有大神知道这题怎么证明吗?求过程。。。

令f(z)=u(x,y)+iv(x,y)
则f'(z)=∂u/∂x+i∂v/∂x
|f(z)|^2=u^2+v^2
|f'(z)|^2=(∂u/∂x)^2+(∂v/∂x)^2
因为f(z)在D内解析，所以∂u/∂x=∂v/∂y，∂u/∂y=-∂v/∂x
所以
(∂^2/∂x^2+∂^2/∂y^2)|f(z)|^2
=(∂^2/∂x^2+∂^2/∂y^2)(u^2+v^2)
=∂^2(u^2)/∂x^2+∂^2(v^2)/∂x^2+∂^2(u^2)/∂y^2+∂^2(v^2)/∂y^2
=∂[2u(∂u/∂x)]/∂x+∂[2v(∂v/∂x)]/∂x+∂[2u(∂u/∂y)]/∂y+∂[2v(∂v/∂y)]/∂y
=2(∂u/∂x)^2+2u[∂^2(u)/∂x^2]+2(∂v/∂x)^2+2v[∂^2(v)/∂x^2]+2(∂u/∂y)^2+2u[∂^2(u)/∂y^2]+2(∂v/∂y)^2+2v[∂^2(v)/∂y^2]
=2(∂u/∂x)^2+2u[∂(∂v/∂y)/∂x]+2(∂v/∂x)^2+2v[∂(-∂u/∂y)/∂x]+2(-∂v/∂x)^2+2u[∂(-∂v/∂x)/∂y]+2(∂u/∂x)^2+2v[∂(∂u/∂x)/∂y]
=4(∂u/∂x)^2+4(∂v/∂x)^2+2u[∂^2(v)/∂x∂y]-2v[∂^2(u)/∂x∂y]-2u[∂^2(v)/∂x∂y]+2v[∂^2(u)/∂x∂y]
=4(∂u/∂x)^2+4(∂v/∂x)^2
=4|f'(z)|^2

太简单了