A generalization of holomorphic functions

Question

Let's fix a matrix $A\in M_{2}(\mathbb{R})$. Assume that the following vector space of smooth functions is closed under complex multiplication:

$$\mathcal{S}_{A}=\{f:\mathbb{C}\to \mathbb{C}\mid Df.A=A.Df \}$$

Here $Df$ is the Jacobian of $f:\mathbb{R}^{2}\to \mathbb{R}^{2}$ (We identify $\mathbb{C}$ with $\mathbb{R}^{2}$).

Does this imply that $A$ is in the form $A=\begin{pmatrix} a&-b\\b&a \end{pmatrix}$?

Note that For $A=\begin{pmatrix} 0&-1\\1&0 \end{pmatrix}$ the relation $Df.A=A.Df$ is equivalent to the Cauchy Riemann equations for $f=u+iv$ so we obtain the class of holomorphic functions.

Answer 1

You're right. But the A is always satisfying Cauchy–Riemann equations.

Let $f = u(x,y)+v(x,y)i$,

By the definition of smoothness: the f must be differentiable which implies f must satisfy Cauchy–Riemann equations.

For simplification: denote $$\frac{\partial{f}}{\partial{x}} = f_x$$

By Cauchy–Riemann equations: $$u_x = v_y \tag{1}$$ and $$u_y = - v_x \tag{2}$$

The Jacobian of f is $\begin{pmatrix} u_x&u_y\\v_x&v_y \end{pmatrix}$ Designate it as B.

Let A = $\begin{pmatrix} a&b\\c&d\end{pmatrix}$

Denote $U[i][j]$ is the entry on $i^{th}$ row $j^{th} $ column where U is a matrix;

By $A\times B =B\times A $

Then $(A\times B)[1][1] = au_x+bv_y=(B\times A)[1][1] = au_x+cu_y $ By $(2)$,$ au_x+bu_y = au_x-bu_y$ forall $f$.

Thus $c = -b$.

Choose $(A\times B)[1][2] = (B\times A)[1][2] $ and using $(1)$, one can get d = a.

Choose $(A\times B)[2][1] = (B\times A)[2][1] $ one can show it has holed.

Choose $(A\times B)[2][2] = (B\times A)[2][2] $ one can show it has holed.

Thus A = $\begin{pmatrix} a&b\\-b&a\end{pmatrix}$

Answer 2

Put $f(z)=z$ Then $f$ belongs to $\mathcal{S}_{A}$. So $f^{2}$ must belong to $\mathcal{S}_{A}$,too. This means that $A$ commutes with all rotations. Hence $A$ has the desired representation as it is pointed out in the comment by user 1952009.