Complex differentiability and differentiability in R2

Solution 1:

A function ${\bf f}:\>{\mathbb R}^2\to{\mathbb R}^2$ is differentiable at $p$ if there is a linear map $A:\>{\mathbb R}^2\to{\mathbb R}^2$ such that $${\bf f}({\bf p}+{\bf h})-{\bf f}({\bf p})=A{\bf h}+o(|{\bf h}|)\qquad({\bf h}\to{\bf 0})\ .\tag{1}$$ This $A$ is then called the derivative of ${\bf f}$ at ${\bf p}$, and is denoted by $d{\bf f}({\bf p})$, or similar. If $d{\bf f}({\bf p})$ exists then the four partial derivatives $f_{i.k}:={\partial f_i\over\partial x_k}({\bf p})$ exist, and the matrix of $d{\bf f}({\bf p})$ with respect to the standard basis in ${\mathbb R}^2$ is the matrix $[f_{i.k}]$.

Let such an ${\bf f}=(f_1,f_2)$ be given. The complexified version $f:=f_1+i f_2$ of this ${\bf f}$ is a differentiable function of the complex variable $z:=x_1+i x_2$ at the "complex point" $p$ iff the matrix of $d{\bf f}({\bf p})$ has the special form $$\left[\matrix{a&-b \cr b& a\cr}\right]\ .$$ The complex number $c:=a+ib$ is then the derivative of $f$ at $p$, and $(1)$ assumes the form $$f(p+h)-f(p)=c\>h+o(|h|)\qquad(h\to0\in{\mathbb C})\ .$$ In short: Complex differentiability is a special case of ${\mathbb R}^2$-differentiability, expressed in terms of other fonts.

Solution 2:

It is not sufficient to make a matrix/vector of partial derivatives to say that a function is differentiable in a point. You need to prove that the limit of the difference quotient is indipendent of the way $h$ approaches the point whatever is the set where the function is defined.