Estimate: $|f^{(3)} (i/3)|$ using Cauchy

Suppose $f:D(0,1) \longrightarrow \mathbb{C}$ is holomorphic, where $$D(0,1) = \{z \in \mathbb{C} \mid |z|<1\},$$ and assume the maximum $|f(z)| \le 2$.

Estimate: $|f^{(3)}(i/3)|$.

I have the solution to this problem which used $R=2/3$.

I am not sure how to find this $R$ because I don't understand the Cauchy Estimate that well. Any help to understand this problem will be real helpful thank you.

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Cauchy estimate is most convenient to use at the center of the disk. For a closed disk centered at $i/3$ and contained in $D(0,1)$, the radius must be $<2/3$. So we use the estimate at $D(i/3,r)$ for $r<2/3$, obtaining $$|f^{(3)}(i/3)|\le \frac{3! \sup|f|}{r^{3+1}} \le \frac{12}{r^4}$$ Since $|f^{(3)}(i/3)| \le 12/r^4$ holds for all $r\in (0,2/3)$, it also holds at $r=2/3$: passing to the limit in a nonstrict inequality preserves the inequality. Hence, $$|f^{(3)}(i/3)|\le \frac{12}{(2/3)^4}$$
Notice that $|\frac{i}{3}|=1/3$, $\frac{i}{3} \in D(0,1)$. In order to apply Cauchy's Theorem and this estimate you need a closed disc. Since $f$ is holomorphic on $D(0,1)$ it is also holomorphic on the closed disc $\overline{D(0, 2/3)}$. Now apply the estimate with $R=2/3$.
I think the reduction of radius from $1$ to $2/3$ had more to do with the center being at $i/3$ than with having to fit a closed disk in $D(0,1)$. –  ˈjuː.zɚ79365 Jun 17 '13 at 15:05