` 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_(2) + p_(4) - . . . = 2^(n//2) cos ( n pi // 4)`

Answer

To solve the problem step by step, we will analyze the provided equations and derive the necessary results.

### Step 1: Understanding the given equations
We have the following equations:
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 result: \( P_0 - P_2 + P_4 - \ldots = 2^{n/2} \cos \left( \frac{n \pi}{4} \right) \)

### Step 2: Verify the first equation
We can verify the first equation:
\[ 1 + i = \sqrt{2} \left( \cos \frac{\pi}{4} + i \sin \frac{\pi}{4} \right) \]
Calculating the right side:
\[ \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}, \quad \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}} \]
Thus,
\[ \sqrt{2} \left( \frac{1}{\sqrt{2}} + i \frac{1}{\sqrt{2}} \right) = 1 + i \]
This confirms that the first equation is true.

### Step 3: Analyze the second equation
The second equation states:
\[ 1 + \omega + \omega^2 = 0 \]
This is the property of the cube roots of unity, where \( \omega = e^{\frac{2\pi i}{3}} \) and \( \omega^2 = e^{\frac{4\pi i}{3}} \). This confirms that the sum of the roots is zero.

### Step 4: Binomial expansion
The binomial expansion of \( (1 + x)^n \) gives us:
\[ (1 + x)^n = \sum_{k=0}^{n} \binom{n}{k} x^k \]
where \( P_k = \binom{n}{k} \).

### Step 5: Calculate \( P_0 - P_2 + P_4 - \ldots \)
To find \( P_0 - P_2 + P_4 - \ldots \), we can use the property of the binomial coefficients:
\[ P_0 = \binom{n}{0}, \quad P_2 = \binom{n}{2}, \quad P_4 = \binom{n}{4}, \ldots \]
The alternating sum of the coefficients can be expressed as:
\[ \sum_{k=0}^{\lfloor n/2 \rfloor} (-1)^k \binom{n}{2k} \]

### Step 6: Use the result
The result states:
\[ P_0 - P_2 + P_4 - \ldots = 2^{n/2} \cos \left( \frac{n \pi}{4} \right) \]
This is derived from the binomial theorem and properties of complex numbers.

### Conclusion
We have verified the equations and derived the necessary results. The statement is true.