Write the additive inverse of `(sqrt(6)+5i)(sqrt(6)-(1)/(5)i)`.

Answer

To find the additive inverse of the expression \((\sqrt{6} + 5i)(\sqrt{6} - \frac{1}{5}i)\), we will first simplify the product and then find its additive inverse.

### Step-by-Step Solution:

1. **Multiply the two complex numbers**:
\[ (\sqrt{6} + 5i)(\sqrt{6} - \frac{1}{5}i) \]
We will use the distributive property (FOIL method):
\[ = \sqrt{6} \cdot \sqrt{6} + \sqrt{6} \cdot \left(-\frac{1}{5}i\right) + 5i \cdot \sqrt{6} + 5i \cdot \left(-\frac{1}{5}i\right) \]

2. **Calculate each term**:
\[ = 6 - \frac{\sqrt{6}}{5}i + 5\sqrt{6}i - 5 \cdot \left(-\frac{1}{5}\right)(i^2) \]
Since \(i^2 = -1\):
\[ = 6 - \frac{\sqrt{6}}{5}i + 5\sqrt{6}i + 1 \]

3. **Combine like terms**:
\[ = (6 + 1) + \left(-\frac{\sqrt{6}}{5} + 5\sqrt{6}\right)i \]
\[ = 7 + \left(-\frac{\sqrt{6}}{5} + \frac{25\sqrt{6}}{5}\right)i \]
\[ = 7 + \left(\frac{24\sqrt{6}}{5}\right)i \]

4. **Final result of the product**:
\[ = 7 + \frac{24\sqrt{6}}{5}i \]

5. **Find the additive inverse**:
The additive inverse \(z_2\) of a complex number \(z_1 = a + bi\) is given by:
\[ z_2 = -a - bi \]
Here, \(z_1 = 7 + \frac{24\sqrt{6}}{5}i\), so:
\[ z_2 = -7 - \frac{24\sqrt{6}}{5}i \]

### Final Answer:
The additive inverse of \((\sqrt{6} + 5i)(\sqrt{6} - \frac{1}{5}i)\) is:
\[ -7 - \frac{24\sqrt{6}}{5}i \]