1.(1)∵z=m(m-1)+(m2+2m-3)i=0,∴,解得m=1; (2)∵z=m(m-1)+(m2+2m-3)i是纯虚数,∴,解得m=0; (3)∵z=m(m-1)+(m2+2m-3)i=2+5i,∴,解得m=2, 2.∵z=m(m-1)+(m2+2m-3)i,且|z|=1,∴m2(m-1)2+(m2+2m-3)2=1 化简得,2m4+2m3-m2-12m+8=0 ①, ∵(3+4i)•z=(3+4i)[m(m-1)+(m2+2m-3)i]=(-m2-11m+12)+(7m2+2m-9)i,且它是纯虚数, ∴-m2-11m+12=0,解得,m=-12或1,代入①式验证也成立,故z=±(+i), 则=z=±(-i). |