How do you simplify #(4+4i)(3+5i)#?
1 Answer
Feb 14, 2016
- 8 + 32i
Explanation:
distribute the brackets usingFOIL ( or any method you have )
hence ( 4 + 4i )(3 + 5i )
# = 12 + 20i + 12i + 20i^2 # [ note:
# i^2 = (sqrt-1)^2 = -1 # ]hence :
# 12 + 20i + 12i - 20 = - 8 +32i#
