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

Options

A. `2^(8)`
B. `2^(5)`
C. `2^(4) cos "" (pi)/(4)`
D. `2^(6) cos "" (pi)/(8)`

Answer

To solve the expression \( (1 + i)^8 + (1 - i)^8 \), we can utilize the polar form of complex numbers and De Moivre's theorem. Here’s a step-by-step solution:

### Step 1: Convert \( 1 + i \) and \( 1 - i \) to Polar Form
The complex number \( 1 + i \) can be expressed in polar form.

1. Calculate the modulus:
\[ r = |1 + i| = \sqrt{1^2 + 1^2} = \sqrt{2} \]

2. Calculate the argument:
\[ \theta = \tan^{-1}\left(\frac{1}{1}\right) = \frac{\pi}{4} \]

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

Similarly, for \( 1 - i \):
1. The modulus is the same:
\[ r = |1 - i| = \sqrt{2} \]

2. The argument is:
\[ \theta = \tan^{-1}\left(\frac{-1}{1}\right) = -\frac{\pi}{4} \]

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

### Step 2: Apply De Moivre's Theorem
Using De Moivre's theorem, we can raise both expressions to the 8th power.

For \( (1 + i)^8 \):
\[ (1 + i)^8 = \left( \sqrt{2} \right)^8 \left( \cos\left(8 \cdot \frac{\pi}{4}\right) + i\sin\left(8 \cdot \frac{\pi}{4}\right) \right) \]
\[ = 8 \cdot 2^4 \left( \cos(2\pi) + i\sin(2\pi) \right) = 16(1 + 0i) = 16 \]

For \( (1 - i)^8 \):
\[ (1 - i)^8 = \left( \sqrt{2} \right)^8 \left( \cos\left(8 \cdot -\frac{\pi}{4}\right) + i\sin\left(8 \cdot -\frac{\pi}{4}\right) \right) \]
\[ = 8 \cdot 2^4 \left( \cos(-2\pi) + i\sin(-2\pi) \right) = 16(1 + 0i) = 16 \]

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

### Final Answer
\[ (1 + i)^8 + (1 - i)^8 = 32 \]

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