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. The quadratic formula states that for any quadratic equation of the form \( ax^2 + bx + c = 0 \), the roots can be found using:

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

### Step 1: Identify coefficients
In our equation \( x^2 - x + (1 - i) = 0 \), we can identify the coefficients:
- \( a = 1 \)
- \( b = -1 \)
- \( c = 1 - i \)

### Step 2: Substitute into the quadratic formula
Now we substitute these values into the quadratic formula:

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

### Step 3: Simplify the expression
This simplifies to:

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

Calculating the term inside the square root:

\[ 1 - 4(1 - i) = 1 - 4 + 4i = -3 + 4i \]

So, we have:

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

### Step 4: Find the square root of the complex number
Next, we need to find \( \sqrt{-3 + 4i} \). Let \( \sqrt{-3 + 4i} = a + bi \). Then:

\[ (a + bi)^2 = -3 + 4i \]

Expanding the left side gives:

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

From here, we can equate the real and imaginary parts:

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

From the second equation, we can express \( ab \):

\[ ab = 2 \implies b = \frac{2}{a} \]

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

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

This simplifies to:

\[ a^2 - \frac{4}{a^2} = -3 \]

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

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

### Step 6: Let \( y = a^2 \)
Let \( y = a^2 \), then we have:

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

### Step 7: Solve the quadratic equation for \( y \)
Using the quadratic formula:

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

This gives us two solutions:

1. \( y = 1 \) (thus \( a^2 = 1 \) or \( a = \pm 1 \))
2. \( y = -4 \) (not valid since \( a^2 \) cannot be negative)

### Step 8: Find \( b \)
If \( a = 1 \), then:

\[ b = \frac{2}{1} = 2 \quad \text{or} \quad b = \frac{2}{-1} = -2 \]

Thus, the square roots are:

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

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

1. For \( \sqrt{-3 + 4i} = 1 + 2i \):

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

2. For \( \sqrt{-3 + 4i} = -1 - 2i \):

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

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

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