I am working on some demanding and revisional exercises to present my work to our Semester Professor and prepare myself for the Semester Exams, this is why I am making such a big post (Any help/tip will be greatly appreciated).
Exercise :
Let :
$$f(z) = \sum_{n=0}^{\infty} c_nz^n = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}z^n$$
the Taylor Expansion of the entire function $f$.
We assume that :
$$|f(z)| \leq Ma^{|z|}, \forall z\in \mathbb C$$
where $a>1$. From the Cauchy-Taylor Theorem for any $r>0$, it is :
$$c_n = \frac{f^{(n)}(0)}{n!} = \frac{1}{2\pi i} \oint_{|z|=r} \frac{f(z)}{z^{n+1}}dz$$
Show that :
(a)$$|c_n| \leq M\frac{a^r}{r^n}$$
(b)
$$|c_n| \leq M\bigg(\frac{e\ln a}{n}\bigg)^n, n \geq 1$$
$$|c_n| \leq M, n=0$$
Attempt :
(a) We have that : $$f(z) = \sum_{n=0}^{\infty} c_nz^n $$
$$|f(z)| \leq Ma^{|z|}, \forall z\in \mathbb C$$
$$|z|=r$$
So, the inequality becomes :
$$|f(z)| \leq Ma^{r}, \forall z\in \mathbb C$$
$$\Leftrightarrow$$
$$\bigg|\sum_{n=0}^{\infty} c_nz^n\bigg| \leq Ma^{r}, \forall z\in \mathbb C$$
Now, I am certainly forgetting something involving Sums and the functional $|.|$, because it is obvious that $|z|^n = r^n$ which easily gives the answer. But I cannot understand how the sum gets out of the way, so I would appreciate some thorough help.
(b)
For $n=0$, it is obvious that $|c_n| \leq M$, from the inequality we have proved. Can't get (until this point), how to prove the fact for $n\geq 1$.