How do you simplify #(7-4i)(4-i)#?

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Answer 1 · 2016-09-09T14:18:02

#24 - 23 i#

We have: #(7 - 4 i) (4 - i)#

Let's expand the parentheses:

#= (7) (4) + (7) (- i) + (- 4 i) (4) + (- 4 i) (- i)#

#= 28 - 7 i - 16 i + 4 i^(2)#

Then, let's apply the fact that #i^(2) = - 1#:

#= 28 + (4 cdot - 1) - 23 i#

#= 28 - 4 - 23 i#

#= 24 - 23 i#