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Answer 1 · 2018-02-12T07:58:33

The answer is #=18+i#

The conjugate of a complex number #z=a+ib# is

#barz=a-ib#

where, # i^2=-1#

The complex number is

#=z_1*z_2#

The conjugate is

#bar((z_1*z_2))=(barz_1)*(barz_2)#

Therefore,

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

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

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

#=6+9i-8i-12i^2#

#=18+i#