`1 + i = sqrt(2) ( cos "" (pi)/( 4) + i sin "" (pi)/( 4)) `
`1 + omega + omega^(2) = 0, omega = ((2 pi i)/3)`
`( 1+ x)^(n) = P_(0) + P_(1) x + P_(2) x^(2) + P_(3) x^(3) + P_(4) x ^(4) + . . . + P_(n) x^(n)`
` p_(0) + p_(4) + p_(8) + . . . = 2 ^(n//2-1) cos "" ( n pi)/( 4) + 2 ^(n-2)`

Answer

To solve the problem step by step, we will analyze the expressions given and derive the required results systematically.

### Step 1: Understand the given expressions

We have the following key expressions:
1. \( 1 + i = \sqrt{2} \left( \cos \frac{\pi}{4} + i \sin \frac{\pi}{4} \right) \)
2. \( 1 + \omega + \omega^2 = 0 \) where \( \omega = e^{\frac{2\pi i}{3}} \)
3. The binomial expansion \( (1 + x)^n = P_0 + P_1 x + P_2 x^2 + P_3 x^3 + P_4 x^4 + \ldots + P_n x^n \)
4. The sum \( P_0 + P_4 + P_8 + \ldots = 2^{\frac{n}{2}-1} \cos \frac{n \pi}{4} + 2^{n-2} \)

### Step 2: Verify the first expression

We start with the expression \( 1 + i \):
- We can express \( 1 + i \) in polar form:
\[ r = |1 + i| = \sqrt{1^2 + 1^2} = \sqrt{2} \]
The angle \( \theta \) is given by:
\[ \theta = \tan^{-1}\left(\frac{1}{1}\right) = \frac{\pi}{4} \]
- Therefore, we can write:
\[ 1 + i = \sqrt{2} \left( \cos \frac{\pi}{4} + i \sin \frac{\pi}{4} \right) \]

### Step 3: Analyze the second expression

The expression \( 1 + \omega + \omega^2 = 0 \) indicates that \( \omega \) is a cube root of unity:
- The roots of the equation \( x^3 - 1 = 0 \) are \( 1, \omega, \omega^2 \).
- Since \( \omega = e^{\frac{2\pi i}{3}} \), we know:
\[ \omega^2 = e^{\frac{4\pi i}{3}} \]
- Thus, \( 1 + \omega + \omega^2 = 0 \) is satisfied.

### Step 4: Binomial expansion

The binomial expansion \( (1 + x)^n \) gives us:
\[ (1 + x)^n = P_0 + P_1 x + P_2 x^2 + P_3 x^3 + P_4 x^4 + \ldots + P_n x^n \]
Where \( P_k = \binom{n}{k} \).

### Step 5: Finding \( P_0 + P_4 + P_8 + \ldots \)

To find \( P_0 + P_4 + P_8 + \ldots \), we can use the fact that:
- \( P_k \) corresponds to the coefficients in the expansion.
- We can evaluate \( (1 + x)^n + (1 - x)^n \) to isolate the even coefficients:
\[ (1 + x)^n + (1 - x)^n = 2(P_0 + P_2 x^2 + P_4 x^4 + \ldots) \]
- Setting \( x = 1 \):
\[ 2^n + 0 = 2(P_0 + P_2 + P_4 + \ldots) \]
This simplifies to:
\[ P_0 + P_2 + P_4 + \ldots = 2^{n-1} \]

### Step 6: Using complex numbers

Next, we can evaluate \( (1 + i)^n + (1 - i)^n \):
- Using the polar form:
\[ (1 + i)^n = \left(\sqrt{2}\right)^n \left( \cos \frac{n\pi}{4} + i \sin \frac{n\pi}{4} \right) \]
\[ (1 - i)^n = \left(\sqrt{2}\right)^n \left( \cos \frac{n\pi}{4} - i \sin \frac{n\pi}{4} \right) \]
- Adding these gives:
\[ 2^{\frac{n}{2}} \cos \frac{n\pi}{4} \]

### Step 7: Final expression

Combining the results:
\[ P_0 + P_4 + P_8 + \ldots = 2^{\frac{n}{2}-1} \cos \frac{n\pi}{4} + 2^{n-2} \]

### Conclusion

Thus, we have verified the statement:
\[ P_0 + P_4 + P_8 + \ldots = 2^{\frac{n}{2}-1} \cos \frac{n\pi}{4} + 2^{n-2} \]
is true.