I seem to be doing it wrong every time. time to learnbn the general formula for mobius transformations.
But what is the general formula?
suppose
Given an Mobius transformation $$ f(Z) \ = \ \frac{AZ+B}{CZ+D} $$
where
- Z is a point (x, y) or, if you prefer the complex number $x+ yi$
- $A = a_r + a_i i$
- $B = b_r + b_i i$
- $C = c_r + c_i i$
- $D = d_r + d_i i$
(so $ Z, A, B, C, D $ are complex numbers, while $x, y, a_r, a_i, b_r, b_i, c_r, c_i, d_r, d_i $ are real numbers)
What are the coordinates of $f(Z) $ expressed as a point $(Re(f(Z) \ , \ Im(f(Z) ) $ and as function of $x, y, a_r, a_i, b_r, b_i, c_r, c_i, d_r, d_i $ ?
start of an answer:
just using elementary complex arithmetic https://en.wikipedia.org/wiki/Complex_number#Multiplication_and_division
$$\frac {(a_r + a_i i) (x + y i) + b_r + b_i i} {(c_r + c_i i) (x + y i) + d_r + d_i i} = $$
$$\frac {(a_r x - a_i y ) + (a_i x + a_r y) i + b_r + b_i i} {(c_r x - c_i y) + (c_i x + c_r y) i + d_r + d_i i} = $$
$$\frac {(a_r x - a_i y+ b_r ) + (a_i x + a_r y + b_i) i } {(c_r x - c_i y+ d_r) + (c_i x + c_r y+ d_i) i} = $$
then division
$$\left( \frac {(a_r x - a_i y+ b_r ) (c_r x - c_i y+ d_r) + (a_i x + a_r y + b_i) (c_i x + c_r y+ d_i) } {(c_r x - c_i y+ d_r)^2 + (c_i x + c_r y+ d_i) ^2} \right) + $$ $$\left( \frac {(a_i x + a_r y + b_i)(c_i x + c_r y+ d_i) -(a_r x - a_i y+ b_r )(c_i x + c_r y+ d_i } {(c_r x - c_i y+ d_r)^2 + (c_i x + c_r y+ d_i)^2 } \right) i = $$
The mistake is in the numerator of the second addend of the answer. Instead of $$\frac{(a_ix+a_ry+b_i)\color{Red}{(c_ix+c_ry+d_i)}-...}{...}$$ there must be $$\frac{(a_ix+a_ry+b_i)\color{Green}{(c_rx−c_iy+d_r)}-...}{...}$$