Assume two functions \(u, v: \RR^2 \to \RR\) have continuous partial derivatives at \((x_0 ,y_0)\). Show that \(f \definedas u + iv\) has derivative \(f'(z_0)\) at \(z_0 = x_0 + iy_0\) if and only if \[ \lim _{r \rightarrow 0} \frac{1}{\pi r^{2}} \int_{\left|z-z_{0}\right|=r} f(z) d z=0 .\]
Solution
Put \(h=z-z_0\). Real differentiability at \((x_0,y_0)\) gives \[ f(z_0+h)=f(z_0)+A h+B\bar h+o(|h|), \] where \[ B={1\over2}\left(u_x-v_y+i(v_x+u_y)\right)_{(x_0,y_0)}. \] On \(|h|=r\) we have \(\bar h=r^2/h\), and therefore \[ \begin{aligned} \int_{|z-z_0|=r}f(z)\,dz &=A\int_{|h|=r}h\,dh +B\int_{|h|=r}\bar h\,dh+o(r^2)\\ &=2\pi i r^2B+o(r^2). \end{aligned} \] Consequently \[ \lim_{r\to0}{1\over\pi r^2} \int_{|z-z_0|=r}f(z)\,dz =2iB. \] This limit is zero exactly when \[ u_x=v_y, \qquad u_y=-v_x \] at \((x_0,y_0)\), i.e. exactly when the Cauchy–Riemann equations hold there. Because the real partial derivatives are continuous at the point, the usual Cauchy–Riemann criterion applies, so this is equivalent to complex differentiability of \(f\) at \(z_0\).