Question

已知复数z1=a+2i(a∈R),z2=3-4i,且z1/z2为纯虚数,求复数z1

Answer 1

z1/z2=(a+2i)/(3-4i)

=[(a+2i)(3+4i)]/[(3-4i)(3+4i)]

=[3a-8+(6+4a)i]/(9+16)

=(3a-8)/25+[(6+4a)/25]i

因为z1/z2为纯虚数

所以(3a-8)/25=0

解得 a=8/3

所以 z1=8/3+2i