设f(z)=u(x,y)+iv(x,y)为z=x+iy的解析函数已知u(x,y)-v(x,y)=x+y求f(z)

Question

设f(z)=u(x,y)+iv(x,y)为z=x+iy的解析函数 已知 u(x,y)-v(x,y)=x+y 求f(z)

Answer 1

设f(z)=u+iv为解析函数,则由
∂v/∂x=-∂u/∂y=-x+2y；
∂v/∂y=∂u/∂x=2x+y.
v=-x^2/2+2xy+y^2/2+C,C为常数.
f(z)=u+iv
=x^2+xy-y^2+i(-x^2/2+2xy+y^2/2+C)
=(1-i/2)(x^2+2ixy-y^2)+iC
=(1-i/2)(x+iy)^2+iC
=(1-i/2)z^2+iC,
f(i)=-1+i代入,得C=1/2,
f(z)=(1-i/2)z^2+i/2