f(z)=2z+z'-3i f(z'+i)=6-3i,则f(-z)=?我知道答案是-6-4i,大概过程我也明白,但没算明白,

来源:学生作业帮助网 编辑:作业帮 时间:2024/05/05 18:48:29

f(z)=2z+z'-3i f(z'+i)=6-3i,则f(-z)=?我知道答案是-6-4i,大概过程我也明白,但没算明白,
f(z)=2z+z'-3i f(z'+i)=6-3i,则f(-z)=?
我知道答案是-6-4i,大概过程我也明白,但没算明白,

f(z)=2z+z'-3i f(z'+i)=6-3i,则f(-z)=?我知道答案是-6-4i,大概过程我也明白,但没算明白,
因为f(z)=2z+z'-3i,把z'+i代入有:f(z'+i)=2(z'+i)+(z-i)-3i=6-3i
所以2z'+z+i=6
所以2z'+z=6-i
所以2z+z'=6+i
f(-z)=-2z-z'-3i=-6-i-3i=-6-4i

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(-z)=-2z+2-(11/3i)