`e^(2 ni cot^(-1)(x)) [ ( x i + 1)/( x i - 1)]^(n)`
where n and x are real numbers, is equal to

Options

A. `(n)/(2)`
B. `((n+1))/(2)`
C. 1
D. `e^(ix)`

Answer

To solve the expression \( e^{2ni \cot^{-1}(x)} \left( \frac{xi + 1}{xi - 1} \right)^{n} \), we will follow these steps:

### Step 1: Substitute \( \theta = \cot^{-1}(x) \)
Let \( \theta = \cot^{-1}(x) \). This implies that \( x = \cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)} \).

**Hint:** Remember that \( \cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)} \).

### Step 2: Rewrite the expression
Now, we can rewrite the expression in terms of \( \theta \):
\[ e^{2ni \theta} \left( \frac{xi + 1}{xi - 1} \right)^{n} \]

### Step 3: Substitute \( x \) in the fraction
Substituting \( x = \cot(\theta) \) into the fraction:
\[ \frac{xi + 1}{xi - 1} = \frac{\cot(\theta)i + 1}{\cot(\theta)i - 1} \]

### Step 4: Simplify the fraction
To simplify, multiply the numerator and denominator by \( \sin(\theta) \):
\[ = \frac{\cos(\theta)i + \sin(\theta)}{\cos(\theta)i - \sin(\theta)} \]

### Step 5: Recognize the form of the expression
This can be recognized as a form of the exponential function:
\[ = \frac{\sin(\theta) + i\cos(\theta)}{\sin(\theta) - i\cos(\theta)} = e^{i\theta} \]

### Step 6: Raise to the power of \( n \)
Now, raise this to the power \( n \):
\[ \left( e^{i\theta} \right)^{n} = e^{in\theta} \]

### Step 7: Combine the exponentials
Now we combine the exponentials:
\[ e^{2ni\theta} \cdot e^{in\theta} = e^{(2n + n)i\theta} = e^{3ni\theta} \]

### Step 8: Substitute back for \( \theta \)
Substituting back \( \theta = \cot^{-1}(x) \):
\[ e^{3ni \cot^{-1}(x)} \]

### Final Result
Thus, the expression simplifies to:
\[ e^{3ni \cot^{-1}(x)} \]