How do you perform the operation and write the result in standard form given #(2-3i)^2#?

2 Answers
Mar 26, 2018

#-5-12i#

Explanation:

#color(orange)"Reminder "color(white)(x)i^2=(sqrt(-1))^2=-1#

#"note that "(2-3i)^2=(2-3i)(2-3i)#

#"expand factors using FOIL"#

#=4-12i+9i^2=-5-12ilarrcolor(red)"in standard form"#

Mar 26, 2018

#-5-12i#

Explanation:

First, multiply using the distributive property:

#(2-3i)^{2} = (2-3i)(2-3i)#

#(2-3i)(2-3i) = 4 - 6i - 6i + 9i^{2}#

Then simplify;

#4 - 6i - 6i + 9i^{2} = 4 - 12i + 9(-1)#

#4 - 12i + 9(-1) = 4 - 12i -9#

#4 - 12i - 9 = -5-12i#