Question

若f(z)=2z+-3i,f(+i)=6-3i，试求f(-z).

Answer 1

解：∵f(z)=2z+-3i,

∴f(+i)=2(+i)+-3i

=2+2i+z-i-3i=2+z-2i.

又知f(+i)=6-3i,

∴2z+z-2i=6-3i.

设z=a+bi(a,b∈R),则=a-bi,

∴2（a-bi)+(a+bi)=6-i.

即3a-bi=6-i.

由复数相等定义解得

∴z=2+i.

故f(-z)=f(-2-i)=2(-2-i)+(-2+i)-3i=-6-4i.