`Re(z^(2)) =1` is represented by

Options

A. The circle `x^(2) + y^(2) =1`
B. The hyperbola `x^(2) - y^(2) =1`
C. Parabola or a circle
D. All of these

Answer

To solve the question `Re(z^(2)) = 1`, we will follow these steps:

### Step 1: Express z in terms of its components
Let \( z = x + iy \), where \( x \) is the real part and \( y \) is the imaginary part of the complex number \( z \).

### Step 2: Calculate \( z^2 \)
Now, we need to compute \( z^2 \):
\[ z^2 = (x + iy)^2 = x^2 + 2xyi - y^2 = (x^2 - y^2) + (2xy)i \]
Here, the real part of \( z^2 \) is \( x^2 - y^2 \).

### Step 3: Set the real part equal to 1
According to the problem, we have:
\[ Re(z^2) = x^2 - y^2 = 1 \]

### Step 4: Rearrange the equation
Rearranging the equation gives us:
\[ x^2 - y^2 = 1 \]

### Step 5: Identify the conic section
The equation \( x^2 - y^2 = 1 \) is the standard form of a hyperbola.

### Conclusion
Thus, the representation of the equation \( Re(z^2) = 1 \) is a hyperbola given by the equation \( x^2 - y^2 = 1 \).

### Final Answer
The correct option is (b) hyperbola \( x^2 - y^2 = 1 \).

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