Geometrically Re `(z^(2)-i)=2`. where `i = sqrt(-1)` and Re is the real part, represents:

Options

A. Circle
B. Ellipse
C. Rectangular hyperbola
D. Parabola

Answer

To solve the problem geometrically represented by the equation \( \text{Re}(z^2 - i) = 2 \), where \( i = \sqrt{-1} \), we will follow these steps:

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

### Step 2: Calculate \( z^2 \)
Now, we calculate \( z^2 \):
\[ z^2 = (x + iy)^2 = x^2 + 2xyi - y^2 = x^2 - y^2 + 2xyi \]

### Step 3: Substitute \( z^2 \) into the equation
Substituting \( z^2 \) into the equation \( \text{Re}(z^2 - i) = 2 \):
\[ \text{Re}(z^2 - i) = \text{Re}(x^2 - y^2 + 2xyi - i) = \text{Re}(x^2 - y^2) + \text{Re}(2xyi - i) \]
The real part of \( -i \) is \( 0 \), so we have:
\[ \text{Re}(z^2 - i) = x^2 - y^2 \]

### Step 4: Set the equation equal to 2
Now, we set the real part equal to 2:
\[ x^2 - y^2 = 2 \]

### Step 5: Rearrange the equation
Rearranging gives us:
\[ \frac{x^2}{2} - \frac{y^2}{2} = 1 \]

### Step 6: Identify the conic section
This equation is in the standard form of a hyperbola:
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]
where \( a^2 = 2 \) and \( b^2 = 2 \). Since \( a^2 = b^2 \), this represents a rectangular hyperbola.

### Conclusion
Thus, geometrically, the equation \( \text{Re}(z^2 - i) = 2 \) represents a rectangular hyperbola.

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