Let `S = {z in C : z^(2) = 4 (i bar(z))^(2)},"then" sum_(z in S)|z+(1)/(2)i|`= ___________.

Answer

To solve the problem, we need to find the set \( S = \{ z \in \mathbb{C} : z^2 = 4 (i \overline{z})^2 \} \) and then compute the sum \( \sum_{z \in S} |z + \frac{1}{2} i| \).

### Step-by-step Solution:

1. **Set up the equation**:
We start with the equation given in the problem:
\[ z^2 = 4 (i \overline{z})^2 \]
where \( z \) is a complex number and \( \overline{z} \) is its conjugate.

2. **Express \( z \) in terms of \( x \) and \( y \)**:
Let \( z = x + iy \), where \( x \) and \( y \) are real numbers. The conjugate \( \overline{z} \) is \( x - iy \).

3. **Substitute \( z \) and \( \overline{z} \)**:
Substitute \( z \) and \( \overline{z} \) into the equation:
\[ (x + iy)^2 = 4 \left( i (x - iy) \right)^2 \]
Simplifying the left side:
\[ x^2 - y^2 + 2xyi \]
And the right side:
\[ 4 \left( i(x - iy) \right)^2 = 4 \left( -y^2 - x^2 + 2xyi \right) = 4(-x^2 - y^2 + 2xyi) \]

4. **Equate real and imaginary parts**:
From the equation:
\[ x^2 - y^2 + 2xyi = 4(-x^2 - y^2 + 2xyi) \]
We can separate the real and imaginary parts:
- Real part:
\[ x^2 - y^2 = -4x^2 - 4y^2 \]
Rearranging gives:
\[ 5x^2 - 5y^2 = 0 \implies x^2 = y^2 \]
- Imaginary part:
\[ 2xy = 8xy \]
This simplifies to:
\[ 6xy = 0 \]

5. **Find solutions for \( x \) and \( y \)**:
From \( x^2 = y^2 \), we have two cases:
- Case 1: \( x = y \)
- Case 2: \( x = -y \)

From \( 6xy = 0 \), we find:
- Either \( x = 0 \) or \( y = 0 \).

This gives us the following solutions:
- If \( x = 0 \), then \( y \) can be any value, leading to \( z = iy \).
- If \( y = 0 \), then \( x \) can be any value, leading to \( z = x \).

Thus, the only solution that satisfies both conditions is \( z = 0 \).

6. **Calculate the sum**:
Now we compute:
\[ \sum_{z \in S} |z + \frac{1}{2} i| = |0 + \frac{1}{2} i| = \left| \frac{1}{2} i \right| = \frac{1}{2} \]

### Final Answer:
\[ \sum_{z \in S} |z + \frac{1}{2} i| = \frac{1}{2} \]