The smallest positive integer n for which `((1+i)/( 1-i))^(n) =1`, is `:`

Options

A. A) 1
B. B) 4
C. C) 8
D. D) 16

Answer

To solve the problem of finding 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:

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

Calculating the denominator:

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

### Step 3: Combine the results

Now we can combine the results from the numerator and denominator:

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

### Step 4: Set up the equation

Now we need to find \( n \) such that

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

### Step 5: Determine the powers of \( i \)

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

- \( i^1 = i \)
- \( i^2 = -1 \)
- \( i^3 = -i \)
- \( i^4 = 1 \)
- \( i^5 = i \) (and so on...)

### Step 6: Find the smallest \( n \)

From the cycle, we see that \( i^4 = 1 \). Therefore, the smallest positive integer \( n \) for which \( (i)^n = 1 \) is

\[ n = 4. \]

### Conclusion

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

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

is

\[ \boxed{4}. \]