How do you simplify #(7-6i)(-8+3i)#?

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Answer 1 · 2016-09-02T14:43:10

#(7-6i)(-8+3i)=-38+69i#

While doing any multiplication or division exercise involving complex numbers, one must always remember that #i^2=-1#.

Hence, #(7-6i)(-8+3i)#

= #7×(-8)+7×3i-6i×(-8)-6i×3i#

= #-56+21i+48i-18i^2#

= #-56+69i-18×(-1)#

= #-56+69i+18#

= #-38+69i#