Solve `x^(2)-x+ (1-i)=0`

Answer

To solve the quadratic equation \( x^2 - x + (1 - i) = 0 \), we will use the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

### Step 1: Identify coefficients
In our equation, we can identify:
- \( a = 1 \) (coefficient of \( x^2 \))
- \( b = -1 \) (coefficient of \( x \))
- \( c = 1 - i \) (constant term)

### Step 2: Calculate the discriminant
Now, we calculate the discriminant \( b^2 - 4ac \):

\[ b^2 = (-1)^2 = 1 \]
\[ 4ac = 4 \cdot 1 \cdot (1 - i) = 4(1 - i) = 4 - 4i \]
\[ b^2 - 4ac = 1 - (4 - 4i) = 1 - 4 + 4i = -3 + 4i \]

### Step 3: Substitute into the quadratic formula
Now we substitute \( a \), \( b \), and the discriminant into the quadratic formula:

\[ x = \frac{-(-1) \pm \sqrt{-3 + 4i}}{2 \cdot 1} = \frac{1 \pm \sqrt{-3 + 4i}}{2} \]

### Step 4: Find the square root of the complex number
Let \( z = \sqrt{-3 + 4i} \). We assume \( z = a + bi \) and square both sides:

\[ z^2 = a^2 + 2abi - b^2 = -3 + 4i \]

This gives us two equations by comparing real and imaginary parts:

1. \( a^2 - b^2 = -3 \) (real part)
2. \( 2ab = 4 \) (imaginary part)

From the second equation, we can express \( b \) in terms of \( a \):

\[ b = \frac{4}{2a} = \frac{2}{a} \]

### Step 5: Substitute \( b \) into the first equation
Substituting \( b \) into the first equation:

\[ a^2 - \left(\frac{2}{a}\right)^2 = -3 \]
\[ a^2 - \frac{4}{a^2} = -3 \]

Multiplying through by \( a^2 \) to eliminate the fraction:

\[ a^4 + 3a^2 - 4 = 0 \]

### Step 6: Let \( u = a^2 \)
Let \( u = a^2 \):

\[ u^2 + 3u - 4 = 0 \]

### Step 7: Solve the quadratic in \( u \)
Using the quadratic formula:

\[ u = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot (-4)}}{2 \cdot 1} = \frac{-3 \pm \sqrt{9 + 16}}{2} = \frac{-3 \pm 5}{2} \]

Calculating the roots:

1. \( u = \frac{2}{2} = 1 \)
2. \( u = \frac{-8}{2} = -4 \) (not valid since \( u = a^2 \) must be non-negative)

Thus, \( a^2 = 1 \) implies \( a = 1 \) or \( a = -1 \).

### Step 8: Find corresponding \( b \)
Using \( b = \frac{2}{a} \):

- If \( a = 1 \), then \( b = 2 \).
- If \( a = -1 \), then \( b = -2 \).

Thus, we have two possible square roots:

\[ \sqrt{-3 + 4i} = 1 + 2i \quad \text{or} \quad -1 - 2i \]

### Step 9: Substitute back into the quadratic formula
Substituting back into the formula for \( x \):

1. For \( z = 1 + 2i \):

\[ x = \frac{1 + (1 + 2i)}{2} = \frac{2 + 2i}{2} = 1 + i \]

2. For \( z = -1 - 2i \):

\[ x = \frac{1 - (1 + 2i)}{2} = \frac{1 - 1 - 2i}{2} = \frac{-2i}{2} = -i \]

### Final Answer
Thus, the solutions to the equation \( x^2 - x + (1 - i) = 0 \) are:

\[ x = 1 + i \quad \text{and} \quad x = -i \]