Suppose that $f$ is analytic on an open set $G$ and that $z_0 \in G$
Let $F(z)=\frac{f(z)-f(z_0)}{z-z_0}$ $s.t.$ $z \neq z_0, F(z_0)=f'(z_0)$.
I want to show that $F$ is analytic on $G$.
My work so far:
$F(z)=\frac{f(z)-f(z_0)}{z-z_0} \Rightarrow F'(z)=\lim_{h \rightarrow 0}\frac{\frac{f(z+h)-f(z_0+h)}{z+h-z_0-h}-\frac{f(z)-f(z_0)}{z-z_0}}{h}=\lim_{h \rightarrow0}\frac{\frac{f(z+h)}{z-z_0}-\frac{f(z_0+h)}{z-z_0}-\frac{f(z)}{z-z_0}+\frac{f(z_0}{z-z_0}}{h}=\lim_{h \rightarrow 0}\frac{\frac{f(z+h)-f(z)}{z-z_0}-\frac{f(z_0+h)-f(z_0)}{z-z_0}}{h}=\frac{f'(z)-f'(z_0)}{h}=\lim_{h \rightarrow } \frac{f'(z)-F(z_0)}{h}$ .
My question is how can I show that $f'(z)=F(z)$ so that I can finish this proof ?
Note: In the first version of this question it only said "differentiable" instead of "analytic" and this is what one of the answers is adressing.
You can't do that directly. An approach such as yours would also work for real functions. But this is false over the reals. For instance, if $f\colon\mathbb{R}\longrightarrow\mathbb{R}$ is the function defined by$$f(x)=\begin{cases}x^2\sin\left(\frac1x\right)&\text{ if }x\neq0\\0&\text{ otherwise,}\end{cases}$$then $f$ is differentiable, but $\frac{f(x)}x$ isn't continuous.