Find the modulus of the following using the property of modulus
`(3+2i)/(2-5i) + (3-2i)/(2+5i)`

Answer

To find the modulus of the expression \((3+2i)/(2-5i) + (3-2i)/(2+5i)\), we can follow these steps:

### Step 1: Define the expression
Let \( Z = \frac{3 + 2i}{2 - 5i} + \frac{3 - 2i}{2 + 5i} \).

### Step 2: Find a common denominator
The common denominator for the two fractions is \((2 - 5i)(2 + 5i)\).

### Step 3: Rewrite the expression with the common denominator
We can rewrite \( Z \) as:
\[ Z = \frac{(3 + 2i)(2 + 5i) + (3 - 2i)(2 - 5i)}{(2 - 5i)(2 + 5i)} \]

### Step 4: Expand the numerators
Now, we will expand the numerators:
1. For \((3 + 2i)(2 + 5i)\):
\[ = 3 \cdot 2 + 3 \cdot 5i + 2i \cdot 2 + 2i \cdot 5i = 6 + 15i + 4i + 10i^2 = 6 + 19i - 10 \quad (\text{since } i^2 = -1) = -4 + 19i \]

2. For \((3 - 2i)(2 - 5i)\):
\[ = 3 \cdot 2 - 3 \cdot 5i - 2i \cdot 2 + 2i \cdot 5i = 6 - 15i - 4i - 10i^2 = 6 - 19i + 10 = 16 - 19i \]

### Step 5: Combine the results
Now we combine both results:
\[ Z = \frac{(-4 + 19i) + (16 - 19i)}{(2 - 5i)(2 + 5i)} \]
\[ = \frac{(-4 + 16) + (19i - 19i)}{(2 - 5i)(2 + 5i)} \]
\[ = \frac{12}{(2 - 5i)(2 + 5i)} \]

### Step 6: Simplify the denominator
The denominator can be simplified using the difference of squares:
\[ (2 - 5i)(2 + 5i) = 2^2 - (5i)^2 = 4 + 25 = 29 \]

### Step 7: Final expression for Z
Thus, we have:
\[ Z = \frac{12}{29} \]

### Step 8: Find the modulus of Z
Since \( Z \) is a real number, the modulus is simply the absolute value:
\[ |Z| = \left|\frac{12}{29}\right| = \frac{12}{29} \]

### Final Answer
The modulus of the given expression is \(\frac{12}{29}\).