Bilinear transformation which maps $z=(1,i,-1)$ and $w=(2,i,-2)$

Question

I have three equations after simplifying this a bit

$a+b-2c-2d=0$

$ai+b+c-di=0$

$-a+b-2c+2d=0$

How do I proceed after obtaining these equations? Please comment after every step or I will be lost.

If you care to know, this is from the chapter Complex Variables in my book.

EDIT- my book explains it like this

a+b-2c-2d=0 ----(1)

ai+b+c-di=0-----(2)

-a+b-2c+2d=0----(3)

(1)+(3) gives 2b-4c *\I get that but want to know why we added 1+3 and not 1+2*

i.e 2b-4c=b-2c

b-2c=0 ----(4)

(2)+i X (3) gives

(1+i)b+(1-2i)c+id=0 ----- (5) *\how does this step work.. totally lost in this step*

Answer 1

The important thing is that you have 3 equations and 4 unknowns. Thus you have to consider one of the unknowns, say $c$, as a parameter, and you will obtain a unique solution in the remaining variables $(a,b,d)$ depending on this parameter. In this way, any method will give you: $$a=6ic, \ \ \ b=2c, \ \ \ d=3ic$$ Plugging these values into $Z=\dfrac{az+b}{cz+d}$, and simplifying by $c$, one gets: $Z=\dfrac{6iz+2}{z+3i}$.

Answer 2

Add the first and the third equations to get

$$(a+b-2c-2d)+(-a+b-2c+2d)=0\implies 2(b-2c)=0\implies b=2c$$

Plugging this in at the first equation, $a-2d=0\implies a=2d$.

Plugging these two into the second equation

$$2di+2c+c-di=0\implies di+3c=0$$