Notice that $\mathbb{C}$ is also a vector space over the real numbers. The standard definition of a derivative of a function $f:U\to {\mathbb{R}}^2$ at $(x_0,y_0)$for $U\subseteq {\mathbb{R}}^2$ is that there exists some matrix $A\in {\mathbb{R}}^{2\times 2}$ such that the following equation holds:

such that:

This means that $f$ has a $\mathbb{R}$-linear approximation at $(x_0,y_0)$. As vector spaces over $\mathbb{R}$, ${\mathbb{R}}^2\cong \mathbb{C}$. But $\mathbb{C}$ has the structure of an algebra, so what if we utilize that here?

We notice that multiplication on the left by a complex number $z=a+bi$ is a $\mathbb{R}$-linear map $M:\mathbb{C}\to \mathbb{C}$. This means that we have a matrix representation of such a map. Let's figure out what that is, exactly. We know the first entry of the matrix will be $\varphi \circ M\circ {\varphi }^{-1}(1,0)=\varphi (M(1))=\varphi (a+bi)=[1,1]$. The second entry will be the same thing evaluated at $[0,1]$, which yields $[-1,1]$. So the matrix representation of $M$ with respect to the standard basis of $\mathbb{C}$ is:

One can show that all of these matrices represent multiplication by a complex number. This suggests that $A$ should be of the form above.

But we also know that $A$ has a special form. Namely, if $f(x,y)=(u(x,y),v(x,y))$, then

From these two observations, we have a sufficient condition to show complex differentiability.