Express \( \frac{(6+i)(2-i)}{(4+3 i)(1-2 i)} \) in the form \( a+i b \)

To express \( \frac{(6+i)(2-i)}{(4+3i)(1-2i)} \) in the form \( a + ib \), let's start by calculating the numerator and denominator separately. **Step 1: Calculate the numerator:** \[ (6+i)(2-i) = 6 \cdot 2 - 6 \cdot i + 2 \cdot i - i^2 \] Since \( i^2 = -1 \), we can simplify this: \[ = 12 - 6i + 2i + 1 = 12 + 1 - 4i = 13 - 4i \] **Step 2: Calculate the denominator:** \[ (4+3i)(1-2i) = 4 \cdot 1 - 4 \cdot 2i + 3i \cdot 1 - 3i \cdot 2i \] Expanding this gives: \[ = 4 - 8i + 3i + 6 = 10 - 5i \] **Step 3: Now, we substitute back into the expression:** \[ \frac{(6+i)(2-i)}{(4+3i)(1-2i)} = \frac{13 - 4i}{10 - 5i} \] **Step 4: To simplify this fraction, we multiply the numerator and denominator by the conjugate of the denominator:** \[ = \frac{(13 - 4i)(10 + 5i)}{(10 - 5i)(10 + 5i)} \] **Calculating the denominator:** \[ (10 - 5i)(10 + 5i) = 10^2 - (5i)^2 = 100 - 25(-1) = 100 + 25 = 125 \] **Calculating the numerator:** \[ (13 - 4i)(10 + 5i) = 13 \cdot 10 + 13 \cdot 5i - 4i \cdot 10 - 4i \cdot 5i \] \[ = 130 + 65i - 40i - 20(-1) = 130 + 65i - 40i + 20 = 130 + 20 + (65i - 40i) = 150 + 25i \] **Step 5: Putting it all together:** \[ \frac{150 + 25i}{125} = \frac{150}{125} + \frac{25i}{125} = 1.2 + 0.2i \] Thus, the final result in the form \( a + ib \) is: \[ \boxed{1.2 + 0.2i} \]