If `z in C`, then `Re(bar(z)^(2))= k^(2), k gt 0`, represents

Options

A. an ellipse
B. a parabola
C. a circle
D. a hyperbola

Answer

To solve the problem, we need to analyze the expression given: \( \text{Re}(\bar{z}^2) = k^2 \), where \( k > 0 \).

### Step-by-Step Solution:

1. **Express \( z \) in terms of its real and imaginary parts**:
Let \( z = x + iy \), where \( x \) and \( y \) are real numbers.

2. **Find the conjugate of \( z \)**:
The conjugate \( \bar{z} \) is given by:
\[ \bar{z} = x - iy \]

3. **Square the conjugate**:
Now, we compute \( \bar{z}^2 \):
\[ \bar{z}^2 = (x - iy)^2 = x^2 - 2x(iy) + (iy)^2 \]
Simplifying this, we have:
\[ \bar{z}^2 = x^2 - 2xyi - y^2 = (x^2 - y^2) - 2xyi \]

4. **Identify the real part**:
The real part of \( \bar{z}^2 \) is:
\[ \text{Re}(\bar{z}^2) = x^2 - y^2 \]

5. **Set the real part equal to \( k^2 \)**:
According to the problem, we have:
\[ x^2 - y^2 = k^2 \]

6. **Rearranging the equation**:
Rearranging gives:
\[ x^2 - k^2 = y^2 \]

7. **Recognize the conic section**:
The equation \( x^2 - y^2 = k^2 \) represents a rectangular hyperbola in the Cartesian plane.

### Conclusion:
Thus, the equation \( \text{Re}(\bar{z}^2) = k^2 \) represents a rectangular hyperbola.