Question regarding complex differentiability I have been introduced to a new understanding of the differentiability of a complex function, which is:

$f$ is $C$-differentiable at $z_0$ iff we canfind a complex number $\alpha$ and a continuous function $R:\mathbb{D}\to\mathbb{C}$, where $D\subset C$ is a suitably small neighborhood of $0$ such that $R(0)=0$ and $f(a+h)=f(a)+\alpha h+hR(h)$.

I want to prove this claim. I can prove that if $f$ is differentiable, it will meet the above condition, but I have trouble proving the converse. Any help or guidance you could give would be greatly appreciated.
 A: What exactly about it do you want to proof? It is a definition, not a claim. It says "if this is true, then $f$ is differentiable at $a$". Which is to say, the "inverse" doesn't need proof, as it is a definition. However I will add this answer as to show why it is the right or natural definition. Recall the most basic definition of differentiability in $\mathbb{R}$. $f$ is said to be differentiable at $a$ if there is a real number $df$ such that it is the limit of:
$$\frac{f(a+h)-f(a)}{h}$$
when $h \rightarrow 0$. If you wish to expand on a definition then you must build on this one. The problem with this form though is that for many objects, the existence of an inverse is not guaranteed. Because of this, we manipulate the expression to give
$$df=\lim_{h\rightarrow 0} \frac{f(a+h)-f(a)}{h} \iff\\
df=\frac{f(a+h)-f(a)}{h}-R(h)\\f(a+h)=f(a)+hdf+hR(h)$$
Where $R$ is some continuous function that vanishes at $h=0$. So as long as this condition is satisfied, your function is differentiable at that point. For complex numbers it's the same. In fact, $\mathbb{C}$ is a field, which guarantees an inverse for every complex number except zero, so you could even use that ratio like definition we are firstly taught.
