What is the smallest positive integer n for which `(1+i)^(2n)=(1-i)^(2n)` ?

Answer

To solve the equation \((1+i)^{2n} = (1-i)^{2n}\), we will follow these steps:

### Step 1: Rewrite the equation
We start with the given equation:
\[ (1+i)^{2n} = (1-i)^{2n} \]
We can rewrite this as:
\[ \frac{(1+i)^{2n}}{(1-i)^{2n}} = 1 \]

### Step 2: Simplify the fraction
This can be simplified to:
\[ \left(\frac{1+i}{1-i}\right)^{2n} = 1 \]

### Step 3: Rationalize the denominator
Next, we rationalize the denominator:
\[ \frac{1+i}{1-i} \cdot \frac{1+i}{1+i} = \frac{(1+i)^2}{(1-i)(1+i)} \]
Calculating the denominator:
\[ (1-i)(1+i) = 1^2 - i^2 = 1 - (-1) = 2 \]
Calculating the numerator:
\[ (1+i)^2 = 1^2 + 2(1)(i) + i^2 = 1 + 2i - 1 = 2i \]
Thus, we have:
\[ \frac{1+i}{1-i} = \frac{2i}{2} = i \]
So, we can rewrite our equation as:
\[ (i)^{2n} = 1 \]

### Step 4: Solve for \(n\)
We know that \(i^4 = 1\). Therefore, \(i^{2n} = 1\) implies that \(2n\) must be a multiple of 4. This can be expressed as:
\[ 2n = 4k \quad \text{for some integer } k \]
Dividing both sides by 2 gives:
\[ n = 2k \]

### Step 5: Find the smallest positive integer \(n\)
The smallest positive integer \(n\) occurs when \(k = 1\):
\[ n = 2 \cdot 1 = 2 \]

Thus, the smallest positive integer \(n\) for which \((1+i)^{2n} = (1-i)^{2n}\) is:
\[ \boxed{2} \]