What does #(1+i)*(6-2i)-4i# equal?

1 Answer
Alan P.
Oct 29, 2015

#(1+i) * (6-2i) = 8#

Explanation:

First evaluate #(color(red)(1+i)) * (color(blue)(6-2i))#
This can be done in various ways: FOIL, or using the distributive property, or tabular multiplication (below)
#color(white)("XXX"){: (xx, color(red)(1),color(red)(+i)), (color(blue)(6),color(orange)(6),color(green)(+6i)), (color(blue)(-2i),color(green)(-2i),color(orange)(+2)), (,"-----","-----"), (,color(orange)(8),color(green)(+4i)) :}#

If #color(cyan)((1+i) * (6-2i)) = color(cyan)(8+4i)#
then #color(cyan)((1+i) * (6-2))-4i = color(cyan)(8+4i)-4i = 8#

Related questions
See all questions in Radical Equations
Impact of this question
1295 views around the world
You can reuse this answer
Creative Commons License