Minimum value of `n` for which `(2i)^n/(1-i)^(n-2)` is positive integer

Answer

To find the minimum value of \( n \) for which \( \frac{(2i)^n}{(1-i)^{n-2}} \) is a positive integer, we can follow these steps:

### Step 1: Simplify the expression
We start with the expression:
\[ \frac{(2i)^n}{(1-i)^{n-2}} \]
To simplify, we can multiply the numerator and denominator by \( (1-i)^2 \):
\[ \frac{(2i)^n (1-i)^2}{(1-i)^{n}} = (2i)^n (1-i)^{2-n} \]

### Step 2: Expand \( (1-i)^2 \)
Calculating \( (1-i)^2 \):
\[ (1-i)^2 = 1 - 2i + i^2 = 1 - 2i - 1 = -2i \]
Thus, we can rewrite our expression as:
\[ (2i)^n \cdot (-2i) \cdot (1-i)^{-n} \]

### Step 3: Combine the terms
Now, substituting \( -2i \) into the expression gives:
\[ (-2i)^{n-2} \cdot (1-i)^{-n} \]
This can be rewritten as:
\[ (-2)^{n-2} \cdot (i)^{n-2} \cdot (1-i)^{-n} \]

### Step 4: Convert to polar form
Next, we convert \( 1-i \) into polar form. The modulus of \( 1-i \) is:
\[ |1-i| = \sqrt{1^2 + (-1)^2} = \sqrt{2} \]
The argument (angle) of \( 1-i \) is:
\[ \theta = \tan^{-1}\left(\frac{-1}{1}\right) = -\frac{\pi}{4} \]
Thus, we can express \( 1-i \) as:
\[ 1-i = \sqrt{2} e^{-i\pi/4} \]
Therefore, \( (1-i)^{-n} \) becomes:
\[ (1-i)^{-n} = \frac{1}{(\sqrt{2})^n} e^{i n \pi/4} \]

### Step 5: Substitute back into the expression
Substituting back gives:
\[ (-2)^{n-2} \cdot (i)^{n-2} \cdot \frac{1}{(\sqrt{2})^n} e^{i n \pi/4} \]
This simplifies to:
\[ \frac{(-2)^{n-2}}{2^{n/2}} \cdot (i)^{n-2} \cdot e^{i n \pi/4} \]
Combining the powers of 2, we have:
\[ \frac{(-1)^{n-2} \cdot 2^{n/2 - 2}}{1} \cdot (i)^{n-2} \cdot e^{i n \pi/4} \]

### Step 6: Analyze the conditions for positivity
For the expression to be a positive integer, the imaginary part must be zero, and the real part must be positive. The imaginary part is given by:
\[ (-1)^{n-2} \cdot 2^{n/2 - 2} \cdot \text{Im}\left((i)^{n-2} e^{i n \pi/4}\right) \]
This means we need to ensure that:
1. \( n \) must be even (to make \( (-1)^{n-2} \) positive).
2. The angle \( \frac{n \pi}{4} \) must correspond to a real number.

### Step 7: Find the minimum \( n \)
To satisfy the conditions, we can test even values of \( n \):
- For \( n = 2 \): The expression evaluates to a negative.
- For \( n = 4 \): The expression evaluates to a positive integer.
- For \( n = 6 \): The expression evaluates to a positive integer.

Thus, the minimum value of \( n \) for which the expression is a positive integer is:
\[ \boxed{4} \]