For all complex numbers z of the form `1 + ialpha, alpha in "R if z"^(2) = x + iy`, then

Options

A. `y^(2) - 4x + 2 = 0`
B. `y^(2) + 4x - 4 = 0`
C. `y^(2) - 4x + 4 = 0`
D. `y^(2) + 4x + 2 = 0`

Answer

To solve the problem step by step, we will start with the given complex number and perform the necessary calculations.

### Step-by-Step Solution:

1. **Define the Complex Number**:
Let \( z = 1 + i\alpha \), where \( \alpha \in \mathbb{R} \).

2. **Square the Complex Number**:
We need to find \( z^2 \):
\[ z^2 = (1 + i\alpha)^2 \]
Using the formula \( (a + b)^2 = a^2 + 2ab + b^2 \):
\[ z^2 = 1^2 + 2(1)(i\alpha) + (i\alpha)^2 \]
\[ = 1 + 2i\alpha + i^2\alpha^2 \]
Since \( i^2 = -1 \):
\[ = 1 + 2i\alpha - \alpha^2 \]
Therefore,
\[ z^2 = (1 - \alpha^2) + 2i\alpha \]

3. **Set Equal to the Given Form**:
We are given that \( z^2 = x + iy \). Thus, we can equate the real and imaginary parts:
\[ 1 - \alpha^2 = x \quad \text{(Real part)} \]
\[ 2\alpha = y \quad \text{(Imaginary part)} \]

4. **Express \( \alpha \) in terms of \( y \)**:
From the equation \( 2\alpha = y \), we can express \( \alpha \):
\[ \alpha = \frac{y}{2} \]

5. **Substitute \( \alpha \) into the Real Part Equation**:
Substitute \( \alpha = \frac{y}{2} \) into the equation \( 1 - \alpha^2 = x \):
\[ 1 - \left(\frac{y}{2}\right)^2 = x \]
Simplifying this:
\[ 1 - \frac{y^2}{4} = x \]
Rearranging gives:
\[ x = 1 - \frac{y^2}{4} \]

6. **Rearranging to Find the Relation**:
To find the relation between \( y^2 \) and \( x \), we can rearrange the equation:
\[ \frac{y^2}{4} + x - 1 = 0 \]
Multiplying through by 4 gives:
\[ y^2 + 4x - 4 = 0 \]

### Final Result:
The required equation is:
\[ y^2 + 4x - 4 = 0 \]