z=x+iy is a complex number. If the imaginary part of `z^(2)` is 32. then Locus of z is a hyperbola of eccentricity-

Options

A. `sqrt2`
B. 2
C. `3//2`
D. `4//3`

Answer

To solve the problem step by step, we need to analyze the given information and derive the required locus of the complex number \( z = x + iy \).

### Step-by-Step Solution:

1. **Understanding the Complex Number**:
We are given that \( z = x + iy \), where \( x \) and \( y \) are real numbers, and \( i \) is the imaginary unit.

2. **Finding \( z^2 \)**:
We need to calculate \( z^2 \):
\[ z^2 = (x + iy)^2 \]
Using the formula \( (a + b)^2 = a^2 + 2ab + b^2 \):
\[ z^2 = x^2 + 2xyi + (iy)^2 \]
Since \( (iy)^2 = -y^2 \):
\[ z^2 = x^2 - y^2 + 2xyi \]

3. **Identifying the Imaginary Part**:
The imaginary part of \( z^2 \) is:
\[ \text{Im}(z^2) = 2xy \]

4. **Setting Up the Equation**:
According to the problem, the imaginary part of \( z^2 \) is given to be 32:
\[ 2xy = 32 \]

5. **Simplifying the Equation**:
Dividing both sides by 2:
\[ xy = 16 \]

6. **Recognizing the Locus**:
The equation \( xy = 16 \) represents a rectangular hyperbola.

7. **Finding the Eccentricity**:
The eccentricity \( e \) of a rectangular hyperbola is known to be \( \sqrt{2} \).

### Final Answer:
The eccentricity of the hyperbola is \( \sqrt{2} \).