$\int_D|f'(z)|^2dxdy=$Area$f(D)$

hbghlyj

Wikipedia

Hence there is a "length-area estimate": $\displaystyle {\int _{0}^{1}\ell (f\circ \gamma _{r})^{2}\,{dr \over r}\leq 2\pi \int _{|z|<1}|f^{\prime }(z)|^{2}\,dx\,dy=2\pi \cdot {\rm {Area}}\,f(D)<\infty .}$

为什么这里 $\int _{|z|<1}|f^{\prime }(z)|^{2}\,dx\,dy={\rm {Area}}\,f(D)$ 呢

Czhang271828

如果 $f:U\to f(U)$ 是单连通开集间的同构, 则直接由
\begin{align*}
\mathrm{Area}(f(U))&=\int_{f(U)}\mathrm dx\wedge \mathrm dy\\
&=\dfrac{i}{2}\int_{f(U)}\mathrm dz\wedge \mathrm d\overline z\\
&=\dfrac{i}{2}\int_{U}\mathrm d f(w)\wedge \mathrm d\overline{f(w)}\\
&=\dfrac{i}{2}\int_Uf'(w)\overline {f'(w)}\mathrm dw\wedge \mathrm d\overline w\\
&=\int_U|f'|^2\mathrm dx\mathrm dy.
\end{align*}
其中 $\mathrm dz\wedge \mathrm d\overline z=(\mathrm dx+i\mathrm dy)\wedge(\mathrm dx-i\mathrm dy)=-2i\mathrm dx\wedge \mathrm dy$.