Let $f: D \subset \mathbb R^2 \to \mathbb R^2$ be a function defined on an open subset $D \subset \mathbb R^2$. Remember that $f$ is said to be differentiable (in the Frechét sense) at (x_0, y_0) \in D if there is a linear map $Df(x_0, y_0) : \mathbb R^2 \to \mathbb R^2$ such what $$\lim_{(h,k) \to (0,0)} \frac{f((x_0,y_0) + (h,k)) – f(x_0, y_0) – Df(x_0, y_0)( h,k)}{\vert\vert(h,k)\vert\vert} = 0$$ Show that $f$ is differentiable in the complex sense at $z_0 = x_0 + iy_0$ if and only if $f$ is differentiable in the Frechét sense and $Df(x_0, y_0)$ defines a linear $\mathbb C$ mapping of $\mathbb C$ into $\mathbb C$ i.e. $Df(x_0, y_0)(z + \lambda z' ) = Df(x_0, y_0)z + \lambda Df(x_0,y_0) z', \forall z, z', \lambda \in \mathbb C$.
$f$ is differentiable in $z_0$ if there is $f'(z_0) = \lim_{h \to 0} \frac{f(z_0 + h) - f(z_0)}{h}$.
$\lim_{h \to 0} \frac{f(z_0 + h) - f(z_0)}{h} - f'(z_0) = \lim_{h \to 0} \frac{f(z_0 + h) - f(z_0) - f'(z_0)h}{h}$.
$f'(z_0)(z + \lambda z') = \lim_{h \to 0} \frac{f(z_0 + h) - f(z_0)}{h} (z + \lambda z')= \lim_{h \to 0} \frac{f(z_0 + h) - f(z_0)}{h}(z + \lambda z') = \lim_{h \to 0} \frac{f(z_0 + h)z +f(z_0 + h)\lambda Z'}{h} - \frac{f(z_0)z + f(z_0)\lambda z'}{h} = \lim_{h \to 0}\frac{f(z_0 + h)z - f(z_0)z}{h} + \frac{f(z_0 + h)\lambda Z'-\lambda z'}{h} = f'(z_0)(z) + f'(z_0)(\lambda z')$
That's what I managed to come up with, is it acceptable?
Grateful for the attention.