Complex differentiability vs differentiability in $\mathbb{R}^2$ I was wondering about the following for quite a while: The euclidean topology on $\mathbb{R}^2$ is homeomorphic with the topology induced by the modulus function on $\mathbb{C}$. Hence limit behavior is the same in both. So, doesn't that mean that the limit which defines the complex derivative converges if and only it converges in $\mathbb{R}^2$?

Clearly, I am missing something, since not every differentiable function $\mathbb{R}^2 \rightarrow \mathbb{R}^2$ is smooth, but over $\mathbb{C}$ this is true.

P.S.: I read the question: Is Complex Analysis equivalent Real Analysis with $f:\mathbb R^2 \to \mathbb R^2$?, it does however not clear up my confusion.
Derivatives are linear maps that approximate the function within $o$. In other words, if $f$ is differentiable at $x$, then $f(x+t)=f(x)+A(t)+o(t)$, where $A$ is a linear map and $\lim_{t\to 0}\frac{o(t)}{|t|}=0$. Here $|t|$ denotes the magnitude of $t$, which, as you pointed out, is the same whether we view $t$ as a complex number or a $2$-dimensional real vector. 

In the 2-dimensional real case, $A$ can be any linear map, ie. any $2\times 2$ matrix with entries in $\mathbb{R}$.

But in the complex case, $A$ is linear map of 1-dimensional complex vector spaces, ie. a $1\times 1$ matrix with entries in $\mathbb{C}$. In other words, $A$ is multiplication by some complex number $z$. If we think of $\mathbb{C}$ as $\mathbb{R}^2$, then $A$ is a rotation (by the argument of $z$) and scaling (by the magnitude of $z$). Not every $2\times 2$ linear map is of this form. 

In sum, you are right to note that the convergence is the same in each space. Real differentiability means that approximations converge to some linear map, while complex differentiability means that approximations converge to a more specific type of linear map, which is why complex differentiability is a stronger condition than differentiability in $\mathbb{R}^2$.
The limit that defines $\mathbb{R}^n$ differentiability is:
$$\lim_{\mathbf{h}\to \mathbf{0}}\frac{\mathbf{f}(\mathbf{a}+\mathbf{h})-\mathbf{f}(\mathbf{a})-T(\mathbf{h})}{\left\|\mathbf{h}\right\|}=\mathbf{0}$$
where $T$ is a linear transform. The limit that defines $\mathbb{C}$ differentiability is not
$$\lim_{h\to 0}\frac{f(z+h)-f(z)-T(h)}{\left|h\right|}=0$$
but rather
$$\lim_{h\to 0}\frac{f(z+h)-f(z)-T(h)}{h}=0$$
In addition, in the complex case we also require $T$ to be a complex linear transform, not just a linear map, that must be "nice" towards complex multiplication.These two facts create the so called Cauchy Riemann Equations. You can read more here pg 111.
 In other words, multiplication and division of complex numbers make $\mathbb{C}$  differentiabilty much stronger than $\mathbb{R}^2$ differentiability.