How do you evaluate #(4- 2i ) - ( 8i )#?

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Answer 1 · 2017-12-22T11:52:35

#-10i+4#

#"simplify by collecting like terms"#

#rArrcolor(purple)(4)color(red)(-2i)color(red)(-8i)#

#=-10i+4larrcolor(red)"in standard form"#

Answer 2 · 2017-12-22T11:53:40

#4-10i#

#z = (4-2i) - 8i#

#z = 4- (2+8)i#

#= 4-10i#