Question

perform the indicated operations.
\\((3 - 4i)(4 - i) - (2 - i)(2 + i)\\)
\\((3 - 4i)(4 - i) - (2 - i)(2 + i) = \square\\)
(simplify your answer. type your answer in the form \\(a + bi\\).)

Explanation:

Step1: Expand first complex product

Multiply $(3-4i)(4-i)$ using FOIL:
$3\times4 + 3\times(-i) -4i\times4 -4i\times(-i) = 12 - 3i -16i +4i^2$
Since $i^2=-1$, substitute:
$12 -19i +4(-1) = 12 -19i -4 = 8 -19i$

Step2: Expand second complex product

Multiply $(2-i)(2+i)$ (difference of squares):
$2^2 - i^2 = 4 - (-1) = 4 +1 = 5$

Step3: Subtract the two results

Subtract the second product from the first:
$(8 -19i) - 5 = 8-5 -19i$

Answer:

$3 - 19i$