v'y=2x,因此u'x=v'y=2x,积分得u=x^2+g(y),又由于u'y=-v'x,所以g'(y)=-2y,g(y)=-y^2+c,故u=x^2-y^2+c,f(z)=x^2-y^2+c+2ixy,所以f(i)=-1+c=-1,故c=0,因此u=x^2-y^2,f(z)=x^2-y^2+2ixy