If ` |z_(1)+z_(2)|^(2) = |z_(1)|^(2) +|z_(2)|^(2) " the " 6/pi amp (z_(1)/z_(2)) ` is equal to ……

Answer

To solve the problem, we need to analyze the given condition and derive the required value step by step.

### Step 1: Understand the Given Condition
We start with the equation:
\[ |z_1 + z_2|^2 = |z_1|^2 + |z_2|^2 \]

### Step 2: Expand the Left Side
Using the property of magnitudes:
\[ |z_1 + z_2|^2 = (z_1 + z_2)(\overline{z_1 + z_2}) = (z_1 + z_2)(\overline{z_1} + \overline{z_2}) \]
Expanding this gives:
\[ |z_1 + z_2|^2 = z_1\overline{z_1} + z_1\overline{z_2} + z_2\overline{z_1} + z_2\overline{z_2} \]
This simplifies to:
\[ |z_1 + z_2|^2 = |z_1|^2 + |z_2|^2 + z_1\overline{z_2} + z_2\overline{z_1} \]

### Step 3: Set the Equation
Now, we set the expanded form equal to the right side:
\[ |z_1|^2 + |z_2|^2 + z_1\overline{z_2} + z_2\overline{z_1} = |z_1|^2 + |z_2|^2 \]
This leads to:
\[ z_1\overline{z_2} + z_2\overline{z_1} = 0 \]

### Step 4: Simplify the Result
From the equation \( z_1\overline{z_2} + z_2\overline{z_1} = 0 \), we can conclude:
\[ z_1\overline{z_2} = -z_2\overline{z_1} \]
This implies that \( z_1\overline{z_2} \) is purely imaginary.

### Step 5: Define the Ratio
Let:
\[ w = \frac{z_1}{z_2} \]
Then:
\[ z_1 = wz_2 \quad \text{and} \quad z_1\overline{z_2} = w|z_2|^2 \]
Since \( z_1\overline{z_2} \) is purely imaginary, we have:
\[ \text{Re}(w) = 0 \]
This means that \( w \) is of the form:
\[ w = i \cdot k \quad \text{for some real number } k \]

### Step 6: Find the Amplitude
The amplitude of \( \frac{z_1}{z_2} \) is given by:
\[ \text{Amplitude}\left(\frac{z_1}{z_2}\right) = \arg\left(\frac{z_1}{z_2}\right) = \frac{\pi}{2} \]

### Step 7: Calculate the Final Value
Now we need to find:
\[ \frac{6}{\pi} \cdot \text{Amplitude}\left(\frac{z_1}{z_2}\right) = \frac{6}{\pi} \cdot \frac{\pi}{2} = 3 \]

### Final Answer
Thus, the value of \( \frac{6}{\pi} \cdot \text{Amplitude}\left(\frac{z_1}{z_2}\right) \) is:
\[ \boxed{3} \]