Let f be an analytic function in the disk ... Prove that Let $f$ be a function analytic in the disk $D={z:|z-1-i| \leq 4}$ that satisfies $|f(z)| \leq 1$ for all $z$ in $D$. 
Prove that $|f^{(3)} (1-i)| \leq \frac{3}{4}$. 
Hint: apply Cauchy's Integral Formula to a circle centered at $1-i$

I know that if a function is analytic then the complex derivative exists at every point.  I also know that the disk is a circle centered at $1-i $ with a radius of $4$. I would like help starting and completing this problem. 
 A: You should be aware that Cauchy's integral formula states
$$f^{(n)}(a) =\frac{n!}{2\pi i}\oint_\gamma\frac{f(z)}{(z-a)^{n+1}}\,\mathrm dz$$
where $\gamma$ is a circle lying in $D$.
The distance between the point of interest $a=1-i$ and the center $1+i$ of the disk $D$ is $2$, hence we can let $\gamma$ have radius $r$ slightly less than $2$ (and consider the limit as $r\to 2$).
We find with unit speed parametrization of $\gamma$
$$| f'''(a)|\le \frac{3}{\pi}\int_{t=0}^{2\pi r}\left|\frac{f(\gamma(t))}{r^4}\right|\,\mathrm dt\le \frac3\pi\frac{2\pi r}{r^4}=\frac 6{r^3}.$$
In the limit $r\to 2$ this gives us $|f'''(a)|\le \frac{6}{2^3}=\frac34$.
A: Cauchy's integral formula is 
$$f^{(n)}(z_0)=\frac{n!}{2\pi i} \int_C \frac{f(z)}{(z-z_0)^{n+1}dz}$$
Now, for this example, $n=3$ and $z_0=1-i$.  If we take the contour to be parameterized by $z=1-i+\rho e^{i\phi}$, $dz=i\rho e^{i\phi} d\phi$, and $0\le \phi \le 2\pi$, then $f$ is analytic for values of $\rho \le 2$.   This is because $f$ is analytic when $|z-(1+i)|\le4$ and the circle $|z-(1-i)|= 2$ is contained in the region of analyticity.  Now, we can write for $\rho \le 2$  
$$\begin{align}
f^{(3)}(1-i)&=\frac{3!}{2\pi i} \int_0^{2\pi} \frac{f(1-i++\rho e^{i\phi})}{\rho^{4}e^{i4\phi}} i\rho e^{i\phi} d\phi\\\\
&=\frac{3}{\pi \rho^3} \int_0^{2\pi} f(1-i++\rho e^{i\phi}) e^{-i3\phi} d\phi\\\\
\end{align}$$
Taking magnitudes of both sides we find
$$\begin{align}
|f^{(3)}(1-i)|&=\left|\frac{3}{\pi \rho^3} \int_0^{2\pi} f(1-i++\rho e^{i\phi})e^{-i3\phi} d\phi\right|\\\\
&\le\frac{3}{\pi \rho^3} \int_0^{2\pi} |f(1-i++\rho e^{i\phi})| d\phi\\\\
&\le \frac{3}{\pi \rho^3} 2\pi \max |f(1-i++\rho e^{i\phi})|\\\\
&\le \frac{6}{\rho^3}\\\\
\end{align}$$
Since this is true for all $\rho \le 2$, then the supremum is of this latter bound is $6/2^3$.
Thus
$$|f^{(3)}(1-i)|\le \frac34$$
as was to be shown!
