How do you divide # (5-2i)+(3-2i)#?

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Answer 1 · 2017-01-05T09:52:03

#19/13+4/13i#

You can multiply both the terms by #3+2i# and get:

#(5-2i)color(red)((3+2i))-:(3-2i)color(red)((3+2i))#

Then, since

#(a+b)(a-b)=a^2-b^2#

expand:

#(15+10i-6i-4i^2)-:(9-4i^2)#

Then, since #i^2=-1#, the expression becomes:

#(15+4i+4)-:(9+4)#

#(19+4i)-:13#

that's, in standard form:

#19/13+4/13i#