Prove that `(3+i)/(1+2i)+(3-i)/(1-2i)` is a real number.

Explanation

To prove that \(\frac{3+i}{1+2i} + \frac{3-i}{1-2i}\) is a real number, we will follow these steps:

### Step 1: Write the expression
We start with the expression:
\[ \frac{3+i}{1+2i} + \frac{3-i}{1-2i} \]

### Step 2: Find a common denominator
The common denominator for the two fractions is \((1+2i)(1-2i)\). Thus, we can rewrite the expression as:
\[ \frac{(3+i)(1-2i) + (3-i)(1+2i)}{(1+2i)(1-2i)} \]

### Step 3: Expand the numerators
Now we will expand the numerators:
1. For \((3+i)(1-2i)\):
\[ = 3 \cdot 1 + 3 \cdot (-2i) + i \cdot 1 + i \cdot (-2i) = 3 - 6i + i - 2i^2 \]
Since \(i^2 = -1\), we have:
\[ = 3 - 6i + i + 2 = 5 - 5i \]

2. For \((3-i)(1+2i)\):
\[ = 3 \cdot 1 + 3 \cdot 2i - i \cdot 1 - i \cdot 2i = 3 + 6i - i - 2i^2 \]
Again, using \(i^2 = -1\):
\[ = 3 + 6i - i + 2 = 5 + 5i \]

### Step 4: Combine the expanded numerators
Now, we combine the two results:
\[ (5 - 5i) + (5 + 5i) = 5 - 5i + 5 + 5i = 10 \]

### Step 5: Calculate the denominator
Next, we calculate the denominator:
\[ (1+2i)(1-2i) = 1^2 - (2i)^2 = 1 - 4(-1) = 1 + 4 = 5 \]

### Step 6: Form the complete expression
Now we can write the complete expression:
\[ \frac{10}{5} = 2 \]

### Conclusion
Since \(2\) is a real number, we have proved that \(\frac{3+i}{1+2i} + \frac{3-i}{1-2i}\) is indeed a real number.