`(1+i)^8+(1-i)^8=?`

Options

A. 16
B. -16
C. 32
D. -32

Answer

To solve the expression \((1+i)^8 + (1-i)^8\), we can follow these steps:

### Step 1: Simplify \((1+i)\) and \((1-i)\)
First, we can express \(1+i\) and \(1-i\) in polar form. The modulus and argument of \(1+i\) can be calculated as follows:
- Modulus: \(|1+i| = \sqrt{1^2 + 1^2} = \sqrt{2}\)
- Argument: \(\tan^{-1}(\frac{1}{1}) = \frac{\pi}{4}\)

Thus, we can write:
\[ 1+i = \sqrt{2} \left( \cos\frac{\pi}{4} + i\sin\frac{\pi}{4} \right) = \sqrt{2} e^{i\frac{\pi}{4}} \]

Similarly, for \(1-i\):
- Modulus: \(|1-i| = \sqrt{1^2 + (-1)^2} = \sqrt{2}\)
- Argument: \(\tan^{-1}(\frac{-1}{1}) = -\frac{\pi}{4}\)

So we can write:
\[ 1-i = \sqrt{2} \left( \cos\left(-\frac{\pi}{4}\right) + i\sin\left(-\frac{\pi}{4}\right) \right) = \sqrt{2} e^{-i\frac{\pi}{4}} \]

### Step 2: Raise to the power of 8
Now we raise both expressions to the power of 8:
\[ (1+i)^8 = \left(\sqrt{2} e^{i\frac{\pi}{4}}\right)^8 = (\sqrt{2})^8 \cdot e^{i\frac{8\pi}{4}} = 16 \cdot e^{2\pi i} = 16 \cdot 1 = 16 \]

\[ (1-i)^8 = \left(\sqrt{2} e^{-i\frac{\pi}{4}}\right)^8 = (\sqrt{2})^8 \cdot e^{-i\frac{8\pi}{4}} = 16 \cdot e^{-2\pi i} = 16 \cdot 1 = 16 \]

### Step 3: Add the results
Now we add the two results together:
\[ (1+i)^8 + (1-i)^8 = 16 + 16 = 32 \]

### Final Answer
Thus, the final answer is:
\[ \boxed{32} \]