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

Options

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

Answer

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

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

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 this, 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**:
Now, we can combine the results of the numerator and denominator:

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

Therefore, we have:

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

5. **Raise to the Power of n**:
Now we need to find \( n \) such that:

\[ (-i)^n \]

is purely imaginary with a positive imaginary part.

6. **Analyze the Powers of -i**:
The powers of \( -i \) can be expressed as follows:

- \( (-i)^1 = -i \) (purely imaginary, negative)
- \( (-i)^2 = -1 \) (not purely imaginary)
- \( (-i)^3 = i \) (purely imaginary, positive)
- \( (-i)^4 = 1 \) (not purely imaginary)

We see that \( (-i)^3 = i \) is the first instance where the result is purely imaginary and positive.

7. **Conclusion**:
The smallest positive integral value of \( n \) for which \( \left(\frac{1-i}{1+i}\right)^n \) is purely imaginary with a positive imaginary part is:

\[ n = 3 \]

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