Question: We have a statement that if *f(z)* is analytic in a domain *D* and $C = \{z : |z-a| = R\}$ is contained in *D*. Then, 

$|f^n(a)|\leq \frac{n!M\_R}{R^n}$ where $M\_R = \max|f(z)|$ on *C*.

What if the given *D* is not a circle? Then how to find the bound? I had problem in solving the following problem:

*D* is the rectangular region $\{(x,y) : |x|\leq4,y\leq3\}$. Let *f* be analytic in *D* and $|f(z)|\leq1$ on $\partial D$. Then how to prove that $|f'(z)|\leq\frac{14}{9\pi}$. Any hint or suggestion will be helpful for me. thanks a lot for help.

Answer: 
Here is how you advance, without loss of generality, we will work on $f'(0)$

$$ f'(0)=\frac{1}{2\pi i}\int\_{C}\frac{f(z)}{z^2}dz \implies |f'(0)|\leq \frac{1}{2\pi}\max \frac{|f(z)|}{|z^2|}\int\_C |dz|. $$

Now, we have to find the max over the given contour

$$ \max \frac{|f(z)|}{|z^2|}\leq \max \frac{1}{|z^2|}=\frac{1}{9}. $$

Also, we know that $ \int\_C |dz| $ gives the length of the contour which is the circumference of the rectangle and equals $28$. Gathering the above gives the desired result

$$ |f'(0)|\leq \frac{1}{2\pi}.\frac{28}{9} = \frac{14}{9\pi}.$$
This theorem holds for any connected region in the complex plane. It's called the maximum modulus principal.

Let me be more precise. Suppose, there exists a maximum value $z\_{0}$ in the interior of $D$. Then since $f$ is analytic, and hence continuous, there is a small open disk $D\_{0}$ of $z\_{0}$ such that such that $\vert f(z) \vert \leq \vert f(z\_{0}) \vert $ for all $z\in D\_{0}$. By the maximum modulus principal, $f$ must be constant on all of $D$