Find a Linear Fractional Transformations (LFT) $w(z)$ 
I have absulotly no idea how to approach this question, Can anyone please provide with a hint or any kinda information so I can solve this question.
Thank you very much for you help
 A: So by definition, a fractional linear transformation (a.k.a. a Möbius transformation) is a complex function of the form $w(z) = \frac{az + b}{cz + d}$, where $a,b,c,d \in \mathbb{C}$ satisfy $ad - bc \neq 0$.
In this particular example, we are looking for a fractional linear transformation $w(z)$ that satisfies $w(1) = 1+i$, $w(i) = 1-i$, and $w(-i) = 2$.
Let's start off by arbitrarily picking a value for one of the four unknowns. (If we have the bad fortune of picking a value that makes the rest of the problem impossible, then we'll need to start over and pick a different arbitrary value.) I'll choose to set $a = 1$.
So $w(z) = \frac{az + b}{cz + d} = \frac{z + b}{cz + d}$. 
Now we apply our constraints:


*

*Plugging in $w(1) = 1+i$ gives us $\frac{1 + b}{c + d} = 1+i.$ Rearranging gives us $-b + (1+i)c + (1+i)d = 1$.

*Plugging in $w(i) = 1-i$ gives us $\frac{i + b}{ci + d} = 1-i.$ Rearranging gives us $-b + (1+i)c + (1-i)d = i$.

*Plugging in $w(-i) = 2$ gives us $\frac{-i + b}{-ci + d} = 2.$ Rearranging gives us $-b -2ic +2d = -i$.


So now we have 3 linear equations in 3 unknowns, namely:


*

*$-b + (1+i)c + (1+i)d = 1$

*$-b + (1+i)c + (1-i)d = i$

*$-b -2ic +2d = -i$


You can solve this using any of the same techniques you would use to solve a 3-by-3 system of linear equations with real coefficients.
When I solved this system, I found $b=-\frac{4}{5}-\frac{3 i}{5}$, $c=\frac{3}{10}+\frac{i}{10}$, and $d=-\frac{1}{2}-\frac{i}{2}$. You should solve the system yourself and make sure you get the same result.
