Is Complex Analysis equivalent Real Analysis with $f:\mathbb R^2 \to \mathbb R^2$?

Solution 1:

No, it isn't. This can most obviously be seen from the fact that the map $z \mapsto \overline z$ is not differentiable as a map from $\mathbb C$ to $\mathbb C$, but the corresponding map $(x,y) \mapsto (x,-y)$ is differentiable as a map from $\mathbb R^2$ to $\mathbb R^2$.

This is because to be complex differentiable you need to have a best complex linear approximation which is a much stronger requirement than to have a best real linear approximation.

Solution 2:

Complex analysis is different from the analysis of functions $R^2 \to R^2$ in that it requires the functions not just to be differentiable, but complex-differentiable, which is a stricter requirement. One way of looking at the requirement is requiring that the functions locally look not only like a linear transformation $R^2 \to R^2$, but like a linear transformation corresponding to multiplication by a complex number, which we can identify with linear transformations of the plane by the obvious multiplication

$$(a,b) \cdot (c,d) = (ac - bd, ad + bc)$$

which is obvious when we write $(a,b) = a + bi$ and $(c,d) = c + di$. We can see this in matrix form if we identify $(a,b) = a+bi$ with the matrix

$$ \begin{bmatrix} a & -b \\ b & a \\ \end{bmatrix}.$$

Such functions have many properties that differentiable functions $R^2 \to R^2$ need not have, which gives complex analysis a much different flavor. For example, if we view complex functions on (simply connected subsets of) the complex plane as vector fields on $R^2$, then complex-differentiable functions will be conservative vector fields. However, if we know something about differentiable functions $R^2 \to R^2$, we then by definition know something about complex-differentiable functions $C \to C$, although we may need to do some translation.