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)

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-...