Coordinates of the real derivative for $f:\mathbb{R}^2\to\mathbb{R}^2$ So. I am trying to see the similarity and difference between the real and complex differentiability.
Now the complex derivative at a point $z$ is defined as $$f'(z)=\lim_{h\rightarrow 0}\frac{f(z+h)-f(z)}{h}.$$ So $f'(z)$ is again a complex number. Precisely if $z=(x+iy), x,y\in\mathbb{R}$ and $f(z)=f(x+iy)=u(x,y)+iv(x,y)$, where $u$ and $v$ are the real and imaginary part of $f$, then $f'(x+iy)$ is calculated to be given as $f'(x+iy)=\frac{\partial u(x,y)}{\partial x}+i\frac{\partial v(x,y)}{\partial x}$.
Now my question is :
If $f:\mathbb{R}^2\longrightarrow\mathbb{R}^2$ is differentiable, then the derivative $f'(x,y)$ at a point $(x,y)$, is it an element of $\mathbb{R}^2$? If not why? If yes then what is the element? Is it because division is not defined in $\mathbb{R}^2$ and hence we have a whole new definition of differentiability for a function $f:\mathbb{R}^n\longrightarrow \mathbb{R}^m, n>1, m\geq 1.$
 A: You can more or less think of "the derivative" of a map $f\colon\Bbb R^m\to\Bbb R^n$ as a function $df\colon\Bbb R^m\to M_{n\times m}(\Bbb R)$, where $M_{n\times m}(\Bbb R)$ is the set of $n\times m$ matrices with real entries. For every point in $\Bbb R^n$, the derivative is another linear map $\Bbb R^m\to\Bbb R^n$. We can write such a map as an $n\times m$ matrix by picking a basis (the standard one in this case).
The first derivatives we encounter are of maps $f\colon\Bbb R\to\Bbb R$. In this case, $f'$ assigns to every point in the domain $\Bbb R$ a linear map $\Bbb R\to\Bbb R$, or a $1\times 1$ real matrix. It makes more sense to just think of the derivative as real-valued in this case.
For complex functions, things are analogous. If $f$ is a function $\Bbb C\to\Bbb C$, then we can write the derivative as a map $f'\colon\Bbb C\to\Bbb C$. However, if we consider $\Bbb C$ to be $\Bbb R^2$ and $f$ as a map $\Bbb R^2\to\Bbb R^2$, then the derivative assigns to each point $(x,y)\in\Bbb R^2$ a $2\times 2$-matrix $df_{(x,y)}$.
These two perspectives might seem incompatible. At each point in the domain, is the derivative a number or a matrix? The connection is that, in some cases, a $2\times 2$ matrix with real entries can be considered a $1\times 1$ matrix with complex entry. The convention is to identify $(1,0)$ with $1\in\Bbb C$ and $(0,1)$ with $i\in\Bbb C$. Then multiplication by $1\in\Bbb C$ (i.e., the $1\times 1$ complex matrix) becomes the $2\times 2$ real matrix $\begin{pmatrix}1 & 0\\0 & 1\end{pmatrix}$. Multiplication by $i$ takes $1$ to $i$ (i.e., $(1,0)$ to $(0,1)$) and $i$ to $-1$ (i.e., $(0,1)$ to $(-1,0)$). Thus $i$ becomes the $2\times 2$ real matrix $\begin{pmatrix}0 & 1\\-1 & 0\end{pmatrix}$.
A map $\Bbb R^2\to\Bbb R^2$ which is real-differentiable is not necessarily going to be complex differentiable when considered as a map $\Bbb C\to\Bbb C$. When it is, it means the derivative $2\times 2$ real matrix can be written $a\begin{pmatrix}1 & 0\\0 & 1\end{pmatrix}+b\begin{pmatrix}0 & 1\\-1 & 0\end{pmatrix}$ for some real $a,b$.
A: We say that $f\colon\Bbb R^n\longrightarrow\Bbb R^m$ is differentiable at a point $a\in\Bbb R^n$ if there is a linear map $L\colon\Bbb R^n\longrightarrow\Bbb R^m$ such that$$\lim_{x\to a}\frac{\|f(x)-f(a)-L(x-a)\|}{\|x-a\|}=0,$$and then $f'(a)$ is, by definition, the linear map $L$ (which is unique). The definition that you mentioned for $\Bbb C$ cannot be used here since there is no division in $\Bbb R^n$.
If $n=m=1$, that linear map $L$ is simply $x\mapsto f'(a)x$, where $f'(a)$ is the usual derivative of $f$ at $a$.
