The multiplicative inverse of a number is the number itself, then its initial value is

Options

A. `-1`
B. i
C. 2
D. 0

Answer

To find the initial value of a complex number \( z \) such that its multiplicative inverse is the number itself, we can follow these steps:

### Step-by-Step Solution:

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

2. **Set Up the Equation**:
The multiplicative inverse of \( z \) is given by \( z^{-1} = \frac{1}{z} = \frac{1}{x + iy} \). According to the problem, we have:
\[ z = z^{-1} \]
This implies:
\[ x + iy = \frac{1}{x + iy} \]

3. **Multiply Both Sides by \( z \)**:
Multiply both sides by \( x + iy \) to eliminate the fraction:
\[ (x + iy)(x + iy) = 1 \]
This simplifies to:
\[ (x + iy)^2 = 1 \]

4. **Expand the Left Side**:
Expanding the left side, we get:
\[ x^2 + 2xyi - y^2 = 1 \]
This can be rewritten as:
\[ x^2 - y^2 + 2xyi = 1 \]

5. **Separate Real and Imaginary Parts**:
From the equation \( x^2 - y^2 + 2xyi = 1 \), we can equate the real and imaginary parts:
- Real part: \( x^2 - y^2 = 1 \)
- Imaginary part: \( 2xy = 0 \)

6. **Analyze the Imaginary Part**:
The equation \( 2xy = 0 \) implies either \( x = 0 \) or \( y = 0 \).

7. **Case 1: \( x = 0 \)**:
If \( x = 0 \):
\[ 0 - y^2 = 1 \implies -y^2 = 1 \implies y^2 = -1 \]
This gives:
\[ y = \pm i \]
Thus, the complex numbers are \( 0 + i \) and \( 0 - i \).

8. **Case 2: \( y = 0 \)**:
If \( y = 0 \):
\[ x^2 - 0 = 1 \implies x^2 = 1 \]
This gives:
\[ x = \pm 1 \]
Thus, the complex numbers are \( 1 + 0i \) and \( -1 + 0i \).

9. **Final Result**:
The possible values of \( z \) are \( 0 + i \), \( 0 - i \), \( 1 + 0i \), and \( -1 + 0i \). Among these, the values \( -1 \) and \( 1 \) are real numbers, and the question asks for the initial value. Therefore, the answer is:
\[ \text{The initial value is } -1 \text{ (option A).} \]