Question

复数z1=3+4i,z2=1+i,i为虚数单位,若z2^2=z×z1,则复数z 需要完整过程

Answer 1

z2=1+i

z2^2

=(1+i)^2

=1+2i-1

=2i

z1=3+4i

z2^2=z×z1

z

=2i/(3+4i)

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

=(6i+8)/(9+16)

=8/25+(6/25)i