因为f(z）=2z+z'-3i,把z'+i代入有：f(z'+i）=2(z'+i)+z'-3i=3z'-i又因为：f(z'+i）=6-3i.令z'=x+yi.x,y是实数,代入上式有：3x+(3y-1)i=6-3i,所以x=2,y=-2/3,z'=2-(2/3)i所以f(z)=2z+2-(2/3)i-3i=2z+2-(11/3)i所以f(-...