= Solution
Consider the <Möbius transformations>
$$
A(z)=z+2,\qquad B(z)=\frac{z}{2z+1},\qquad
A=\begin{pmatrix}1&2\\0&1\end{pmatrix},\quad B=\begin{pmatrix}1&0\\2&1\end{pmatrix}.
$$
Their <group> lies in $\operatorname{PSL}_2(\mathbb Z)$, which is a <discrete subgroup> of the <Möbius group>: integer entries cannot approach the identity matrix except by eventually equalling it, and passing to the quotient by $\{I,-I\}$ preserves this property.
For freeness let $X=\{x\in\mathbb R:|x|>1\}$ and $Y=\{x\in\mathbb R:|x|<1\}$. For every nonzero integer $n$, $A^n(Y)=(2n-1,2n+1)\subset X$ and
$$
B^n(x)=\frac{x}{2nx+1},\qquad |B^n(x)|=\frac1{|2n+1/x|}<1\quad(x\in X).
$$
A reduced word involving both generators can be conjugated to a cyclically reduced word beginning with $A^n$ and ending with a nonzero power of $B$, after interchanging the generators if necessary. Repeated use of these inclusions shows that this word sends $X$ into the proper subset $A^n(Y)$ of $X$, so it cannot be the identity. Nonzero pure powers of either generator are also nonidentity. This is the <ping-pong lemma> in this particular action, and proves that $\boxed{\langle A,B\rangle\cong F_2}$ is a <free group> of rank two as well as a <discrete subgroup> of $\operatorname{PSL}_2(\mathbb C)$.
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