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

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(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
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