翻翻书,就有了：记
f(z) = u(x,y) +iv(x,y),
有
u(x,y) = x^3 - 3xy^2,v(x,y) = 3yx^2 - y^3,
且
Du(x,y)/Dx = 3x^2 - 3y^2 = Dv(x,y)/Dy,
即满足 Cauchy-Riemann 条件,因此
f'(z) = Du(x,y)/Dx +i[Dv(x,y)/Dx]
= (3x^2 - 3y^2) + i6xy
= 3z^2.
这个函数怎么对求导
翻翻书,就有了：记
f(z) = u(x,y) +iv(x,y),
有
u(x,y) = x^3 - 3xy^2,v(x,y) = 3yx^2 - y^3,
且
Du(x,y)/Dx = 3x^2 - 3y^2 = Dv(x,y)/Dy,
即满足 Cauchy-Riemann 条件,因此
f'(z) = Du(x,y)/Dx +i[Dv(x,y)/Dx]
= (3x^2 - 3y^2) + i6xy
= 3z^2.