Let $f$ be analytic on $G= \{z:\Re(z)>0\}$,one one,with $\Re f(z)>0$ for all $z$ in $G$,and $f(a)=a$ for some real number a.Show that $|f'(a)|\leq 1$.
This is the exercise related to Riemann Mapping Theorem. I feel somehow we have to use unique-ness of Riemann map.
Thanks in advance
As suggested by Brevan Ellefsen, you can proceed this way:
The map $\varphi \colon \mathbb{D} \rightarrow \left\lbrace z \in \mathbb{C} : \Re(z) > 0 \right\rbrace$ given by $$\varphi(z) = a \frac{1 -z}{1 +z}$$ is a biholomorphism that satisfies $\varphi(0) = a$. Its inverse $\varphi^{-1} \colon \left\lbrace z \in \mathbb{C} : \Re(z) > 0 \right\rbrace \rightarrow \mathbb{D}$ is given by $$\varphi^{-1}(w) = \frac{a -w}{a +w} \, \text{.}$$ Now consider $g = \varphi^{-1} \circ f \circ \varphi \colon \mathbb{D} \rightarrow \mathbb{D}$. It satisfies $g(0) = 0$ and $g^{\prime}(0) = f^{\prime}(a)$. Thus, by Schwarz lemma, we have $\left\lvert f^{\prime}(a) \right\rvert \leq 1$.
Edit: Note that it was not necessary to write the map $\varphi$ explicitly. Alternatively, we could just have said that, by the Riemann mapping theorem, there exists a biholomorphism $\varphi \colon \mathbb{D} \rightarrow \left\lbrace z \in \mathbb{C} : \Re(z) > 0 \right\rbrace$ that satisfies $\varphi(0) = a$ and continue the same way.