Sequential approach to differentiability of complex valued function Let $f$ be analytic function on an open set $G$. Let $z\in G$. Let $\{z_n\}$ and $\{w_n\}$ be two sequences in $G$ which converges to $z$. Then show that $lim_{n \to \infty} \frac{f(z_n)-f(w_n)}{z_n-w_n}=f'(z)$.
It feels like this should follow just by the definition of differentiability. But can we do the proof in $\epsilon$ and $n$ argument?
I also thought of keeping one n fixed say for $\{z_n\}$ and letting another n tend to infinity we may get $\frac{f(z_n)-f(z)}{z_n-z}$ which will converge to $f'(z)$. But I could not write proper argument for this. Any hint would be helpful.
 A: Let's use the Taylor expansion of $f$. Suppose $w_n,z_n \to z_0$ and let $\sum_{n \ge 0} a_n (z-z_0)^n$ be the Taylor expansion of $f$ around $z_0$. Then
$$
f(z_n) - f(w_n) = \sum_{n \ge 0} a_n \left( (z_n-z_0)^n - (w_n-z_0)^n \right) \\
= \sum_{n \ge 1} a_n \left( (z_n-w_n) \left(\sum_{i=0}^{n-1} (z_n-z_0)^i (w_n-z_0)^{n-1-i} \right) \right) \\
= a_1(z_n-w_n) + (z_n - w_n) \left( \sum_{n \ge 2} a_n \left(\sum_{i=0}^{n-1} (z_n-z_0)^i (w_n-z_0)^{n-1-i} \right) \right).
$$
Dividing by $z_n - w_n$ and letting $n \to \infty$ proves that
$$
\lim_{n \to \infty} \frac{f(z_n) - f(w_n)}{z_n-w_m} = a_1 = f'(z_0)
$$
since all the terms in the big sum go to zero (because $z_n,w_n \to z_0$). For details, if $|z_n-z_0|,|w_n-z_0| < \varepsilon$, then the big sum's absolute value is $\le \sum_{n \ge 2} |a_n| \varepsilon^{n-1}$ which goes to $0$ with $\varepsilon$ since this is equal to $g(\varepsilon)$ where
$$
g(w) = \frac{f(z_0+w) - a_0 - a_1w}{w}
$$
which is analytic when $f$ is analytic.
Hope that helps,
