(5+2 i)(4-3i) - (5-2yi)(4-3i) in a + bi form, where y is a real number.

(5+2 i)(4-3i) - (5-2yi)(4-3i)

Factorize out (4 -3i)

(4 -3i)( (5 +2i) - (5 -2yi) )

=  (4 -3i)(5 +2i - 5 + 2yi)

=  (4 -3i)(5 - 5 + 2i + 2yi)

= (4 -3i)(2i + 2yi)

= (4 - 3i)(2 + 2y)i.    Let's multiply the first two.

(4 - 3i)(2 + 2y) = 2*(4 -3i) + 2y*(4 - 3i)           
                      =  8 - 6i + 8y - 6yi
                      =  8 + 8y - 6i - 6yi

(4 - 3i)(2 + 2y)i = (8 + 8y - 6i - 6yi)i                Note: i² = -1
                       =  8i  + 8yi - 6i² - 6yi²
                       =  8i  +  8yi - 6(-1) - 6y(-1)  
                       =  8i  +  8yi  + 6 + 6y
                       =  6 + 6y + 8i + 8yi
                       =  (6 + 6y) + (8 + 8y)i    In the form a + bi

Iwould like to think so. what’s the question you need with?