The least positive integer n for which `((1+i)/(1-i))^n= 2/pi sin^-1 ((1+x^2)/(2x))`, where `x> 0 and i=sqrt-1` is

Explanation

To solve the equation \(\left(\frac{1+i}{1-i}\right)^n = \frac{2}{\pi} \sin^{-1}\left(\frac{1+x^2}{2x}\right)\) where \(x > 0\) and \(i = \sqrt{-1}\), we will follow these steps:

### Step 1: Simplify the Left Side
We start with the expression \(\frac{1+i}{1-i}\). To simplify this, we can multiply the numerator and denominator by the conjugate of the denominator:

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

Calculating the numerator and denominator separately:

- Numerator:
\[ (1+i)^2 = 1^2 + 2(1)(i) + i^2 = 1 + 2i - 1 = 2i \]

- Denominator:
\[ (1-i)(1+i) = 1^2 - i^2 = 1 - (-1) = 2 \]

Thus, we have:
\[ \frac{1+i}{1-i} = \frac{2i}{2} = i \]

### Step 2: Raise to the Power of \(n\)
Now, we raise \(i\) to the power of \(n\):
\[ \left(\frac{1+i}{1-i}\right)^n = i^n \]

### Step 3: Set the Equation
Now we set our equation:
\[ i^n = \frac{2}{\pi} \sin^{-1}\left(\frac{1+x^2}{2x}\right) \]

### Step 4: Analyze the Right Side
We need to analyze the right side, \(\frac{2}{\pi} \sin^{-1}\left(\frac{1+x^2}{2x}\right)\). The expression \(\frac{1+x^2}{2x}\) can be rewritten as:
\[ \frac{1+x^2}{2x} = \frac{1}{2x} + \frac{x}{2} \]

This expression is valid for \(x > 0\) and can be shown to be equal to 1 when \(x = 1\):
\[ \frac{1+1^2}{2 \cdot 1} = \frac{2}{2} = 1 \]

Thus, \(\sin^{-1}(1) = \frac{\pi}{2}\).

### Step 5: Substitute Back
Substituting back, we have:
\[ \frac{2}{\pi} \cdot \frac{\pi}{2} = 1 \]

### Step 6: Equate Powers of \(i\)
Now we have:
\[ i^n = 1 \]

### Step 7: Determine \(n\)
The powers of \(i\) cycle every 4:
- \(i^0 = 1\)
- \(i^1 = i\)
- \(i^2 = -1\)
- \(i^3 = -i\)
- \(i^4 = 1\)

Thus, \(i^n = 1\) when \(n\) is a multiple of 4. The least positive integer \(n\) that satisfies this is:
\[ n = 4 \]

### Final Answer
The least positive integer \(n\) for which \(\left(\frac{1+i}{1-i}\right)^n = \frac{2}{\pi} \sin^{-1}\left(\frac{1+x^2}{2x}\right)\) is:
\[ \boxed{4} \]