If `|z|= "min" {|z-1|, |z+1|}`, then

Options

A. `|z+ bar(z)|=(1)/(2)`
B. `z + bar(z) =1`
C. `|z + bar(z) | =1`
D. None of these

Answer

To solve the problem \( |z| = \min \{ |z - 1|, |z + 1| \} \), we will analyze the conditions under which the modulus of \( z \) is minimized compared to the moduli of \( z - 1 \) and \( z + 1 \).

### Step-by-Step Solution:

1. **Understanding the Condition**:
We need to find the points \( z \) in the complex plane such that the modulus of \( z \) is the minimum of the distances from \( z \) to the points 1 and -1 on the real axis.

2. **Setting Up the Equations**:
We can express \( z \) in terms of its real and imaginary parts:
\[ z = x + iy \]
where \( x \) is the real part and \( y \) is the imaginary part.

3. **Expressing the Moduli**:
The moduli can be expressed as:
\[ |z| = \sqrt{x^2 + y^2} \]
\[ |z - 1| = |(x - 1) + iy| = \sqrt{(x - 1)^2 + y^2} \]
\[ |z + 1| = |(x + 1) + iy| = \sqrt{(x + 1)^2 + y^2} \]

4. **Setting Up the Equations**:
Since \( |z| \) is the minimum of \( |z - 1| \) and \( |z + 1| \), we consider two cases:
- Case 1: \( |z| = |z - 1| \)
- Case 2: \( |z| = |z + 1| \)

5. **Case 1: \( |z| = |z - 1| \)**:
\[ \sqrt{x^2 + y^2} = \sqrt{(x - 1)^2 + y^2} \]
Squaring both sides:
\[ x^2 + y^2 = (x - 1)^2 + y^2 \]
Simplifying:
\[ x^2 = x^2 - 2x + 1 \]
\[ 2x = 1 \implies x = \frac{1}{2} \]

6. **Case 2: \( |z| = |z + 1| \)**:
\[ \sqrt{x^2 + y^2} = \sqrt{(x + 1)^2 + y^2} \]
Squaring both sides:
\[ x^2 + y^2 = (x + 1)^2 + y^2 \]
Simplifying:
\[ x^2 = x^2 + 2x + 1 \]
\[ 0 = 2x + 1 \implies x = -\frac{1}{2} \]

7. **Conclusion**:
From the two cases, we find that \( x \) can be either \( \frac{1}{2} \) or \( -\frac{1}{2} \). Thus, the real part of \( z \) can take these values, while the imaginary part \( y \) can be any real number.

8. **Final Result**:
Therefore, the points \( z \) that satisfy the original condition are:
\[ z = \frac{1}{2} + iy \quad \text{or} \quad z = -\frac{1}{2} + iy \quad \text{for any } y \in \mathbb{R} \]