证明函数f(z)=x^2+2xy-y^2-i(x^2-2xy-y^2)在复平面内处处解析并求其导数证明函数f(z)=x^2+2xy-y^2-i(x^2-2xy-y^2)在复平面内处处解析并求其导数,

Question

证明函数f(z)=x^2+2xy-y^2-i(x^2-2xy-y^2)在复平面内处处解析并求其导数
证明函数f(z)=x^2+2xy-y^2-i(x^2-2xy-y^2)在复平面内处处解析并求其导数,

Answer 1

方法一是用轲西黎曼方程.
方法二是直接配成z=x+iy的函数
f(z)=(x+yi)^2-2xyi+2xy-i[(x+yi)^2-2xyi-2xy]
=(x+yi)^2-2xyi+2xy-i(x+yi)^2-2xy+2xyi
=(x+yi)^2-i(x+yi)^2
=(1-i)z^2
因此它是复平面处处解析的函数
f'(z)=2(1-i)z