Solve the Equation:
`x^(2)-x+1+i=0`.

Answer

To solve the equation \( x^2 - x + 1 + i = 0 \), we can follow these steps:

### Step 1: Rearranging the Equation
First, we can rewrite the equation in standard quadratic form:
\[ x^2 - x + (1 + i) = 0 \]
Here, we identify \( a = 1 \), \( b = -1 \), and \( c = 1 + i \).

### Step 2: Applying the Quadratic Formula
We will use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Substituting the values of \( a \), \( b \), and \( c \):
\[ x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot (1 + i)}}{2 \cdot 1} \]
This simplifies to:
\[ x = \frac{1 \pm \sqrt{1 - 4(1 + i)}}{2} \]

### Step 3: Simplifying the Discriminant
Now, we simplify the discriminant:
\[ 1 - 4(1 + i) = 1 - 4 - 4i = -3 - 4i \]
Thus, we have:
\[ x = \frac{1 \pm \sqrt{-3 - 4i}}{2} \]

### Step 4: Finding the Square Root of a Complex Number
Let \( z = x + iy \) such that \( z^2 = -3 - 4i \). Expanding \( z^2 \):
\[ (x + iy)^2 = x^2 - y^2 + 2xyi = -3 - 4i \]
From this, we can equate the real and imaginary parts:
1. \( x^2 - y^2 = -3 \)
2. \( 2xy = -4 \)

### Step 5: Solving the System of Equations
From the second equation, we can express \( y \) in terms of \( x \):
\[ y = \frac{-4}{2x} = \frac{-2}{x} \]
Substituting \( y \) into the first equation:
\[ x^2 - \left(\frac{-2}{x}\right)^2 = -3 \]
This simplifies to:
\[ x^2 - \frac{4}{x^2} = -3 \]
Multiplying through by \( x^2 \) to eliminate the fraction:
\[ x^4 + 3x^2 - 4 = 0 \]

### Step 6: Letting \( u = x^2 \)
Let \( u = x^2 \). The equation becomes:
\[ u^2 + 3u - 4 = 0 \]
Using the quadratic formula:
\[ u = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot (-4)}}{2 \cdot 1} = \frac{-3 \pm \sqrt{9 + 16}}{2} = \frac{-3 \pm 5}{2} \]
This gives us:
\[ u = 1 \quad \text{or} \quad u = -4 \]
Since \( u = x^2 \), we discard \( u = -4 \) (as \( x^2 \) cannot be negative) and take \( u = 1 \):
\[ x^2 = 1 \implies x = 1 \quad \text{or} \quad x = -1 \]

### Step 7: Finding Corresponding \( y \) Values
Using \( y = \frac{-2}{x} \):
1. If \( x = 1 \): \( y = -2 \)
2. If \( x = -1 \): \( y = 2 \)

### Step 8: Finding Values of \( z \)
Thus, the values of \( z \) are:
1. \( z = 1 - 2i \)
2. \( z = -1 + 2i \)

### Step 9: Final Values of \( x \)
Substituting back into our expression for \( x \):
\[ x = \frac{1 \pm z}{2} \]
This gives us:
1. \( x = \frac{1 + (1 - 2i)}{2} = 1 - i \)
2. \( x = \frac{1 + (-1 + 2i)}{2} = i \)

### Conclusion
The solutions to the equation \( x^2 - x + 1 + i = 0 \) are:
\[ x = 1 - i \quad \text{and} \quad x = i \]