I am just learning about the group of LFTs under composisiton. An LFT is a map on $\mathbb C$ so that $z\mapsto \frac{az+b}{cz+d}$ where $a,b,c,d\in \mathbb R$ are fixed.
However, I am finding basic computations with this group very tricky and would really appreciate some help with the following problems.
(1) Is there an LFT $L:\mathbb C \to \mathbb C$ such that $L(i)=-1+4i$, $L(1+i)=2+4i$, and $L(1-i)=2-4i$? Is it unique?
Plugging in, this would give $\frac{ai+b}{ci+d}=-1+4i$, $\frac{a+ai +b}{c+ci+d}=2+4i$, and $\frac{a-ai+b}{c-ci+d}=2-4i$. But once I expand out, I'm having trouble solving for $a,b,c,d$. Is there a good strategy for these computations? I am familiar with linear-algebraic methods, if applicable.
(2) Is there an LFT $L:\mathbb C \to \mathbb C$ such that $L(i)=-1+4i$ and $L(1+i)=2+4i$? Is it unique?
Same question but one fewer image point specified. Would my approach for this problem be similar to the first one? And is uniqueness affected?
First, a refinement: to make $G = \{f : z \mapsto (az + b)/(cz + d) \, | \, a, b, c, d \in \mathbb{R} \}$ into a group under composition $\circ$, we need each function to at least have an inverse. Thus we need each map to be invertible, so additionally require the condition $ad - bc \neq 0$. Thus we can view this group $(G, \circ)$ as a subgroup of the Mobius group $\mathcal{M}$. (It can be shown that $G$ is indeed closed under inverses and composition, and hence is indeed a subgroup of $\mathcal{M}$.) Further, if we require $f \in G$ to be a function $\mathbb{C} \to \mathbb{C}$, then it can be shown that $f$ is linear (i.e., $f(z) = az + b$ with $a \in \mathbb{R}\setminus\{0\}$ and $b \in \mathbb{R}$), so instead let us extend both the domain and range to $\mathbb{C}_\infty$ to allow "true" rational functions.
In terms of computing, it would be helpful to make use of the cross ratio. For three distinct complex numbers $a_1$, $a_0$ and $a_\infty$ and another complex number $z$, we define the cross ratio to be $$[a_1, a_0, a_\infty, z] = g_{(a_1, a_0, a_\infty)}(z) = \frac{a_1 - a_\infty}{a_1 - a_0} \frac{z - a_0}{z - a_\infty}.$$ (Notice that $[a_1, a_0, a_\infty, a_\kappa] = \kappa$, which explains the strange indexing.) These are straightforward to write down, so to construct a map that sends $a_\kappa \mapsto b_\kappa$ for $\kappa \in \{1,2,3\}$ with the $a_\kappa$'s and $b_\kappa$'s distinct, one can compute ${g_{(b_1,b_2,b_3)}}^{-1} \circ g_{(a_1, a_2, a_3)} \in \mathcal{M}$, and finally check whether this is indeed in $G$. Since Mobius maps are uniquely determined by three points, the map is unique, up to re-scaling each of $a$, $b$, $c$ and $d$ by some constant $\zeta$.
Hopefully this outline helps.