证明：因为f(z)解析,所以f'(z)=du/dx+idv/dx且du/dx和dv/dx不同时为0由隐函数求导法曲线u(x,y)=c1的斜率k1=-(du/dx)/(du/dy)同理导法曲线u(x,y)=c1的斜率k2=-(dv/dx)/(dv/dy)若du/dx和dv/dx都不为0,根据C-R方程du/dx...