Hello,

I'm trying to show,

Let $f:\mathcal{O} \to \mathbb{C}$ be real differentiable at $a\in\mathcal{O}$. Show that $$Df(a) = \frac{\partial f}{\partial z}(a)I+\frac{\partial f}{\partial \bar{z}}(a)\bar{I}.$$

I have it already that $T(z)= w_1 I +w_2 \bar{I}$ with $w_1=1/2(T(1)-T(i))$ and $w_2=1/2(T(1)+T(i))$.

If $f$ is real differentiable at $z$ then $Df(z)$ corresponds to multiplication by the matrix $$\begin{pmatrix}u_x(z) & u_y(z) \\ v_x(z) & v_y(z)\end{pmatrix}\begin{pmatrix}\Delta x \\ \Delta y \end{pmatrix}.$$ This we write out as $(u_x+u_y)+i(v_x+v_y)$.

The CR operators are defined as: $\frac{\partial}{\partial z}=1/2 (\frac{\partial}{\partial x} - i \frac{\partial}{\partial y})$, and $\frac{\partial}{\partial \bar{z}}= 1/2 (\frac{\partial}{\partial x} + i \frac{\partial}{\partial y})$ so I'm trying to show that $$1/2 (\frac{\partial}{\partial x} - i \frac{\partial}{\partial y})f(a)(z)+ 1/2 (\frac{\partial}{\partial x} + i \frac{\partial}{\partial y})\bar{f}(a)(z)=Df(a)(z)=(u_x x + u_y y) + i(v_x x + v_y y).$$

The last definition is $f\sim \begin{pmatrix} u \\ v \end{pmatrix}$ i.e., $f(z)=u(z)+iv(z)$.

I think all I'm trying to do now is figure out what $\frac{\partial}{\partial x}f(a)(z)$ means. It should be the partial derivative wrt x of the function $f$ at $a$ evaluated at $z$. Which I guess I can write out as $(u_x-iv_y)(z)$. Is this correct?