设Z=a+bi,则：Z-4=(a-4)+bi 因为arg（Z-4)=π/4 所以a-4=b |Z|=2倍根号10,故：a^2+b^2=40 a^2+(a-4)^2=40 2a^2-8a+16=40 a^2-4a-12=0 (a-6)(a+2)=0 a=-2或6 b=a-4=-6或2 故：Z=-2-6i或6+2i