Using the maximum modulus principle to show an inequality Let $f$ be a holomorphic function defined on a neighborhood of $\bar{D}$, where $D$ is the unit disc, and suppose that $f(0) = 0$.
(a) Show that $g(z) = f(z)/z$ is holomorphic on a neighborhood of $\bar{D}$.
(b) Suppose further that $f(D) ⊂ D$. 
Show that $|f(z)| ≤ |z|$ for
all $z ∈ D$, and deduce that $|f'(0)| ≤ 1$. 
(Hint: apply the maximum
principle to $g$).
My attempt:
a) I found lim$_{h \to 0}$ $\frac{g(z_0 + h) - g(z_0)}{h} = f'(z_0)/z_0 - f(z_0)/z_0^2$
And hence $g$ is holomorphic on a neighborhood of $\bar{D}$ 
b) Since $g$ is holomorphic on a neighborhood of $\bar{D}$. Then if there is a point $z_0$ in $\bar{D}$ such that $|g(z_0)|\geq |g(z)|$ for all $z \in \bar{D}$, then $g$ is constant. In other words,
max_$D|g|$ $=$ max_${C}|g|$, where $C = bD$ (boundary of $D$) but how to continue from here? Any help please? 
 A: a) The function $g$ is analytic on $D\setminus\{0\}$ since it's the quotient of two holomorphic funtions there. In the neighborhood of $0$, you can express $f(z)$ as a power series $a_1z+a_2z^2+a_3z^3+\cdots$, and therefore $g(z)=\frac{f(z)}z=a_1+a_2z+a_3z^2+\cdots$. So, $g$ is holomorphic too.
b) Suppose that $|f(z_0)|>|z_0|$, for some $z_0\in D$. Then $z_0\ne0$. On the other hand, $|f(z_0)|>|z_0|\iff|g(z_0)|>1$. But $|z|=1\implies|g(z)|=|f(z)|\leqslant1$. This goes against the maximum principle, since the maximum of $|g|$ should be attained at the boundary of the disk. And, of course, if you always have $|f(z)|\leqslant|z|$, then you always have $\left|\frac{f(z)-f(0)}z\right|\leqslant1$, and therefore $|f'(0)|\leqslant1$.
A: The only possible problem with $g$ is at $0$; and your limit does not show that things work at $z_0=0$. But on a small disk that surrounds $0$, you have that $f(z)=\sum_{k=1}^\infty a_nz^n$, so $g(z)=\sum_{k=1}^\infty a_nz^{n-1}$ is also holomorphic. 
Now for part b) you have, since $|f(z)|\leq1$ for all $z\in D$, 
$$
|f(z)|=|zg(z)|=|z|\,|g(z)|\leq|z|\,|g(z_0)|=|z|\,\frac{|f(z_0)|}{|z_0|}=|z|\,|f(z_0)|\leq|z|.
$$
Then
$$
\left|\frac{f(z)-f(0)}{z-0}\right|=\frac{|f(z)|}{|z|}\leq1,
$$
so $|f'(0)|\leq1$. 
