If `xgt0`, the least value of `n in N` such that `((1+i)/(1-i))^(n)=(2)/(pi)sin^(-1)((1+x^(2))/(2x))` is :

Options

A. 2
B. 4
C. 8
D. 32

Explanation

To solve the problem, we need to find the least value of \( n \in \mathbb{N} \) such that

\[ \left(\frac{1+i}{1-i}\right)^n = \frac{2}{\pi} \sin^{-1}\left(\frac{1+x^2}{2x}\right) \]

### Step 1: Simplifying the Left Side

First, we simplify the left side:

\[ \frac{1+i}{1-i} \]

To simplify this, we can multiply the numerator and denominator by the conjugate of the denominator:

\[ \frac{(1+i)(1+i)}{(1-i)(1+i)} = \frac{1 + 2i - 1}{1 + 1} = \frac{2i}{2} = i \]

Thus, we have:

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

### Step 2: Analyzing the Right Side

Next, we analyze the right side:

\[ \frac{2}{\pi} \sin^{-1}\left(\frac{1+x^2}{2x}\right) \]

We need to find the range of the expression \( \frac{1+x^2}{2x} \). Using the AM-GM inequality:

\[ \frac{1+x^2}{2} \geq \sqrt{1 \cdot x^2} = x \]

Thus,

\[ \frac{1+x^2}{2x} \geq 1 \]

This means that \( \sin^{-1}\left(\frac{1+x^2}{2x}\right) \) will be defined for values greater than or equal to 1. The maximum value of \( \sin^{-1}(y) \) is \( \frac{\pi}{2} \) when \( y = 1 \).

### Step 3: Finding the Minimum Value

The minimum value of \( \frac{2}{\pi} \sin^{-1}\left(\frac{1+x^2}{2x}\right) \) occurs when \( x \to 0 \), which gives:

\[ \sin^{-1}(1) = \frac{\pi}{2} \]

Thus,

\[ \frac{2}{\pi} \cdot \frac{\pi}{2} = 1 \]

### Step 4: Equating Both Sides

Now we equate both sides:

\[ i^n = 1 \]

### Step 5: Finding the Least Value of \( n \)

The powers of \( i \) are:

- \( i^0 = 1 \)
- \( i^1 = i \)
- \( i^2 = -1 \)
- \( i^3 = -i \)
- \( i^4 = 1 \)

The least \( n \) such that \( i^n = 1 \) is \( n = 4 \).

### Conclusion

Thus, the least value of \( n \in \mathbb{N} \) such that

\[ \left(\frac{1+i}{1-i}\right)^n = \frac{2}{\pi} \sin^{-1}\left(\frac{1+x^2}{2x}\right) \]

is

\[ \boxed{4} \]