Suppose that f is differentiable on an open set G and $z_0 ∈ G$.
Let
$F(z) = \frac{f(z)-f(z_0)}{z-z_0},$ $(z \neq z_0), F(z_0) = f ′ (z_0)$
Prove that F is differentiable on G.
I wanted to use an argument which relied on the definition of differentiability to show that $F'(z)=lim_{h \to 0}\frac{F(z+h)-F(z)}{h}$. But apparently you can't use arguments like that here , because then it would imply that it is true for the reals also. Now I'm thinking that if f is differentiable it has a taylor series so if I can show that F(z) has a taylor series then this will imply that F is differentiable.
Is this the right approach ? and if not is- there any suggestions as to what approach\proofs might prove more fruitful ?
You're close with the Taylor series argument. Here are the details.
For the point $z_0$, it is $z_0\in G$ therefore there exists $\varepsilon>0$ such that $D(z_0,\varepsilon)\subset G$. One can write $f$ as a power series in this disc, centered at $z_0$: it is $f(z)=f(z_0)+f'(z_0)(z-z_0)+\frac{f''(z_0)}{2!}(z-z_0)^2+\dots$ Therefore $f(z)-f(z_0)=f'(z_0)(z-z_0)+\frac{f''(z_0)}{2!}(z-z_0)^2+\dots$ Now we can write $f(z)-f(z_0)=g(z)(z-z_0)$, with $g(z)=\displaystyle{\sum_{k=0}^{\infty}\frac{f^{(k+1)}(z_0)}{(k+1)!}(z-z_0)^k}$ and $g(z_0)=f'(z_0)$. It is for $z\neq z_0$, $g(z)=\frac{f(z)-f(z_0)}{z-z_0}$. Since $g$ is being represented by a power series on the disc $D(z_0,\varepsilon)$, it is holomorphic there. Now for $z\in G\setminus D(z_0,\varepsilon)$, it is obvious that $F$ is holomorphic on every such $z$, since it is a product of holomorphic functions $(z\mapsto f(z)-f(z_0, z\mapsto\frac{1}{z-z_0})$. By all above, $F$ is holomorphic on its domain.