As we all know, all the basic properties of holomorphic functions (i.e. functions which are differentiable in the complex sense) can be deduced from Cauchy's formula. Moreover, Cauchy's formula itself can be viewed as a rather simple consequence of the Green-Riemann formula, provided that the holomorphic function you have at hand is assumed to have a continuous derivative.
Of course, Cauchy's formula holds without assuming continuity of the derivative, and it yields continuity of the derivative and much more since it implies power series expansion. But Cauchy's formula (or, if you prefer, Cauchy's theorem) without assuming continuity of the derivative is a rather subtle thing, and this gives a rather "indirect" proof of the fact that holomorphic functions are in fact $\mathcal C^1$.
So my question is the following: does anybody know a direct proof of the fact that if a function $f$ defined on an open subset of $\mathbb C$ is differentiable in the complex sense, then its derivative $f'$ is continuous?
I'm pretty sure I am not the first one and will not be the last one to ask this question, at least for himself (or herself). So please feel free to close it if it has indeed been asked previously on this site.
Edit. Perhaps I should say a few more words about what I mean by a "direct proof". Anything that relies in one way or another to Cauchy's formula or Cauchy's theorem is not considered as a direct argument. A direct proof should somehow establish "from scratch", or "from very basic principles" that holomorphic implies $\mathcal C^1$.
Best Answer
There is an efficient and straightforward proof of the Cauchy-Goursat theorem, by a `lion-hunting' argument based only on differentiability at every point in $\Omega$, available at this link.
In an honors undergraduate analysis class that has dealt with path integrals, one can use it to provide an account of the Cauchy integral formula, analyticity and the classification of zeros and poles in only a couple of weeks.