答设z=a+bi则x=a-bixz-3ix=1+3ia²+b²-3i(a-bi)=1+3ia²+b²-3ai-3b=1+3i(a²+b²-3b)-3ai=1+3i∴a²+b²-3b=1-3a=3a=-1∴b²-3b=0b=0或3∵z∈C∴b=3∴z=-1+3i如果你认可我的...