Number of complex numbers such that |z| = 1 and `z = 1 - 2 bar(z)` is

Options

A. 0
B. 1
C. 2
D. infinite

Answer

To solve the problem, we need to find the number of complex numbers \( z \) such that \( |z| = 1 \) and \( z = 1 - 2\overline{z} \).

### Step-by-step Solution:

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

2. **Write the conjugate of \( z \)**:
The conjugate \( \overline{z} \) is given by \( \overline{z} = x - iy \).

3. **Substitute \( z \) and \( \overline{z} \) into the equation**:
The equation \( z = 1 - 2\overline{z} \) becomes:
\[ x + iy = 1 - 2(x - iy) \]
Simplifying the right side:
\[ x + iy = 1 - 2x + 2iy \]

4. **Separate the real and imaginary parts**:
Equating the real parts and the imaginary parts:
- Real part: \( x = 1 - 2x \)
- Imaginary part: \( y = 2y \)

5. **Solve for \( x \)**:
From the real part equation:
\[ x + 2x = 1 \implies 3x = 1 \implies x = \frac{1}{3} \]

6. **Solve for \( y \)**:
From the imaginary part equation:
\[ y = 2y \implies y - 2y = 0 \implies -y = 0 \implies y = 0 \]

7. **Form the complex number**:
Thus, we have:
\[ z = x + iy = \frac{1}{3} + 0i = \frac{1}{3} \]

8. **Check the modulus condition**:
We need to check if \( |z| = 1 \):
\[ |z| = \left| \frac{1}{3} + 0i \right| = \sqrt{\left(\frac{1}{3}\right)^2 + 0^2} = \frac{1}{3} \]
Since \( \frac{1}{3} \neq 1 \), the condition \( |z| = 1 \) is not satisfied.

9. **Conclusion**:
Since there are no complex numbers \( z \) that satisfy both conditions, the answer is that there are **no such complex numbers**.