Complex numbers `z_(1)` and `z_(2)` satisfy `|z_(1)|=2` and `|z_(2)|=3`. If the included angle of their corresponding vectors is `60^(@)`, then the value of `19|(z_(1)-z_(2))/(z_(1)+z_(2))|^(2)` is

Options

A. `5`
B. `6`
C. `7`
D. `8`

Explanation

To solve the problem, we need to find the value of \( 19 \left| \frac{z_1 - z_2}{z_1 + z_2} \right|^2 \) given that \( |z_1| = 2 \), \( |z_2| = 3 \), and the angle between the vectors corresponding to \( z_1 \) and \( z_2 \) is \( 60^\circ \).

### Step 1: Understanding the Magnitudes
We know:
- \( |z_1| = 2 \)
- \( |z_2| = 3 \)

### Step 2: Using the Law of Cosines
To find \( |z_1 + z_2| \) and \( |z_1 - z_2| \), we can use the Law of Cosines.

#### For \( |z_1 + z_2| \):
The angle between \( z_1 \) and \( z_2 \) is \( 60^\circ \). Therefore, we can write:
\[ |z_1 + z_2|^2 = |z_1|^2 + |z_2|^2 + 2 |z_1| |z_2| \cos(60^\circ) \]
Substituting the values:
\[ |z_1 + z_2|^2 = 2^2 + 3^2 + 2 \cdot 2 \cdot 3 \cdot \frac{1}{2} \]
Calculating this gives:
\[ |z_1 + z_2|^2 = 4 + 9 + 6 = 19 \]
Thus, \( |z_1 + z_2| = \sqrt{19} \).

#### For \( |z_1 - z_2| \):
The angle between \( z_1 \) and \( z_2 \) is \( 60^\circ \) as well, but we need to consider the angle for the subtraction, which will be \( 120^\circ \):
\[ |z_1 - z_2|^2 = |z_1|^2 + |z_2|^2 - 2 |z_1| |z_2| \cos(120^\circ) \]
Substituting the values:
\[ |z_1 - z_2|^2 = 2^2 + 3^2 - 2 \cdot 2 \cdot 3 \cdot \left(-\frac{1}{2}\right) \]
Calculating this gives:
\[ |z_1 - z_2|^2 = 4 + 9 + 6 = 19 \]
Thus, \( |z_1 - z_2| = \sqrt{7} \).

### Step 3: Calculating the Expression
Now we need to compute:
\[ 19 \left| \frac{z_1 - z_2}{z_1 + z_2} \right|^2 \]
This can be rewritten using the magnitudes we found:
\[ 19 \left| \frac{\sqrt{7}}{\sqrt{19}} \right|^2 = 19 \cdot \frac{7}{19} = 7 \]

### Final Answer
The value of \( 19 \left| \frac{z_1 - z_2}{z_1 + z_2} \right|^2 \) is \( \boxed{7} \).