How was this result obtained?

Question

In Stein’s and Sharakachi’s Complex Analysis (Chapter 1), the authors state that, to prove that $F$ (where $F: (x,y)\mapsto(u(x,y),v(x,y))$) is differentiable it suffices to observe that $H=(h_1,h_2)$ and $h=h_1+i\,h_2$ then the Cauchy-Riemann equations imply:

$$J_F(x_0,y_0)(H)=\left(\frac{\partial u}{\partial x}-i\frac{\partial u}{\partial y}\right)(h_1+i\,h_2)=f'(z_0)h.$$

However, I can't see how this result is obtained, as I understood $J.H$ being a matrix multiplication:

$$J_F(x_0,y_0)(H)=\begin{bmatrix}\frac{\partial u}{\partial x} & \frac{\partial u}{\partial y}\\ -\frac{\partial u}{\partial y} & \frac{\partial u}{\partial x}\end{bmatrix}\begin{bmatrix}h_1\\h_2\end{bmatrix}=\begin{bmatrix}h_1\frac{\partial u}{\partial x}+h_2\frac{\partial u}{\partial y}\\h_2\frac{\partial u}{\partial x}-h_1\frac{\partial u}{\partial y}\end{bmatrix}$$

which does not conclude anything.

Answer 1

The essential fact: a $\Bbb R$-linear function form $\Bbb R^2$ to $\Bbb R^2$ is $\Bbb C$-linear iff its matrix in the canonical base has the form $$\pmatrix{a & b\cr -b &a}.$$ Edit:

$$(a + bi)(x + yi) = (ax - by) + (ay + bx)i,$$ $$\pmatrix{a & b\cr -b &a}\pmatrix{x & y\cr -y &x} = \cdots$$

Answer 2

This is a little late, but I was working to understand the same proof and think I can clarify. We use both the Cauchy-Riemann equations and the fact that we've already proven that $f'(z_0) = 2\partial_z u(z_0)$. So starting where you left off and using the association of $\mathbb{C}$ and $\mathbb{R}^2$: \begin{align*} \begin{bmatrix} h_1 u_x +h_2 u_y\\ h_1 v_x + h_2 v_y \end{bmatrix} ={}& h_1 u_x + h_2u_y + i(h_1 v_x +h_2 v_y)\\ ={}& h_1u_x + h_2 u_y +i(-h_1 u_y + h_2 u_x)\\ ={}& (u_x - iu_y)(h_1 + ih_2)\\ ={}& (u_x + \frac{1}{i}u_y)(h_1+ih_2)\\ ={}& f'(z_0)h. \end{align*} I left out the earlier parts in each line that we're evaluating this at $(x_0,y_0)$.