I have two definitions of the Riemann mapping theorem
1.Suppose that $G$ is a simply connected domain in the complex plane, $G\neq \mathbb{C}$, and that $z_0$ is a point of $G$. There exists a unique conformal mapping f of $G$ onto the open unit disk $\mathbb{D}$ satisfying the conditions $f(z_0)=0$ and $f'(z_0)>0$.
2.$G$ is a non-empty simply connected Domain of the complex number plane $\mathbb{C}$ which is not all of $\mathbb{C}$, then there exists a biholomorphic mapping $f$ from $G$ onto the open unit disk $\mathbb{D}$.
Regarding that $conformal \Leftrightarrow biholomorphic$ it follows: $1 \Rightarrow 2$. Is $2 \Rightarrow 1$ also true ?
Best Answer
The answer is yes.
So let $f: G\to \mathbb{D}$ be the biholomorphic map you're given. Let $w=f(z_0)$.
Then, $T_w(z)=\frac{z-w}{1-\overline{w}z}$ is a conformal equivalence of the unit disk onto itself mapping $w$ to $0$.
Hence, $g=\frac{1}{f'(z_0)T_w'(w))}T_w\circ f$ is a conformal equivalence with the desired property. This gives you existence.
Must it be unique?
Well, assume $g$ and $\tilde{g}$ are both biholomoprhic maps onto $\mathbb{D}$ mapping $z_0$ to $0$ and such that their derivatives at $z_0$ are positive. Assume without loss of generality that $g'(z_0)\geq \tilde{g}'(z_0)$. Then, $$ (g\circ \tilde{g}^{-1})'(0)=\frac{g'(z_0)}{\tilde{g}'(z_0)}\geq 1 $$ but by the Schwarz lemma, $|(g\circ \tilde{g}^{-1})'(0)|\leq 1$ so, in fact it must be equal to 1. Thus, by the second part of the Schwarz lemma, $g\circ \tilde{g}^{-1}$ must be a rotation and, since it's derivative is positive, it must be the identity. Hence, $g=\tilde{g}$.