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I searched for this complex analysis problem in both Mathematics Stack Exchange and Math Overflow and I didn't find anything. Here is the problem:

Suppose that $f$ is an holomorphic function on the unit disc $D$ and conformal (holomorphic and one-to-one) on the punctured disc $D\backslash\{0\}$.
Show that $f$ is conformal on the whole disc $D$.

1st thought: Taking into account the hypothesis, I tried to construt a sequence of conformal functions on $D$ that converge to $f$ uniformly in each compact subset of $D$. Then I would apply Hurwitz's Theorem to get that $f$ is conformal on $D$.
I thought of creating the sequence by taking the products of $f$ with characteristic functions of the annuli $A_n := \{z: \frac{1}{n}<|z|<1\}$. Unfortunately these products don't necessarily define holomorphic functions on $D$.

2nd thought: Suppose the opposite and reach a contradiction, but I did't know how to proceed with a right idea.

Maybe theorems about conformal mappings, such the Riemann's Uniformization Theorem, can be used.

What do you think I might do?

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    If you go with the second thought, you assume there is a $w \in D\setminus \{0\}$ with $f(w) = f(0)$. You want to conclude that $f$ isn't injective on the punctured disk. Look at the local mapping behaviour of non-constant holomorphic functions. – 2017-01-02
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    I don't know that definition of conformal. Do you just need to show $f$ is 1-1 on the whole disk? That should be easy, right? If $f(0)=f(a)$, then what must happen in a small neighborhood of each of these points? – 2017-01-02
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    @Daniel Fischer Thanks for the responce. What do you mean by local mapping behaviour of holomorphic functions? – 2017-01-02
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    The title is misleading as it suggests $f$ is only assumed holomorphic in the punctured disc. Reading the body of the Question we are told $f$ is assumed holomorphic "on the unit disc $D$", so apply the local analysis proposed by @MPW. – 2017-01-02
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    How does a holomorphic function behave on small neighbourhoods of $z_0$? Stuff like "If $f$ attains the value $w_0$ with multiplicity $k$ at $z_0$, then there are neighbourhoods $U$ of $z_0$ and $V$ of $w_0$ such that every $w\in V\setminus \{w_0\}$ is attained at $k$ distinct points of $U$". – 2017-01-02
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    @Daniel Fischer Thanks for the answer. This is what I thought you meant. I used it for the function $g(z)=f(z)-f(0)$, but only on a small neighbourhood of 0 and that's the reason I stucked. – 2017-01-02

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Suppose that $f(0) =f(v) =u$ for some $v\in D.$ Since the set $W=f(D\setminus \{0\} )$ is open and $u\in W $ then there exists open sets $U, V$ such that $0\in U, v\in V $ such that $f(U\setminus \{0\}), f(V)$ are open and $f(U\setminus \{0\})\cap f(V)\neq\emptyset .$ Taking any $s\in f(U\setminus \{0\})\cap f(V)$ we get $f(d) =f(e) =s $ for some $d\in U\setminus \{0\} ,e\in V.$