`(2^(n))/( (1 - i)^(2 n)) + ((1 + i )^(2 n))/( 2^(n)) , n in I` is equal to

Options

A. 2
B. 0
C. `[1 + (-1)^(n)]i^(n)`
D. `(2)/((-i)^(n))`

Answer

To solve the expression

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

we will simplify each term step by step.

### Step 1: Simplifying the first term

We start with the first term:

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

To simplify \((1 - i)^{2n}\), we can express \(1 - i\) in polar form. The modulus of \(1 - i\) is:

\[ |1 - i| = \sqrt{1^2 + (-1)^2} = \sqrt{2} \]

The argument (angle) \(\theta\) of \(1 - i\) is:

\[ \theta = \tan^{-1}\left(\frac{-1}{1}\right) = -\frac{\pi}{4} \]

Thus, we can write:

\[ 1 - i = \sqrt{2} \left(\cos\left(-\frac{\pi}{4}\right) + i \sin\left(-\frac{\pi}{4}\right)\right) \]

Using De Moivre's theorem, we have:

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

### Step 2: Substituting back into the first term

Now substituting back into the first term:

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

### Step 3: Simplifying the second term

Now we simplify the second term:

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

Similarly, we can express \(1 + i\) in polar form. The modulus of \(1 + i\) is:

\[ |1 + i| = \sqrt{2} \]

The argument \(\theta\) of \(1 + i\) is:

\[ \theta = \tan^{-1}\left(\frac{1}{1}\right) = \frac{\pi}{4} \]

Thus, we can write:

\[ 1 + i = \sqrt{2} \left(\cos\left(\frac{\pi}{4}\right) + i \sin\left(\frac{\pi}{4}\right)\right) \]

Using De Moivre's theorem, we have:

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

### Step 4: Substituting back into the second term

Now substituting back into the second term:

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

### Step 5: Combining both terms

Now we combine both terms:

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

Using the fact that \(\cos(-x) = \cos(x)\) and \(\sin(-x) = -\sin(x)\):

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

### Final Step: Simplifying the expression

This can be further simplified, but the exact form will depend on the value of \(n\). However, the expression can be simplified to:

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

### Final Answer

Thus, the expression simplifies to:

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