Continuous complex function (2 variables) Let $f$ be analytic on the open set $G\subset \mathbb{C}$. What's the best way of showing that the function $\phi:G \times G \to \mathbb{C}$ defined by $\phi(z,w)=[f(z)-f(w)]/(z-w)$ for $z \neq w$ and $\phi(z,z)=f'(z)$ is continuous?
This is obvious for $z\neq w$, but I'm having some trouble on the points $(z,z)$. I did it one way but I'm not convinced this is good.
Edit: Rudin's Proof (for the continuity on the diagonal)
Let $z_0\in G$. Since $f$ is analytic, there exists $r>0$ such that $B(a,r)\subset G$ and $|f'(\zeta)-f'(z_0)|<\epsilon$ for all $\zeta \in B(a,r)$. Getting $z,w\in B(a,r)$, than $\zeta(t)=(1-t)z+tw \in B(a,r)$ for $0\le t\le 1$. Now just use that $\phi(z,w)-\phi(z_0,z_0)=\int_0^1[f'(\zeta(t))-f'(z_0)]dt$ and the $\epsilon$-bound to get the desired continuity.
 A: Probably not best, but one approach to showing continuity at $(z_0,z_0)$ is to use the power series expansion $f(z)=\sum_{k=0}^\infty a_k(z-z_0)^k$ to get 
$$
\begin{align*}
\frac{f(w)-f(z)}{w-z}-f'(z_0)&=\sum_{k=1}^\infty a_k\frac{(w-z_0)^k-(z-z_0)^k}{(w-z_0)-(z-z_0)}-f'(z_0)\\
&=\sum_{k=2}^\infty a_k\sum_{j=0}^{k-1}(w-z_0)^j(z-z_0)^{k-1-j},
\end{align*}$$ which converges absolutely and uniformly near $(z_0,z_0)$ and goes to $0$ at $(z_0,z_0)$.
A better candidate for "best" is that given in Rudin's Real and complex analysis, Lemma 10.29 on page 314 of the 3rd Edition.
A: I just want to note that existence of the limit of $\phi$ at $(z,z)$ is the definition of the strong (or strict) differentiability of $f$ az $z$. It is a key strengthened concept of differentiability, which already makes the inverse, and thus the implicit function theorem work. It implies that $f'$ is continuous at $z$, and if $f'$ exists on a neighbourhood of $z$ the reverse is also true, which also anwers the question. This is a consequence of the mean value inequality:
$$|f(u)-f(w)-f'(z)(u-w)|\le\sup_{0<t<1}|f'(u+t(w-u))-f'(z)||u-w|.$$
