This question is a question that I found here on MSE but it was deleted by the OP. Since I already wrote an answer for it, I will repost (a paraphrase of) the question and the answer here:
Let $F:\mathbb H\to\mathbb D$ be the standard conformal map (here, $\mathbb H:=\{z\in\mathbb C\mid \operatorname{Im}z\geq 0\}$ and $\mathbb D:=\{z\in\mathbb C\mid |z|< 1\}$.)
For $M=\begin{pmatrix}a & b \\ c & d\end{pmatrix}\in SL(2, \mathbb R)$, define $f_M(z):=\dfrac{a\cdot z + b}{c \cdot z + d}$ where $z\in\mathbb C\setminus\{-\frac dc\}$.
In the proof of the Theorem, Stein/Shakarchi say in the proof of Theorem 2.4 of chapter 8 that for $\theta\in\mathbb R$ and $$M_\theta=\begin{pmatrix}\cos\theta & -\sin\theta \\ \sin\theta & \cos\theta\end{pmatrix},$$ we have $F\circ f_{M_\theta}\circ F^{-1}$ corresponds to the rotation by $-2\theta$ in the plane. Why is this true?
We want to show that $$(F\circ f_{M_\theta}\circ F^{-1})(z)=\exp(-2i\theta)\cdot z.$$ If we can show that $(F\circ f_{M_\theta})(z)=\exp(-2i\theta)\cdot F(z)$, then the above would immediately follow (substitute $F^{-1}(z)$ for $z$.)
Lemma. For all $z$ in the domain of $f_{M_\theta}$ such that $f_{M_\theta}(z)\neq i$, $$(F\circ f_{M_\theta})(z)=\exp(-2i\theta)\cdot F(z).$$ Proof. The standard conformal map is explicitly given by $$F(z)=\frac{z-i}{z+i}$$ so that $$F(f_{M_\theta}(z))=\frac{\frac{\cos(\theta) \cdot z-\sin(\theta)}{\sin(\theta)\cdot z+\cos(\theta)}-i}{\frac{\cos(\theta) \cdot z-\sin(\theta)}{\sin(\theta)\cdot z+\cos(\theta)}+i}=\frac{(-i + z) (\cos(\theta) - i \sin(\theta))}{(i + z) (\cos(\theta) + i \sin(\theta))}=\frac{z-i}{z+i}\frac{\exp(-i\theta)}{\exp(i\theta)}.$$
This proves the Lemma and thus also the original claim.