How to prove the associated real functions of a holomorphic complex function is differentiable? I see this on a note of complex analysis
Let $f: \Omega \to C$ be any complex function. Since we can identify $\mathbb{C}$ with $\mathbb{R}^2$, any such function automatically induces a related function $f: \Omega \to \mathbb{R}^2$, where now we think of $\Omega$ as being a subset of $\mathbb{R}^2$ instead of $\mathbb{C}$. If $f$ is holomorphic, does this imply anything about real differentiability of its associated real function? As a matter of fact, yes, it does, and it turns out that this connection is not merely a curiosity; many deep theorems about certain real functions can be proven by appealing to this connection!
But I don't quite understand how to prove the real associative functions of a holomorphic complex function is differentiable. For instance, in the picture below, how do I know that $u(x,y), v(x,y)$ is differentiable.
Here is a screenshot of the note which contains the necessary information
 A: The key thing to notice is that multiplication by a complex constant is a linear map if $\mathbb C$ is considered an $\mathbb R$ vector space. We have
$$(a+\mathrm ib)(x+\mathrm iy)=ax-by+\mathrm i(ay+bx).$$
Written as vectors, the map $x+\mathrm iy\mapsto (a+\mathrm ib)(x+\mathrm iy)$ thus becomes
$$\begin{pmatrix}x\\y\end{pmatrix}\mapsto\begin{pmatrix}ax-by\\bx+ay\end{pmatrix}=\begin{pmatrix}a&-b\\ b&a\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}$$
So complex multiplication is an $\mathbb R$-linear map. And here comes the connection between complex and real differentiability. A function is complex differentiable if it can be approximated by a $\mathbb C$-linear map (so a complex multiplication). It is real differentiable if it can be approximated by an $\mathbb R$-linear map (which complex multiplication is!). More formally, a function is complex differentiable in $z_0$ iff there exists a complex constant $c$ such that
$$\lim_{z\to z_0}\frac{f(z)-f(z_0)-c(z-z_0)}{\vert z-z_0\vert}=0.$$
You can easily obtain this condition by manipulating the difference quotient definition ($c$ is its limit). Also, a function is real differentiable in $z_0$ iff there exists an $\mathbb R$-linear map $L$ such that
$$\lim_{z\to z_0}\frac{f(z)-f(z_0)-L(z-z_0)}{\vert z-z_0\vert}=0.$$
And since multiplication by $c$ is an $\mathbb R$-linear map, this is trivially true if the above condition for complex differentiability is fulfilled.
