Why is this condition required? (Schwarz's lemma) Good afternoon!
I  was having a look at some complex analysis, and encountered the following lemma (Schwarz's):
If $f$ is holomorphic in the unit disc, and $f(0) = 0$, then $|f(z)| \leq |z|$ for any $z$ in the unit disc $\mathbb{D}$.
The proof is quite straightforward:

Let $g$ be the function defined by $g(z) = f(z)/z$ if $z \ne 0$, and $g(0) = f'(0)$. Then, a simple application of the maximum modulus principle implies $|g(z)| \leq 1$, which finishes the proof.

However, I do not quite understand why we would need the condition $f(0) = 0$. Wikipedia says this enforces differentiability of $g$ at $z = 0$. But why is that so? It looks like some application of L'Hopital's rule but this is leaving me quite puzzled.
And even so, why couldn't we just define $g$ on the unit disc without $0$? The maximum modulus principle only requires the domain to be open and connected, which $\mathbb{D}\backslash\{0\}$ is. So why would this not work?
Thanks for your time :)
 A: Since, $f$ is holomorphic, it is analytic. So, near $0$, and since $f(0)=0$, you have$$f(z)=a_1z+a_2z^2+a_3z^3+\cdots$$So,$$g(z)=\frac{f(z)}z=a_1+a_2z+a_3z^2+\cdots,$$and therefore $g$ is analytic too. In particular, it is differentiable at $0$.
Concerning your other question, it's up to you to provide a proof of Schwarz's lemma if we remove $0$ from the picture. I don't see how to do it.
A: If $f$ has a zero, say $f(a)=0$, we can reduce it to the use of the Schwarz's lemma. Consider the function $\varphi_a\colon\Delta\rightarrow\Delta$ given by
$$
  \varphi_a(z)=\frac{z-a}{1-\bar{a}z}
  .
$$
(Can you prove that, indeed, $|\varphi_a(z)|<1$ for every $|z|<1$?) Here $\Delta$ denotes the unit disk. 
Then $h=f\circ\varphi_a$ is an analytic function from $\Delta$ to itself which satisfies $h(0)=0$. We apply Schwarz's lemma to get 
$$
  |h(z)|\leq|z| \quad\Rightarrow\quad |f(\varphi_a(z))|\leq|z|\quad\forall\:|z|\leq1
  .
$$
Now, these $\varphi_a$'s are invertible: 
$\varphi_{-a}\circ\varphi_a=\varphi_a\circ\varphi_{-a}=\mathrm{id}_\Delta$ (I leave the computations to you!). Then, from the inequality above we obtain
$$
  |f(z)|\leq|\varphi_{-a}(z)|=\left|\frac{z+a}{1+\bar{a}z}\right|
  .
$$
Maybe not a satisfactory inequality since for $z=a$ we have the upper bound $2|a|/(1+|a|^2)$ for zero... But at least this ilustrates some use of the $\varphi_a$'s.
