Step 2: Calculate the denominator
Calculating the denominator:

\[ (\sqrt{3})^2 - (i)^2 = 3 - (-1) = 3 + 1 = 4 \]

### Step 3: Calculate the numerator
Now, we calculate the numerator:

\[ (\sqrt{3} + i)^2 = (\sqrt{3})^2 + 2(\sqrt{3})(i) + (i)^2 = 3 + 2\sqrt{3}i - 1 = 2 + 2\sqrt{3}i \]

### Step 4: Combine the results
Now we can combine the results:

\[ \frac{(\sqrt{3} + i)^2}{4} = \frac{2 + 2\sqrt{3}i}{4} = \frac{1 + \sqrt{3}i}{2} \]

### Step 5: Express in exponential form
Next, we express \(\frac{1 + \sqrt{3}i}{2}\) in exponential form. We can recognize this as a complex number in polar form.

The modulus \(r\) is:

\[ r = \sqrt{\left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} = \sqrt{\frac{1}{4} + \frac{3}{4}} = \sqrt{1} = 1 \]

The argument \(\theta\) is:

\[ \theta = \tan^{-1}\left(\frac{\sqrt{3}/2}{1/2}\right) = \tan^{-1}(\sqrt{3}) = \frac{\pi}{3} \]

Thus, we can write:

\[ \frac{1 + \sqrt{3}i}{2} = e^{i\frac{\pi}{3}} \]

### Step 6: 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 7: Evaluate \(e^{i2\pi}\)
Using Euler's formula, we know:

\[ e^{i2\pi} = \cos(2\pi) + i\sin(2\pi) = 1 + 0i = 1 \]

### Final Answer
Thus, the final answer is:

\[ \left(\frac{\sqrt{3} + i}{\sqrt{3} - i}\right)^{6} = 1 \]

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