Find the conjugate of `((3-2i)(2+3i))/((1+2i)(2-i))`.

Explanation

To find the conjugate of the complex number \(\frac{(3-2i)(2+3i)}{(1+2i)(2-i)}\), we will follow these steps:

### Step 1: Calculate the numerator
The numerator is \((3-2i)(2+3i)\). We will use the distributive property (FOIL method) to expand this.

\[ (3-2i)(2+3i) = 3 \cdot 2 + 3 \cdot 3i - 2i \cdot 2 - 2i \cdot 3i \]
Calculating each term:
- \(3 \cdot 2 = 6\)
- \(3 \cdot 3i = 9i\)
- \(-2i \cdot 2 = -4i\)
- \(-2i \cdot 3i = -6i^2\) (Recall that \(i^2 = -1\), so this becomes \(+6\))

Combining these results:
\[ 6 + 9i - 4i + 6 = 12 + 5i \]

### Step 2: Calculate the denominator
The denominator is \((1+2i)(2-i)\). Again, we will use the distributive property.

\[ (1+2i)(2-i) = 1 \cdot 2 + 1 \cdot (-i) + 2i \cdot 2 + 2i \cdot (-i) \]
Calculating each term:
- \(1 \cdot 2 = 2\)
- \(1 \cdot (-i) = -i\)
- \(2i \cdot 2 = 4i\)
- \(2i \cdot (-i) = -2i^2 = +2\)

Combining these results:
\[ 2 - i + 4i + 2 = 4 + 3i \]

### Step 3: Form the complex number
Now we can form the complex number:
\[ z = \frac{12 + 5i}{4 + 3i} \]

### Step 4: Multiply by the conjugate of the denominator
To simplify, we multiply the numerator and denominator by the conjugate of the denominator, which is \(4 - 3i\).

\[ z = \frac{(12 + 5i)(4 - 3i)}{(4 + 3i)(4 - 3i)} \]

Calculating the denominator:
\[ (4 + 3i)(4 - 3i) = 4^2 - (3i)^2 = 16 - 9(-1) = 16 + 9 = 25 \]

Calculating the numerator:
\[ (12 + 5i)(4 - 3i) = 12 \cdot 4 + 12 \cdot (-3i) + 5i \cdot 4 + 5i \cdot (-3i) \]
Calculating each term:
- \(12 \cdot 4 = 48\)
- \(12 \cdot (-3i) = -36i\)
- \(5i \cdot 4 = 20i\)
- \(5i \cdot (-3i) = -15i^2 = +15\)

Combining these results:
\[ 48 + 15 - 36i + 20i = 63 - 16i \]

### Step 5: Final expression for \(z\)
Thus, we have:
\[ z = \frac{63 - 16i}{25} = \frac{63}{25} - \frac{16}{25}i \]

### Step 6: Find the conjugate of \(z\)
The conjugate of a complex number \(x + yi\) is \(x - yi\). Therefore, the conjugate of \(z\) is:
\[ \overline{z} = \frac{63}{25} + \frac{16}{25}i \]

### Final Answer
The conjugate of \(\frac{(3-2i)(2+3i)}{(1+2i)(2-i)}\) is:
\[ \overline{z} = \frac{63}{25} + \frac{16}{25}i \]