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: Simplify the Numerator
First, we need to calculate the numerator \((3-2i)(2+3i)\).

\[ (3-2i)(2+3i) = 3 \cdot 2 + 3 \cdot 3i - 2i \cdot 2 - 2i \cdot 3i \]
\[ = 6 + 9i - 4i - 6i^2 \]
Since \(i^2 = -1\), we substitute:
\[ = 6 + 9i - 4i + 6 = 12 + 5i \]

### Step 2: Simplify the Denominator
Now, we calculate the denominator \((1+2i)(2-i)\).

\[ (1+2i)(2-i) = 1 \cdot 2 + 1 \cdot (-i) + 2i \cdot 2 + 2i \cdot (-i) \]
\[ = 2 - i + 4i - 2i^2 \]
Substituting \(i^2 = -1\):
\[ = 2 - i + 4i + 2 = 4 + 3i \]

### Step 3: Combine the Results
Now we have:
\[ \frac{(3-2i)(2+3i)}{(1+2i)(2-i)} = \frac{12 + 5i}{4 + 3i} \]

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

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

### Step 5: Simplify the Numerator
Calculating the numerator:
\[ (12 + 5i)(4 - 3i) = 12 \cdot 4 + 12 \cdot (-3i) + 5i \cdot 4 + 5i \cdot (-3i) \]
\[ = 48 - 36i + 20i - 15i^2 \]
Substituting \(i^2 = -1\):
\[ = 48 - 36i + 20i + 15 = 63 - 16i \]

### Step 6: Simplify the Denominator
Calculating the denominator:
\[ (4 + 3i)(4 - 3i) = 4^2 - (3i)^2 = 16 - 9(-1) = 16 + 9 = 25 \]

### Step 7: Final Result
Now we can write the simplified form:
\[ \frac{63 - 16i}{25} = \frac{63}{25} - \frac{16}{25}i \]

### Step 8: Find the Conjugate
The conjugate of a complex number \(a + bi\) is \(a - bi\). Therefore, the conjugate of \(\frac{63}{25} - \frac{16}{25}i\) is:
\[ \frac{63}{25} + \frac{16}{25}i \]

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