Use Cauchy Inequalities to find an upper bound for $|f^4(i)|$ and $|f^4(0)|$ Let's suppose that $f$ is differentiable on a disk $B_{10}(r)$, and $|f(z)| \leq 54$ for $z$ on the circle $|z-i| = 3$. My goal is to use Cauchy Inequalities to find an upper bound for $|f^4(i)|$ and $|f^4(0)|$.
The first of these is simply a direct application of the Cauchy Inequality formula. However, the hint I was given for $|f^4(0)|$ is this statement: 

A strict maximum of the absolute value of an analytic function can't be attained at the interior point of a set. 

Using this information, my guess was that because $0$ is in the interior of the circle $|z-i| = 3$, we cannot find an upper bound for it. Hence, finding an upper bound for $|f^4(0)|$ is not possible.
This sounds fine at first, but I run into a question: isn't $i$ also an interior point of $|z-i| = 3$? If that's true, why does a contradiction not arise with this point as was the case for $z_0 = 0$?
 A: Looking for $\boldsymbol{f^{(4)}(z)}$
If $\left|w-i\right|=3$, then
$$
\begin{align}
\left|f^{(4)}(i)\right|
&=\frac{4!}{2\pi}\left|\oint\frac{f(w)\,\mathrm{d}w}{(w-i)^5}\right|\\
&\le\frac{4!}{2\pi}\oint\frac{\left|f(w)\right|\,\left|\mathrm{d}w\right|}{\left|w-i\right|^5}\\
&\le\frac{24}{2\pi}\frac{54\cdot6\pi}{3^5}\\[6pt]
&=16
\end{align}
$$
If $\left|w-i\right|=3$, then $\left|w-0\right|\ge2$
$$
\begin{align}
\left|f^{(4)}(0)\right|
&=\frac{4!}{2\pi}\left|\oint\frac{f(w)\,\mathrm{d}w}{(w-0)^5}\right|\\
&\le\frac{4!}{2\pi}\oint\frac{\left|f(w)\right|\,\left|\mathrm{d}w\right|}{\left|w-0\right|^5}\\
&\le\frac{24}{2\pi}\frac{54\cdot6\pi}{2^5}\\[6pt]
&=\frac{243}2
\end{align}
$$

Looking for $\boldsymbol{f^4(z)}$
If $\left|w-i\right|=3$, then
$$
\begin{align}
\left|f(i)\right|
&=\frac1{2\pi}\left|\oint\frac{f(w)\,\mathrm{d}w}{w-i}\right|\\
&\le\frac1{2\pi}\oint\frac{\left|f(w)\right|\,\left|\mathrm{d}w\right|}{\left|w-i\right|}\\
&\le\frac1{2\pi}\frac{54\cdot6\pi}{3}\\[6pt]
&=54
\end{align}
$$
Therefore,
$$
\left|f^4(i)\right|\le54^4
$$
If $\left|w-i\right|=3$, then $\left|w-0\right|\ge2$
$$
\begin{align}
\left|f(0)\right|
&=\frac1{2\pi}\left|\oint\frac{f(w)\,\mathrm{d}w}{w-0}\right|\\
&\le\frac1{2\pi}\oint\frac{\left|f(w)\right|\,\left|\mathrm{d}w\right|}{\left|w-0\right|}\\
&\le\frac1{2\pi}\frac{54\cdot6\pi}{2}\\[6pt]
&=81
\end{align}
$$
But the maximum of $\left|f(z)\right|$ on $\left|z-i\right|\le3$ is attained on the boundary $\left|z-i\right|=3$, where $\left|f(z)\right|\le54$. Therefore,
$$
\left|f^4(0)\right|\le54^4
$$
