Find the multiplicative inverse of each of the following complex numbers when it exists.
`((2 + 3i) (3 + 2i)i)/(5+ i)`

Answer

To find the multiplicative inverse of the complex number given by the expression \(\frac{(2 + 3i)(3 + 2i)i}{5 + i}\), we will follow a systematic approach.

### Step-by-step Solution:

1. **Define the Complex Number**:
Let \( z = \frac{(2 + 3i)(3 + 2i)i}{5 + i} \).

2. **Multiply the Numerator**:
First, we need to simplify the numerator \((2 + 3i)(3 + 2i)i\).
\[ (2 + 3i)(3 + 2i) = 2 \cdot 3 + 2 \cdot 2i + 3i \cdot 3 + 3i \cdot 2i \]
\[ = 6 + 4i + 9i + 6i^2 \]
Since \( i^2 = -1 \):
\[ = 6 + 13i - 6 = 13i \]
Now, multiplying by \( i \):
\[ 13i \cdot i = 13i^2 = 13(-1) = -13 \]

3. **Substitute Back into \( z \)**:
Now substituting back, we have:
\[ z = \frac{-13}{5 + i} \]

4. **Rationalize the Denominator**:
To rationalize the denominator, multiply the numerator and denominator by the conjugate of the denominator:
\[ z = \frac{-13(5 - i)}{(5 + i)(5 - i)} = \frac{-65 + 13i}{25 + 1} = \frac{-65 + 13i}{26} \]

5. **Simplify \( z \)**:
We can express \( z \) as:
\[ z = -\frac{65}{26} + \frac{13}{26}i = -\frac{5}{2} + \frac{1}{2}i \]

6. **Find the Multiplicative Inverse**:
The multiplicative inverse of \( z \) is given by:
\[ z^{-1} = \frac{1}{z} = \frac{\overline{z}}{|z|^2} \]
where \( \overline{z} \) is the conjugate of \( z \) and \( |z|^2 \) is the modulus squared.

7. **Calculate the Conjugate**:
The conjugate \( \overline{z} \) is:
\[ \overline{z} = -\frac{5}{2} - \frac{1}{2}i \]

8. **Calculate the Modulus Squared**:
The modulus squared \( |z|^2 \) is:
\[ |z|^2 = \left(-\frac{5}{2}\right)^2 + \left(\frac{1}{2}\right)^2 = \frac{25}{4} + \frac{1}{4} = \frac{26}{4} = \frac{13}{2} \]

9. **Substitute to Find the Inverse**:
Now substituting into the formula for the inverse:
\[ z^{-1} = \frac{-\frac{5}{2} - \frac{1}{2}i}{\frac{13}{2}} = \frac{-5 - i}{13} \]

### Final Answer:
Thus, the multiplicative inverse of the complex number is:
\[ z^{-1} = -\frac{5}{13} - \frac{1}{13}i \]