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Suppose that $f \in H(\Omega)$ is holomorphic in an open set $\Omega \subseteq \mathbb{C}$. Suppose further that $\overline{B(z_0,r)} \subseteq \Omega$. I want to show that $$f(z_0) = \frac{1}{\pi r^2} \iint_{\overline{B(z_0, r)}} f(x+iy)dxdy.$$

I have attempted the following: Since $f$ is holomorphic in $B(z_0,r)$, we may write $f$ as the Cauchy integral $$f(z) = \frac{1}{2\pi i} \int_{\partial B(z_0,r)} \frac{f(\zeta)}{\zeta - z} d\zeta.$$ Stoke's theorem then allows us to convert this integral over the boundary, to an integral over the ball, so $$f(z) = \frac{1}{2\pi i} \iint_{\overline{B(z_0,r)}} \frac{\partial}{\partial \zeta} \left( \frac{f(\zeta)}{\zeta - z} \right) dxdy.$$ I don't think computing this partial derivative is worthwhile. But have not been able to proceed beyond this point.

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  • $\begingroup$ Change of variable $x+iy = R e^{it}, dxdy = R dR dt$ and try to recognize the Cauchy integral formula, or assume $f$ is analytic and given by a power series to solve it directly. $\endgroup$ – reuns Sep 12 '17 at 2:08
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Hint: Assume without loss of generality that $z_0$ is $0$. Then multiply both numerator and denominator by $\overline{\zeta}$. Then open everything up, and use Green's theorem.

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  • $\begingroup$ So you assume $f$ has continuous partial derivatives, and you are reproving Cauchy's integral theorem $\endgroup$ – reuns Sep 12 '17 at 2:26
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Hint: Polar coordinates gives

$$\int_{B(z_0,r)} f \, dA = \int_0^r\int_0^{2\pi}f(z_0 + se^{it})\,dt\,sds.$$

Cauchy's formula shows the inner integral on the right equals $2\pi f(z_0)$ for all $s,$ and the result falls right out.

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HINT:

Write $$f(z_0 + \rho e^{i\theta})= \sum_{n\ge 0} a_n \rho^n e^{i n \theta}$$

Now to integrate, use polar coordinates and the fact that $\int_{0}^{2\pi} e^{in \theta} d \theta = 0$ for $n>0$.

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