The real part of $(1+i)^{n}$ is:

Question

Accepted Answer

To determine the real part of $(1+i)^n$, we can use the polar form of complex numbers and De Moivre's Theorem. Let's break down the solution step by step:

1. **Convert $1+i$ to Polar Form:**
   The complex number $1+i$ can be expressed in polar form. The modulus $r$ of $1+i$ is calculated as:
   $$
   r = \sqrt{1^2 + 1^2} = \sqrt{2}
   $$
   The argument $\theta$ is:
   $$
   \theta = \tan^{-1}\left(\frac{1}{1}\right) = \frac{\pi}{4}
   $$
   Therefore, in polar form, $1+i = \sqrt{2} \left(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\right)$.

2. **Apply De Moivre's Theorem:**
   Using De Moivre's Theorem, $(1+i)^n$ can be expressed as:
   $$
   (\sqrt{2})^n \left(\cos\left(n\frac{\pi}{4}\right) + i\sin\left(n\frac{\pi}{4}\right)\right)
   $$
   Simplifying $(\sqrt{2})^n$, we get:
   $$
   (\sqrt{2})^n = 2^{n/2}
   $$
   Thus, $(1+i)^n = 2^{n/2} \left(\cos\left(n\frac{\pi}{4}\right) + i\sin\left(n\frac{\pi}{4}\right)\right)$.

3. **Extract the Real Part:**
   The real part of $(1+i)^n$ is:
   $$
   2^{n/2} \cos\left(n\frac{\pi}{4}\right)
   $$

Therefore, the correct answer is Option 4: $2^{n/2} \cos\left(\frac{n\pi}{4}\right)$.

**Explanation of Incorrect Options:**

- **Option 1: $2^{n/2} \cos\frac{n\pi}{2}$**
  - This option incorrectly uses $\cos\frac{n\pi}{2}$ instead of $\cos\frac{n\pi}{4}$. The angle should be $n\frac{\pi}{4}$ as derived from the polar form of $1+i$.

- **Option 2: $2^{n} \cos\frac{n\pi}{4}$**
  - This option incorrectly uses $2^n$ instead of $2^{n/2}$. The modulus of $1+i$ is $\sqrt{2}$, and raising it to the power $n$ gives $2^{n/2}$, not $2^n$.

- **Option 3: $2^{-n/2} \cos\frac{n\pi}{4}$**
  - This option incorrectly uses $2^{-n/2}$ instead of $2^{n/2}$. The modulus calculation should result in a positive exponent, as shown in the derivation.

In conclusion, Option 4 is the correct answer because it accurately reflects the real part of $(1+i)^n$ using the correct modulus and angle derived from the polar form and De Moivre's Theorem.