How do you multiply #(4-2i)^2#?

2 Answers
Sep 30, 2016

#(4-2i)^2=12-16i#

Explanation:

#(4-2i)^2#

= #(4-2i)(4-2i)#

= #4(4-2i)-2i(4-2i)#

= #16-8i-8i+4i^2# and as #i^2=-1#

= #16-16i-4#

= #12-16i#

Sep 30, 2016

#12-16i#

Explanation:

distribute the brackets using, for example, the FOIL method.

#rArr(4-2i)^2=(4-2i)(4-2i)=16-8i-8i+4i^2#

#color(orange)"Reminder " color(red)(bar(ul(|color(white)(a/a)color(black)(i^2=(sqrt(-1))^2=-1)color(white)(a/a)|)))#

#16-8i-8i+4i^2=16-16i-4#

#rArr(4-2i)^2=12-16i#