How do you simplify #(4-2i) (3 + 2i)#?

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Answer 1 · 2016-06-14T04:54:45

#(4-2i)(3+2i)=16+2i#

To simplify #(4-2i)(3+2i)#, just multiply them like two binomials, but remember #i^2=-1#.

Hence, #(4-2i)(3+2i)#

= #4(3+2i)-2i(3+2i)#

= #12+8i-6i-4i^2#

= #12+2i-4(-1)#

= #12+2i+4#

= #16+2i#