The smallest positive integral value of n for which `((1-i)/(1+i))^(n)` is purely imaginary with positive imaginary part is

Options

A. 1
B. 3
C. 4
D. 5

Explanation

To solve the problem, we need to find the smallest positive integral value of \( n \) such that

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

is purely imaginary with a positive imaginary part.

### Step-by-Step Solution:

1. **Simplify the Expression**:
We start with the expression \(\frac{1-i}{1+i}\). To simplify it, we can multiply the numerator and the denominator by the conjugate of the denominator:

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

2. **Calculate the Numerator**:
The numerator becomes:

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

3. **Calculate the Denominator**:
The denominator becomes:

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

4. **Combine the Results**:
Thus, we have:

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

5. **Raise to the Power of n**:
Now we need to raise this result to the power of \( n \):

\[ \left(-i\right)^n \]

6. **Determine When it is Purely Imaginary**:
The expression \((-i)^n\) can be expressed in terms of \( i \):

- If \( n = 1 \), \((-i)^1 = -i\) (purely imaginary, negative part).
- If \( n = 2 \), \((-i)^2 = -1\) (not purely imaginary).
- If \( n = 3 \), \((-i)^3 = i\) (purely imaginary, positive part).
- If \( n = 4 \), \((-i)^4 = 1\) (not purely imaginary).
- If \( n = 5 \), \((-i)^5 = -i\) (purely imaginary, negative part).
- If \( n = 6 \), \((-i)^6 = -1\) (not purely imaginary).
- If \( n = 7 \), \((-i)^7 = i\) (purely imaginary, positive part).

7. **Find the Smallest Positive Integral n**:
From the above calculations, we can see that the smallest positive integral value of \( n \) for which \((-i)^n\) is purely imaginary with a positive imaginary part is:

\[ n = 3 \]

### Final Answer:
The smallest positive integral value of \( n \) is \( \boxed{3} \).