If the number `((1 - i)^(n))/( (1 + i)^(n - 2))` is real and positive , then n is
Put the following numbers in trigonometrical form, that is, int he form `( r, theta)` where r is a positive real number and ` - pi lt theta le pi`

Options

A. any integer
B. any even integer
C. any odd integer
D. none of these

Answer

To solve the problem, we need to determine the values of \( n \) for which the expression

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

is real and positive. We will analyze the expression step by step.

### Step 1: Convert \( 1 - i \) and \( 1 + i \) to Trigonometric Form

First, we need to express \( 1 - i \) and \( 1 + i \) in polar (trigonometric) form.

1. **For \( 1 - i \)**:
- The modulus \( r \) is given by:
\[ r = \sqrt{1^2 + (-1)^2} = \sqrt{2} \]
- The argument \( \theta \) is:
\[ \theta = \tan^{-1}\left(\frac{-1}{1}\right) = -\frac{\pi}{4} \]
- Therefore, in trigonometric form:
\[ 1 - i = \sqrt{2} \left( \cos\left(-\frac{\pi}{4}\right) + i \sin\left(-\frac{\pi}{4}\right) \right) \]

2. **For \( 1 + i \)**:
- The modulus \( r \) is:
\[ r = \sqrt{1^2 + 1^2} = \sqrt{2} \]
- The argument \( \theta \) is:
\[ \theta = \tan^{-1}\left(\frac{1}{1}\right) = \frac{\pi}{4} \]
- Therefore, in trigonometric form:
\[ 1 + i = \sqrt{2} \left( \cos\left(\frac{\pi}{4}\right) + i \sin\left(\frac{\pi}{4}\right) \right) \]

### Step 2: Substitute into the Expression

Now we substitute these forms into the original expression:

\[ \frac{(1 - i)^n}{(1 + i)^{n - 2}} = \frac{\left(\sqrt{2} \text{cis}\left(-\frac{\pi}{4}\right)\right)^n}{\left(\sqrt{2} \text{cis}\left(\frac{\pi}{4}\right)\right)^{n - 2}} \]

This simplifies to:

\[ \frac{(\sqrt{2})^n \text{cis}\left(-\frac{n\pi}{4}\right)}{(\sqrt{2})^{n - 2} \text{cis}\left(\frac{(n - 2)\pi}{4}\right)} = \frac{(\sqrt{2})^n}{(\sqrt{2})^{n - 2}} \cdot \text{cis}\left(-\frac{n\pi}{4} - \frac{(n - 2)\pi}{4}\right) \]

### Step 3: Simplify the Expression

The modulus simplifies to:

\[ (\sqrt{2})^2 = 2 \]

The argument simplifies to:

\[ -\frac{n\pi}{4} - \frac{(n - 2)\pi}{4} = -\frac{n\pi + (n - 2)\pi}{4} = -\frac{2n - 2}{4}\pi = -\frac{(n - 1)\pi}{2} \]

Thus, we have:

\[ 2 \text{cis}\left(-\frac{(n - 1)\pi}{2}\right) \]

### Step 4: Condition for Real and Positive

For the expression to be real and positive, the argument must be \( 0 \) or \( 2k\pi \) for some integer \( k \). This gives us:

\[ -\frac{(n - 1)\pi}{2} = 2k\pi \]

Solving for \( n \):

\[ n - 1 = -4k \implies n = 1 - 4k \]

### Step 5: Determine Possible Values of \( n \)

From this equation, we see that \( n \) can take values of the form \( 1, -3, -7, \ldots \) (i.e., \( n \) can be any integer of the form \( 1 - 4k \)).

### Conclusion

Thus, \( n \) can be any integer of the form \( 1 - 4k \), where \( k \) is an integer.