The equation of the smallest degree in the real coefficients having `1-i` as one of its roots is

Options

A. `x^(2)+2x-2=0`
B. `x^(2)-2x+2=0`
C. `x^(2)+2x+2=0`
D. `x^(2)-2x-2=0`

Answer

To find the equation of the smallest degree with real coefficients that has \(1 - i\) as one of its roots, we will follow these steps:

### Step 1: Identify the Complex Conjugate Root
Since \(1 - i\) is a root and the coefficients of the polynomial are real, the complex conjugate \(1 + i\) must also be a root.

**Hint:** Remember that complex roots come in conjugate pairs when dealing with polynomials with real coefficients.

### Step 2: Write Down the Roots
The roots of the polynomial are:
- \(r_1 = 1 - i\)
- \(r_2 = 1 + i\)

### Step 3: Calculate the Sum of the Roots
The sum of the roots \(S\) is given by:
\[ S = r_1 + r_2 = (1 - i) + (1 + i) = 2 \]

**Hint:** The sum of the roots can be calculated simply by adding the two roots together.

### Step 4: Calculate the Product of the Roots
The product of the roots \(P\) is given by:
\[ P = r_1 \cdot r_2 = (1 - i)(1 + i) = 1^2 - i^2 = 1 - (-1) = 2 \]

**Hint:** Use the difference of squares formula \(a^2 - b^2\) to simplify the product of complex conjugates.

### Step 5: Formulate the Quadratic Equation
Using the sum and product of the roots, we can write the quadratic equation in the standard form \(x^2 - Sx + P = 0\):
\[ x^2 - 2x + 2 = 0 \]

**Hint:** The standard form of a quadratic equation is derived from the roots using the relationships between the coefficients and the roots.

### Final Answer
The equation of the smallest degree with real coefficients having \(1 - i\) as one of its roots is:
\[ \boxed{x^2 - 2x + 2 = 0} \]