Complex Differentiation - why treat z as a single variable Why does one treat in Complex Analysis in the case of a function of a complex variable $z=x+i y$ the complex variable $z$ as a single variable and not the imaginary part just as another axis? In this way one could define a directional derivative and have more freedom?
 A: This can be understood in terms of linear algebra. On the one hand we have ${\mathbb R}$-linear maps
$$A:\quad {\mathbb R}^2\to{\mathbb R}^2,\qquad(x,y)\mapsto (u,v):=(ax+by,cx+dy)\ ,\tag{1}$$
encoded in a matrix $\left[\matrix{a&b\cr c&d \cr}\right]$ with real entries $a$, $b$, $c$, $d$,  and on the other hand we have ${\mathbb C}$-linear maps
$$T:\quad {\mathbb C}\to{\mathbb C},\qquad z\mapsto w:=r\, z\tag{2}$$
characterized by a single complex constant $r=p+iq\in{\mathbb C}$. 
The identification of $z=x+iy\in {\mathbb C}$ with $(x,y)\in{\mathbb  R}^2$, and similarly of $w=u+iv\in {\mathbb C}$ with $(u,v)\in{\mathbb R}^2$ allows to view any map $T$ of type $(2)$ as a map $A$ of type $(1)$ with matrix $$M_r:=\left[\matrix{p&-q\cr q&p \cr}\right]\ .\tag{3}$$ 
The real matrices $M_r$ arising in this way are special, and you certainly have less freedom. Geometrically, such a matrix describes a rotation by the angle ${\arg}(p,q)$, followed by a scaling with the factor $\sqrt{p^2+q^2}$.
Concerning maps $f:\>{\mathbb R}^2\to{\mathbb R}^2$ we can learn the following from these observations: If the  matrices of the Jacobian maps $df(x,y)$ have the property $(3)$ at all points $(x,y)$ in the domain of $f$, then such an $f$ is very special, and can be realized in a $1$-dimensional complex environment. If not, such an $f$ is just differentiable "in the real sense".
