Describing automorphisms of a punctured unit disc Here is my attempt :

Let $f \in \text {Aut}(D(0,1) \setminus \{0\})$.
Since $f$ is bounded on a deleted neighborhood of $0$, function $f$
  has a removable singularity at $0$. Since $f$ is bijective, we must
  have $f(0)=0.$
By Schwarz's lemma we get $|f(z)| \le |z|.$
But $f^{-1} \in \text {Aut}(D(0,1) \setminus \{0\}) \; \text {and}
 f^{-1}(0)=0 \Rightarrow |z|=|f^{-1}(f(z))| \le |f(z)|.$
Thus $|f(z)|=|z| \Rightarrow f(z)=\lambda z \;$ for some $\lambda \in
 \Bbb C$ and $|\lambda|=1$.
We infer that all the elements of $\text {Aut}(D(0,1) \setminus \{0\})$ are
  rotations.


Are my arguments correct?
 A: The important missing detail is where you say "since $f$ is bijective, $f(0)=0$".
To explain what the problem is we need to  be a little more careful with the notation (in fact you made the error because you're actually using the notation "$f$" for two different things. Not to say your notation was bad, it's fairly standard, but that's where the error came from).
Say $D$ is the disk and $D'$ is the punctured disk. You're given $f:D'\to D'$. Now $f$ has a removable singularity at the origin, great. So $f$ extends to a function $F:D\to\Bbb C$.
Now when you say "since $f$ is a bijection..." you're actually referring to the new function $F$. You're given that $f:D'\to D'$ is a bijection - how do you know that $F$ is a bijection?
For that matter, how do you even know that $F(0)\in D$?
Hints: A certain very basic result that you use all the time shows that $|F(0)|<1$. So you have $F:D\to D$. Now you can use the Open Mapping Theorem to show that since $f$ is a bijection we must have $F(0)=0$.
A: In my opinion one argument is missing: we have $|f(z)|=|z|$ for all $z \in D(0,1)$. Why we get $f(z)=\lambda z \;$ for some $ \lambda$ with $|\lambda|=1$ ?
Define $g : D(0,1) \to \mathbb C$ by $g(z)=f(z)/z$ if $z \ne 0$ and $g(0)=f'(0)$. Then $g$ is holomorphic and $|g(z)|=1$ for all $z \in D(0,1)$.
Hence $g(D(0,1))$ is not open, thus $g$ is constant.
Can you proceed ?
