For an open set $C\subset\mathbb{C}$ and a fixed $z_0\in C$, we are asked to prove that $f$ is complex differentiable at $z_0$ if and only if there exists $\delta>0$ and complex linear function $L$ such that $f(z_0+h)=f(z_0)+L(h)+o(h)$
where $o$ is a function that satisfies $o(h)/h\rightarrow 0$ as $h\rightarrow 0$.
Attempt: the one direction yields, as $f'(z_0)=\lim_{h\to 0}\frac{f(z_0+h)-f(z_0)}{h}=\lim_{h\to 0} L(h)/h+o(h)/h$ which goes to $L(h)/h$ as h goes to 0. Which I am not sure where to go form here. The other direction I am not sure where to start.
Since $L$ is a $\mathbb{C}$-linear operator on $\mathbb{C}$ and $\mathbb{C}$ is one-dimensional, $L$ must be of the form $x\mapsto cx$ for some $c\in\mathbb{C}$.
If $f(z_0 +h) = f(z_0) + L(h) + o(h)$, we have $(f(z_0+h) - f(z_0))/ h - c \to 0$ as $h\to 0$. Hence, $f$ is complex-differentialbe at $z_0$.
Conversely, if $f$ is complex-differentiable at $z_0$, take $L(h)=f'(z_0)h$.