How do you simplify #(3i)*2+(2i)#?

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Answer 1 · 2016-08-06T15:37:34

#8i#

Complex numbers are just like any other numbers -- just treat #i# like a variable.

The first thing to do, according to the order of operations, is multiply.

#overbrace((3i)*2)^("multiply first")+(2i)#

Since #3i# is #3*i#, #(3i)*2=(3*2)*i=6i#.

#=6i+2i#

From here, add this like you would for #6x+2x#. Another way to envision this is to factor out #i# first:

#=i(6+2)#

#=i*8#

#=8i#