The complex number `2^n/(1+i)^(2n)+(1+i)^(2n)/2^n , n epsilon I` is equal to

Options

A. 0
B. 2
C. `{ 1 + (-1)^(n)} 1^(n)`
D. None of these

Answer

To solve the problem, we need to simplify the expression:

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

where \( n \in \mathbb{I} \) (the set of imaginary numbers).

### Step 1: Simplify the expression

We start with the given expression:

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

### Step 2: Rationalize the denominator

To simplify further, we can rationalize the denominator of the first term. The conjugate of \( 1+i \) is \( 1-i \). Thus, we can multiply the numerator and denominator by \( (1-i)^{2n} \):

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

### Step 3: Simplify the denominator

Using the property \( (a+b)(a-b) = a^2 - b^2 \), we have:

\[ (1+i)^{2n} (1-i)^{2n} = |1+i|^{4n} = (2^{1/2})^{4n} = 2^{2n} \]

Thus, the denominator simplifies to:

\[ 2^{2n} \]

### Step 4: Combine the fractions

Now we can rewrite the expression as:

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

This simplifies to:

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

### Step 5: Factor out \( \frac{1}{2^n} \)

Factoring out \( \frac{1}{2^n} \):

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

### Step 6: Evaluate \( (1-i)^{2n} + (1+i)^{2n} \)

Using the polar form of complex numbers, we can express \( 1+i \) and \( 1-i \):

- \( 1+i = \sqrt{2} e^{i\pi/4} \)
- \( 1-i = \sqrt{2} e^{-i\pi/4} \)

Thus,

\[ (1+i)^{2n} = (\sqrt{2})^{2n} e^{i\frac{n\pi}{2}} = 2^n e^{i\frac{n\pi}{2}} \]
\[ (1-i)^{2n} = (\sqrt{2})^{2n} e^{-i\frac{n\pi}{2}} = 2^n e^{-i\frac{n\pi}{2}} \]

Adding these gives:

\[ (1-i)^{2n} + (1+i)^{2n} = 2^n \left( e^{-i\frac{n\pi}{2}} + e^{i\frac{n\pi}{2}} \right) = 2^n \cdot 2 \cos\left(\frac{n\pi}{2}\right) \]

### Step 7: Substitute back into the expression

Now substituting back, we get:

\[ \frac{1}{2^n} \cdot 2^n \cdot 2 \cos\left(\frac{n\pi}{2}\right) = 2 \cos\left(\frac{n\pi}{2}\right) \]

### Final Result

Thus, the final result is:

\[ 2 \cos\left(\frac{n\pi}{2}\right) \]