Question

Perform the indicated operation
(1 – 3i)(5 + 5i)(4 – 1i) =

Answer 1

(1 - 3i)(5 + 5i)(4 - 1i)

= (5 + 5i - 15i - 15i²)(4 - 1i)

= (5 - 10i - 15i²)(4 - 1i)

= 20 - 40i - 60i² - 5i + 10i² + 15i³

= 20 - 45i - 50i² + 15i³

Answer 2

70-60i

Explanation:

(1 – 3i)(5 + 5i)(4 – 1i) =

taking the first two, before we do the last one

(1 – 3i)(5 + 5i)

1(5+5i)-3i(5+5i)

1(5)+1(5i)-3i(5)-3i(5i)

5+5i-15i-15i²

from the miltiplication law, i² is -1

5-10i-15i² = 5-10i-15(-1) = 5-10i+15 = 5+15-10i = 20-10i

then multiply (20-10i) (4 – 1i)

20(4 – 1i)-10i(4 – 1i)

20(4)-20(1i)-10i(4)-10i(-1i)

80-20i-40i+10i²

80-60i+10(-1)

80-60i-10

80-10-60i

(1 – 3i)(5 + 5i)(4 – 1i) =70-60i