z=-1+2i,则:
(5i)/z
=(5i)/(-1+2i)
=[(5i)×(-1-2i)]/[(-1+2i)×(-1-2i)]
=[-(5i)×(1+2i)]/[(-1)²-(2i)²]
=[-(5i)×(1+2i)]/[5]
=-i(1+2i)
=2-i
z=-1+2i,则:
(5i)/z
=(5i)/(-1+2i)
=[(5i)×(-1-2i)]/[(-1+2i)×(-1-2i)]
=[-(5i)×(1+2i)]/[(-1)²-(2i)²]
=[-(5i)×(1+2i)]/[5]
=-i(1+2i)
=2-i