Prove that $\lim_{r \to 0} \frac{1}{r^2} \int_{|z| =r} f(z)dz = 2\pi i\frac{\partial f}{\partial \overline z}(0) $. Given a Frechet differentiable function $f:U\to \mathbb{C}$, where $U$ is a open disk centered at 0 with some radius. Prove that $$\lim_{r \to 0} \frac{1}{r^2} \int_{|z| =r} f(z)dz = 2\pi i\frac{\partial f}{\partial \overline z}(0) $$
I know that the integral inside can be expressed in a parameterized form $z = re^{i\theta}$, which makes the integral $\int_{0}^{2\pi} f(re^{i\theta})rie^{i\theta}d\theta$. After that I have no idea  how to use the Frechet differentiability. Any help will be appreciated.
 A: This expression can be used to define a more general derivative (see aereolar derivative).
Recall the definition of Wirtinger derivatives. Let $\mathrm{U}$ be an open subset of $\mathbb{C}$ and $f:\mathrm{U}\to\mathbb{C}$ be Fréchet differantiable. The Wirtinger derivatives with respect to $z=x + i y$ are:
$$ \frac{\partial f}{\partial z} = \frac{1}{2} \left( \frac{\partial f}{\partial x} + \frac{1}{i}\frac{\partial f}{\partial y} \right)\, ,$$
$$ \frac{\partial f}{\partial \bar{z}} = \frac{1}{2} \left( \frac{\partial f}{\partial x} - \frac{1}{i}\frac{\partial f}{\partial y} \right)\, .$$
For an Holomorphic function holds $\frac{\partial f}{\partial \bar{z}} =0$ (as well as $\frac{\partial f}{\partial z} = f'$) and the integral $\int_{\gamma} f(z)dz$ is zero by Cauchy's theorem. The problem is trivial in this situation.
In a general case, write $f=u+iv$. The Wirtinger derivative is
$$ \frac{\partial f}{\partial \bar{z}} = \frac{1}{2} \left( \frac{\partial u}{\partial x} - \frac{\partial v}{\partial y} \right)+\frac{i}{2} \left( \frac{\partial v}{\partial x} - \frac{\partial u}{\partial y} \right)\, .$$
Let $\Gamma$ be an open circle centered at the origin and with radius $r$ and $R$ be the disk that is bounded by $\Gamma$. The integral of $f$ in $\Gamma$ can be re-written using Stokes theorem
\begin{align}
\int_{\Gamma} f(z)dz &= \int_{\Gamma} (u dx - v dy ) \; + \; i \int_{\Gamma} (v dx + u dy ) \\[6pt]
&= \int_{R} \left[  - \left( \frac{\partial v}{\partial x} - \frac{\partial u}{\partial y} \right) + i \left( \frac{\partial u}{\partial x} - \frac{\partial v}{\partial y} \right)
\right] dx dy\\[6pt]
&= 2 i   \int_{R}  \frac{\partial f}{\partial \bar{z}} dx dy \; .
\end{align}
Since $\frac{\partial f}{\partial \bar{z}}$ is continuous at $z=0$, as $r \to 0$ the right-hand-side of the last equation goes to $2 i  \cdot \frac{\partial f}{\partial \bar{z}}(0) \cdot \pi r^2$. Thus, dividing by $2\pi i r^2$ and taking the limit we obtain:
$$  \frac{\partial f}{\partial \bar{z}}(0) = \frac{1}{2 \pi i} \lim_{r \to 0} \frac{1}{r^2} \int_{\vert z \vert = r} f(z)dz  \; .$$
References:

*

*Henrici, Peter [1986], Applied and Computational Complex Analysis Volume 3 - Section 15.10.
A: All we need is $$f(z)= f(0)+x\partial_x f(0)+y\partial_y f(0)+o(|z|)$$
$$ = f(0)+x(\partial_z f(0)+\partial_\overline{z} f(0))+iy(\partial_z f(0)-\partial_\overline{z} f(0))+o(|z|)$$
$$=f(0)+z\, \partial_z f(0)+\overline{z}\, \partial_{\overline{z}} f(0)+o(|z|)$$
Then $$r^{-2}\int_{|z|=r}\overline{z}\ \partial_{\overline{z}} f(0)dz=2i\pi \partial_{\overline{z}} f(0)$$
$$r^{-2}\int_{|z|=r} f(0)dz= 0,\qquad r^{-2}\int_{|z|=r} z \partial_z f(0)dz= 0,\qquad r^{-2}\int_{|z|=r} o(|z|)dz= o(1)$$
