Past exam of the algebra, analysis and differential equations II course of the University of Oxford / 2006 / Analysis / 4

A function is holomorphic on an open set if it is complex differentiable at every point of that set. At $z_0=x_0+i{y}_0$, take difference quotients first along real increments. Their limit is $f^{\prime }(z_0)$, so $u_x=\mathrm{Re}{f}^{\prime }(z_0)$ and $v_x=\mathrm{Im}{f}^{\prime }(z_0)$. Along imaginary increments,so $u_y=-\mathrm{Im}{f}^{\prime }(z_0)$ and $v_y=\mathrm{Re}{f}^{\prime }(z_0)$. These limits prove existence of all four partial derivatives and the Cauchy-Riemann equations $u_x=v_y$, $u_y=-v_x$.

For the first example, the identity function and the reciprocal function are holomorphic away from zero, so $f$ is holomorphic on the given domain, with

Write $r=\sqrt{x^2+y^2}>0$. For $u=x/(1+r)$, $v=y/(1+r)$,The two Cauchy-Riemann equations would require $x^2=y^2$ and $xy=0$ simultaneously, impossible when $r>0$. Thus $f$ is not complex differentiable at any nonzero point.

At zero, however,So $f$ is complex differentiable at zero, with derivative one, but is not holomorphic on $\mathbb{C}$ or on any nonempty open subset. Complex differentiability at a single point is weaker than holomorphy in a neighborhood.