How do you simplify #(3-1i)(4+6i)#?

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Answer 1 · 2016-09-24T14:55:05

#(3-4i)(4+6i)=36+2i#

While multiplying complex numbers, one must always remember that #i^2=-1#

Hence #(3-4i)(4+6i)#

= #3(4+6i)-4i(4+6i)#

= #3xx4+3xx6i-4ixx4-4ixx6i#

= #12+18i-16i-24i^2#

= #12+18i-16i-24xx(-1)#

= #12+18i-16i+24#

= #36+2i#