Local epxansion of a holomorphic function into taylor series I am currently taking a beginners course in complex analysis and there it was mentioned that every holomorphic function can be locally expanded into a taylor series. 
From real analysis I know that a smooth function can be expanded into its corresponding taylor series, which does not need to converge to the function. It does not even need to converge at all.
What is the big difference to real analysis then? In the complex case it does not need to converge, too, right? Is it just the taylor coefficients that are granted to exist in a given expansion point, because the function is holomorphic and therefore smooth?
Could someone elaborate the exact differences between the real and complex case? I have a feeling, that I am missing something important, because the seemingly important differences seem trivial to me.
 A: This is one of the amazing things  about complex analysis: If $f$ is complex-differentiable in $D(0,r)$ then there exist $a_n$ so $$f(z)=\sum_{n=0}^\infty a_nz^n\quad(|z|<r).$$Amazing because as you note it's far from true in real analysis.
A: There are several stages that the complex analysis course typically goes through to arrive at this result:


*

*) the special form of complex differentiability as a case of 2D real differentiability,

*) the Cauchy-Riemann differential equations as defining property of analytic/holomorphic functions

*) The Goursat theorem about integrals over the sides of a triangle,

*) the existence of anti-derivatives

*) the Cauchy integral theorem and Cauchy integral formula, and 

*) with the help of those one finally arrives at the Taylor expansion via geometric series,
$$ \frac1{ζ−z}=\frac1ζ\,\frac{1}{1-\frac{z}{ζ}}=\sum_{k\ge 0}\frac{z^k}{ζ^{k+1}}$$ as follows.
For any $R>r>0$ so that $f$ is holomorphic on $B(0,R)$ and any $|z|<r$ 
$$
f(z)=\frac1{2\pi i}\oint_{|\zeta|=r}\frac{f(\zeta)}{\zeta - z}d\zeta
=\sum_{k=0}^\infty \left[\frac1{2\pi i}\oint_{|\zeta|=r}\frac{f(\zeta)}{\zeta^{k+1}}d\zeta\right]\;z^k
$$ 
and the integral in the coefficients is $$\frac1{2\pi r^k}\int_0^{2\pi}f(re^{it})e^{-ikt}dt$$ thus well-defined.
A: With functions of a real variable $z$, in the expression
$$
f'(z) = \lim_{\Delta z \to 0} \frac{f(z+\Delta z) - f(x)}{\Delta z}
$$
the bound variable $\Delta z$ can approach $0$ from either of two directions.
But if $z$ is a complex variable, it can approach from any of many directions. That means far fewer functions will qualify as differentiable: only those for which the limit exists and is the same number regardless of how $\Delta z$ approaches $0.$
And "holomorphic at $z$" must be construed not simply as "differentiable at $z$" but rather as "differentiable everywhere in some open neighborhood of $z$." By that definition, the function $z\mapsto |z|^2$ is differentiable at $0$ but is not holomorphic.
