Find the smallest positive integer n, for which `((1+i)/(1-i))^(n)=1`

Answer

To find the smallest positive integer \( n \) such that

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

we can follow these steps:

### Step 1: Simplify the expression \(\frac{1+i}{1-i}\)

To simplify \(\frac{1+i}{1-i}\), we can multiply the numerator and the denominator by the conjugate of the denominator:

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

### Step 2: Calculate the numerator and denominator

Calculating the numerator:

\[ (1+i)(1+i) = 1^2 + 2 \cdot 1 \cdot i + i^2 = 1 + 2i - 1 = 2i. \]

Calculating the denominator:

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

Thus, we have:

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

### Step 3: Substitute back into the equation

Now substituting back into the equation, we have:

\[ (i)^n = 1. \]

### Step 4: Determine when \( i^n = 1 \)

The powers of \( i \) cycle every 4:

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

This means \( i^n = 1 \) when \( n \) is a multiple of 4.

### Step 5: Find the smallest positive integer \( n \)

The smallest positive integer \( n \) that satisfies this condition is:

\[ n = 4. \]

### Conclusion

Thus, the smallest positive integer \( n \) such that

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

is

\[ \boxed{4}. \]

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