Show that $\lim_{r\to 0} \frac{1}{r^2}\int_{C_{r}}f(z)dz = 2 \pi i\frac{\partial f}{\partial \bar{z}}(z_0)$ Well, after spending hours on this problem, I'm still stuck, so I thought I'd turn to you guys. The problem statement is as follows.
Let $f$ be a complex-valued function that is $C^1$ in the disk $|z - z_0| < R$ for some $R>0$, and let $C_r$ be the circle $|z - z_0| = r$. Show that $$\lim_{r\to 0} \frac{1}{r^2}\int_{C_{r}}f(z)dz = 2 \pi i\frac{\partial f}{\partial \bar{z}}(z_0).$$
I tried parametrizing the circle in the usual way and differentiating to see if I get something useful, but no luck. I just can't get the LHS to give me something to do with $\frac{\partial f}{\partial \bar{z}}$, so that I could then put the RHS under the integral and show that the difference between the two can be less than arbitrary $\epsilon > 0$ by the continuity of partial derivatives.
Thanks in advance for your help!
 A: It is convenient to use little-o notation. Apply the Taylor estimate for complex functions. If $f$ is $C^1$ in a neighborhood of $z_0$ then $$f(z) = f(z_0) + \frac{\partial f}{\partial  z}(z_0) (z - z_0) + \frac{\partial f}{\partial \bar z} (z_0) (\bar z - \bar z_0) + o(|z - z_0|).$$
According to Cauchy's theorem you have $$\int_{C_r} f(z) \, dz = \frac{\partial f}{\partial \bar z} (z_0) \int_{C_r} (\bar z - \bar z_0) \, dz + \int_{C_r} o(|z - z_0|) dz.$$
The latter integrals can be parameterized using $z = z_0 + re^{it}$.  Then $\bar z - \bar z_0 = r e^{-it}$ and $dz = ire^{it}$ so that
$$\int_{C_r} (\bar z - \bar z_0) \, dz = \int_0^{2\pi} re^{-it} i re^{it} \, dt = 2\pi i r^2$$
and
$$\left| \int_{C_r} o(|z - z_0|) dz \right|\le \int_{C_r} o(|z - z_0|) |dz| = \int_0^{2\pi} o(r) r \,dt = 2\pi r^2 \frac{o(r)}{r}.$$
Thus
$$\frac{1}{r^2} \int_{C_r} f(z) \, dz = 2\pi i \frac{\partial f}{\partial \bar z} (z_0) + 2\pi \frac{o(r)}{r}.$$

Comment on notation: we can express points $z \in \mathbb C$ as $z = x + iy$.  Then $\bar z = x - iy$, $x = \frac 12 (z + \bar z)$ and $y = -\frac i2 (z - \bar z)$. The $z$ and $\bar z$ derivatives of $f$ are defined formally via the chain rule: $$\frac{\partial f}{\partial z} = \frac{\partial f}{\partial x} \frac{\partial x}{\partial z} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial z} = \frac 12[ f_x - i f_y]$$
$$\frac{\partial f}{\partial \bar z} = \frac{\partial f}{\partial x} \frac{\partial x}{\partial \bar z} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial \bar z} = \frac 12[ f_x + i f_y].$$
If $f$ is analytic then $\dfrac{\partial f}{\partial z} = f'(z)$ and $\dfrac{\partial f} {\partial \bar z} = 0$.
