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Assuming that $f$ has $C^1$ real and imaginary parts, prove that:

$$\displaystyle \lim_{r \to 0} \frac{1}{r^2} \int_{C_r}f(z)dz = 2\pi i\frac{\partial{f}}{\partial{\bar{z}}}(z_0).$$

Additionally, show that the above equality implies that $f$ is differentiable at $z_0$ iff the limit on the left hand side vanishes.

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1 Answer 1

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I think $C_r$ should be $\{z\in\mathbb{C}:|z-z_0|=r\}$ and is anticlockwise oriented. Then for $D_r=\{z\in\mathbb{C}:|z-z_0|\le r\}$, $C_r=\partial D_r$. Since $f$ is $C^1$, by Stokes' theorem(or Green's theorem in particular), $$\int_{C_r}fdz=\int_{D_r}df\wedge dz=\int_{D_r}\frac{\partial f}{\partial \bar{z}}d \bar{z} \wedge dz=2i\int_{D_r}\frac{\partial f}{\partial \bar{z}}dx\wedge dy.$$ The conclusion follows from continuity of $\frac{\partial f}{\partial \bar{z}}$ and the fact $\int_{D_r}dx\wedge dy=\pi r^2$.

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