Prove that $f$ is differentiable in a whole $\mathbb{C}$ set.

Question

If $f:\mathbb{C}→\mathbb{C}$ is a continuous function in a whole $\mathbb{C}$ set and differentiable in a set $\mathbb{C}$ \ $\mathbb{R}$. Prove that $f$ is differentiable in a whole $\mathbb{C}$ set. Proof must be based on Morera's theorem.

Answer 1

Here is a sketch of a straight forward approach. There might be easier possibilities, though:

We are only interested in integration paths that "cut" the real axis. Let $a$, $b$ be points on the real axis, $\gamma_1$ be a path from $b$ to $a$ in the upper half plane (imaginary part $>0$) and $\gamma_2$ a path from $a$ to $b$ in the lower half plane. Both $\gamma_1$ and $\gamma_2$ must not cut the real axis in any other point.

The goal is to show that the integral of $f$ vanishes along the union $\gamma_1 \cap \gamma_2$. Consider a path $\beta_1$ from $a$ to $b$ in the upper half plane. The union $\gamma_1 \cap \beta_1$ is then a closed curve in the upper half plane along which the integral of $f$ vanishes by assumption. Construct a path $\beta_2$ in the lower half plane accordingly.

The sum of integrals along $\gamma_1 \cap \beta_1$ and $\gamma_2 \cap \beta_2$ vanishes. Now find a way to construct series of paths $\beta^i_1,\;\beta^i_2$ so that $\beta^i_1$ and $\beta^i_2$ close in on each other for $i\to\infty$. In the limit the integral of $f$ along the union $\beta_1^i \cap \beta_2^i$ will vanish because of the continuity of $f$ and the paths becoming close to each other with opposite (!) orientation (mind that when constructing them!).

Due to the integral along the $\beta$ cancelling out in the limit, we will be left with the integral along $\gamma_1 \cap \gamma_2$ being zero (qed).

The non-trivial part is constructiong the $\beta$ and applying the continuity arguments. This can be tedious depending on what has already been shown.