Let $f:\mathbb{C}\rightarrow \mathbb{C}$ be entire. Let $w_1,w_2$ be any two distinct complex numbers. Must there exist $c\in \overline{B_{|w_2-w_1|}(w_1)}$ such that $f'(c)=\frac{f(w_2)-f(w_1)}{w_2-w_1}$ ?
Note:
I have shown the folowing: Let $[w_1, w_2]$ be the line segment with endpoints $w_1,w_2$. Then there exists $c_1,c_2\in [w_1,w_2]$ such that: $$Re(f'(c_1))+Im(f'(c_2))i=\frac{f(w_2)-f(w_1)}{w_2-w_1} ...(1)$$
I was wondering whether allowing $c_1,c_2$ to lie in $\overline{B_{|w_2-w_1|}(w_1)}$ (instead of just the line segment joining $w_1,w_2$) enables us to choose $c_1,c_2$ such that $(1)$ holds and $c_1=c_2$.
Thank you
No, in general such a $c$ does not exist. The classical counterexample is $f(z) = e^z$, with $w_2 = w_1 + 2\pi i$. Then
$$\frac{f(w_2) - f(w_1)}{w_2-w_1} = 0,$$
but $f'(c) \neq 0$ for all $c\in\mathbb{C}$.
What you have in terms of the mean value theorem for holomorphic functions is the estimate
$$\lvert f(w_2) - f(w_1)\rvert \leqslant \lvert w_2-w_1\rvert\cdot \max \left\lbrace \lvert f'(\zeta)\rvert : \zeta \in [w_1,w_2]\right\rbrace,$$
and your equation $(1)$ with generally different points for the real and imaginary part.