From the polynomial equation whose roots are 1+I,1-I,1+I,1-I

Answer

To find the polynomial equation whose roots are \(1 + i\), \(1 - i\), \(1 + i\), and \(1 - i\), we can follow these steps:

### Step 1: Identify the Roots
The roots of the polynomial are given as:
- \( \alpha = 1 + i \)
- \( \beta = 1 - i \)
- \( \gamma = 1 + i \)
- \( \delta = 1 - i \)

### Step 2: Calculate the Sum of the Roots
The sum of the roots can be calculated as:
\[ \text{Sum} = \alpha + \beta + \gamma + \delta = (1 + i) + (1 - i) + (1 + i) + (1 - i) \]
Simplifying this:
\[ \text{Sum} = (1 + 1 + 1 + 1) + (i - i + i - i) = 4 + 0 = 4 \]

### Step 3: Calculate the Product of the Roots
The product of the roots can be calculated as:
\[ \text{Product} = \alpha \cdot \beta \cdot \gamma \cdot \delta = (1 + i)(1 - i)(1 + i)(1 - i) \]
Calculating \( (1 + i)(1 - i) \):
\[ (1 + i)(1 - i) = 1^2 - i^2 = 1 - (-1) = 2 \]
Thus, the product becomes:
\[ \text{Product} = 2 \cdot 2 = 4 \]

### Step 4: Form the Quadratic Polynomial for Each Pair of Roots
Using the roots \( \alpha \) and \( \beta \):
\[ x^2 - (\text{Sum of roots}) \cdot x + (\text{Product of roots}) = 0 \]
This gives:
\[ x^2 - 2x + 2 = 0 \]
Since the roots are repeated, we will use this polynomial twice.

### Step 5: Form the Final Polynomial
Now, we multiply the two quadratic equations:
\[ (x^2 - 2x + 2)(x^2 - 2x + 2) \]
Using the distributive property (FOIL):
\[ = x^4 - 2x^3 + 2x^2 - 2x^3 + 4x^2 - 4x + 2x^2 - 4x + 4 \]
Combining like terms:
\[ = x^4 - 4x^3 + 8x^2 - 8x + 4 \]

### Final Polynomial
Thus, the polynomial equation whose roots are \(1 + i\), \(1 - i\), \(1 + i\), and \(1 - i\) is:
\[ x^4 - 4x^3 + 8x^2 - 8x + 4 = 0 \]