`I f|z|=max{|z-1|,|z+1|}`, then

Options

A. `|z+barz|=1/2`
B. `z+barz=1`
C. `|z+barz|=1`
D. z-barz=5`

Explanation

To solve the problem \( |z| = \max\{|z - 1|, |z + 1|\} \), we will follow these steps:

### Step 1: Define the complex number
Let \( z = a + ib \), where \( a \) and \( b \) are real numbers, and \( i \) is the imaginary unit.

### Step 2: Express the modulus of \( z \)
The modulus of \( z \) is given by:
\[ |z| = \sqrt{a^2 + b^2} \]

### Step 3: Express \( |z - 1| \) and \( |z + 1| \)
Now, we calculate \( |z - 1| \) and \( |z + 1| \):
\[ |z - 1| = |(a - 1) + ib| = \sqrt{(a - 1)^2 + b^2} \]
\[ |z + 1| = |(a + 1) + ib| = \sqrt{(a + 1)^2 + b^2} \]

### Step 4: Set up the equation
According to the problem, we have:
\[ |z| = \max\{|z - 1|, |z + 1|\} \]
This means:
\[ \sqrt{a^2 + b^2} = \max\{\sqrt{(a - 1)^2 + b^2}, \sqrt{(a + 1)^2 + b^2}\} \]

### Step 5: Analyze the two cases
We will analyze two cases based on the maximum value.

**Case 1:** \( |z - 1| \geq |z + 1| \)

In this case:
\[ \sqrt{a^2 + b^2} = \sqrt{(a - 1)^2 + b^2} \]
Squaring both sides, we get:
\[ a^2 + b^2 = (a - 1)^2 + b^2 \]
This simplifies to:
\[ a^2 + b^2 = a^2 - 2a + 1 + b^2 \]
Cancelling \( a^2 \) and \( b^2 \) from both sides:
\[ 0 = -2a + 1 \implies 2a = 1 \implies a = \frac{1}{2} \]

**Case 2:** \( |z + 1| \geq |z - 1| \)

In this case:
\[ \sqrt{a^2 + b^2} = \sqrt{(a + 1)^2 + b^2} \]
Squaring both sides, we get:
\[ a^2 + b^2 = (a + 1)^2 + b^2 \]
This simplifies to:
\[ a^2 + b^2 = a^2 + 2a + 1 + b^2 \]
Cancelling \( a^2 \) and \( b^2 \) from both sides:
\[ 0 = 2a + 1 \implies 2a = -1 \implies a = -\frac{1}{2} \]

### Step 6: Conclusion
From both cases, we find that \( a \) can either be \( \frac{1}{2} \) or \( -\frac{1}{2} \).

### Step 7: Find \( z + z^* \)
Now, we will find \( z + z^* \):
\[ z + z^* = (a + ib) + (a - ib) = 2a \]
Substituting \( a = \frac{1}{2} \):
\[ z + z^* = 2 \cdot \frac{1}{2} = 1 \]
Thus, \( |z + z^*| = |1| = 1 \).

### Final Answer
The correct option is that \( |z + z^*| = 1 \).

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