Show that $\sum_{n \geqslant 0} |a_n z^n|<1$ with $|z|<1/3$ . 
Assume that $\sum_{n \geqslant 0} a_n z^n$ converges for $|z|<1$ and $$\left|\sum_{n \geqslant 0} a_n z^n \right|<1$$ 
  Show that for all $z \in \mathbb{C}$, with $|z|<1/3$ we have $$\sum_{n \geqslant 0} |a_n z^n|<1$$ 

I am stuck with this exercise, could someone give me some hint please ?
 A: This (beautiful) result is usually called Bohr's inequality.
An internet search will show you that there are lots of proofs for this inequality. Here is the simplest I know (unfortunately, I don't remember where I saw it; it is definitely not mine!).
The key point is the following 
Fact. We have $\vert a_m\vert\leq 2(1-\vert a_0\vert)\;$ for all $m\geq 1$.
To prove this, we may assume (multiplying $f(z)=\sum_0^\infty a_nz¨n$ by some constant $\omega$ with $\vert\omega\vert =1$) that $a_0\in\mathbb R^+$.
Define $g(z):=1-f(z)$, where $f(z)=\sum_0^\infty a_n z^n$. Note that ${\rm Re}(g)\geq 0$ because $\vert f\vert\leq 1$. Also,
$$g(z)=\sum_{m=0}^\infty b_m z^m\, ,$$
where $b_0=1-c_0$ and $b_m=-a_m$, $m\geq 1$. 
Since $g$ is holomorphic we have $$\int_0^{2\pi} \overline{g(e^{i\theta})}e^{-im\theta}\, d\theta=0$$ for all $m\geq 1$, and
$$ b_m=\frac1{2\pi}\int_0^{2\pi} g(e^{i\theta})e^{-im\theta}\, d\theta\, .$$
So we may write (for all $m\geq 1$)
$$b_m=\frac1{2\pi}\int_0^{2\pi}\left( g(e^{i\theta})+\overline{g(e^{i\theta})}\right)e^{-im\theta}\, d\theta $$
It follows (remembering that ${\rm Re}(g)\geq 0$ and using the mean value formula for the function ${\rm Re}(g)$) that for all $m\geq 1$ :
$$\vert a_m\vert=\vert b_m\vert\leq \frac1{2\pi}\int_0^{2\pi} 2{\rm Re}(g(e^{i\theta}))\, d\theta=2{\rm Re}(g(0))=2(1-a_0) \, . $$
This proves the Fact.
Now, by the Fact, we have for all $r>0$ :
$$\sum_{m=0}^\infty \vert a_m\vert r^m\leq \vert a_0\vert+2(1-\vert a_0\vert)\sum_{m=1}^\infty r^m\, . $$
The sum in the right hand side is equal to $\frac12$ if $r=\frac13$; so we get
$$\sum_{m=0}^\infty \vert a_m\vert \left(\frac13\right)^m\leq \vert a_0\vert +(1-\vert a_0\vert)=1.$$
Hence, $\sum_0^\infty\vert a_mz^m\vert\leq 1$ for all $z\in\mathbb C$ such that $\vert z\vert\leq 1/3$.
A: This is a partial answer. I will prove that for $|z|<\frac{1}{3}$ we have
$$
\sum_{n=0}^\infty|a_n z^n|<\frac{3}{2\sqrt{2}}\approx 1,06066
$$
Indeed, for $0<r<1$, the Fourier series of $t\to f(re^{it})$ is given by
$f(re^{it})=\sum_{n=0}^\infty a_nr^ne^{int}$. Using Perceval's formula we conclude that
$$
\sum_{n=0}^\infty|a_n|^2r^{2n}=\frac{1}{2\pi}\int_0^{2\pi}|f(re^{it})|^2dt\leq1
$$
letting $r$ tend to $1^{-}$ we get
$$
\sum_{n=0}^\infty|a_n|^2 \leq1
$$
Now, if $|z|<\frac{1}{3}$ we have
$$\eqalign{
\sum_{n=0}^\infty|a_n z^n|&<\sum_{n=0}^\infty|a_n| 3^{-n}\leq \sqrt{\sum_{n=0}^\infty|a_n|^2}\cdot\sqrt{\sum_{n=0}^\infty 3^{-2n}}<\sqrt{\frac{9}{8}}\cr
}
$$
Which is the announced result. $\qquad\square$
A: Use the radius $r=1-ε$ for some small $ε>0$. Then from the Cauchy integral formula for the Taylor series coefficients it follows that $$|a_n|\le (1-ε)^{-n}$$ for all $n$, which can also be seen more directly by just manipulating the series $f(z)$ to get
$$
a_n=\frac1{2\pi i}\int_{|z|=1-ε} \frac{f(z)}{z^{n+1}} dz\implies 
|a_n|\le\frac1{2\pi}\,\frac{\max\{ |f(z)|:|z|=1-ε\}}{(1-ε)^{n+1}}\,2\pi(1-ε).
$$

You can get a similar result much more elementary from the theorem of Cauchy-Hadamard. Since the radius of convergence is larger than $1$, then 
$$
\limsup_{n\to\infty}\sqrt[n]{|a_n|}\le 1
$$
and thus $|a_n|\le (1+ε)^n$ for almost all $n\in \Bbb N$ for all $ε>0$.
