Show that `(√5+i √2)/(√5-i√2) + (√5-i√2)/(√5+i√2) ` is real.

Answer

To show that the expression

\[ \frac{\sqrt{5} + i \sqrt{2}}{\sqrt{5} - i \sqrt{2}} + \frac{\sqrt{5} - i \sqrt{2}}{\sqrt{5} + i \sqrt{2}} \]

is real, we will follow these steps:

### Step 1: Find a common denominator
The common denominator for the two fractions is \((\sqrt{5} - i \sqrt{2})(\sqrt{5} + i \sqrt{2})\).

### Step 2: Rewrite the expression
We can rewrite the expression as:

\[ \frac{(\sqrt{5} + i \sqrt{2})^2 + (\sqrt{5} - i \sqrt{2})^2}{(\sqrt{5} - i \sqrt{2})(\sqrt{5} + i \sqrt{2})} \]

### Step 3: Calculate the numerator
Now we will calculate the numerator:

1. Calculate \((\sqrt{5} + i \sqrt{2})^2\):
\[ (\sqrt{5} + i \sqrt{2})^2 = 5 + 2i\sqrt{10} - 2 = 3 + 2i\sqrt{10} \]

2. Calculate \((\sqrt{5} - i \sqrt{2})^2\):
\[ (\sqrt{5} - i \sqrt{2})^2 = 5 - 2i\sqrt{10} - 2 = 3 - 2i\sqrt{10} \]

3. Add these results:
\[ (3 + 2i\sqrt{10}) + (3 - 2i\sqrt{10}) = 6 \]

### Step 4: Calculate the denominator
Now we calculate the denominator:

\[ (\sqrt{5} - i \sqrt{2})(\sqrt{5} + i \sqrt{2}) = (\sqrt{5})^2 - (i \sqrt{2})^2 = 5 - (-2) = 5 + 2 = 7 \]

### Step 5: Combine the results
Now we can combine the results from the numerator and denominator:

\[ \frac{6}{7} \]

### Conclusion
Since \(\frac{6}{7}\) is a real number, we conclude that the original expression is indeed real.