# Find Bilinear Transformation that maps the points $-i, 0, 2+i$ from $z$ plane on to the points $0, -2i, 4$ of the $ω-$plane?

The solution is supposed to be

$$w = \frac{2(z+i)}{z-1}$$

Also, is it Okay to ask such questions here? It's not homework, it's the last few sets of problems I can't get on my own. Some belong to Complex Variables & Conformal Mapping and some to Vector Calculus.

Here is a method using "points" $$0,1,\infty$$.

Let us consider the Möbius transformation :

$$f_{a,b,c}(z):=\dfrac{c(a-b)z+a(b-c)}{(a-b)z+(b-c)} \ \ \ \leftrightarrow \begin{pmatrix} \ \ \ c(a-b)&a(b-c)\\ \ (a-b)&(b-c)\end{pmatrix}\tag{1}$$

It is such that $$f_{a,b,c}(0)=a, \ \ \ f_{a,b,c}(1)=b, \ \ \ f_{a,b,c}(\infty)=c.$$

Using $$0,1,\infty$$ as intermediate values, the final transform can be expressed as the composition :

$$f=f_{0,-2i,4}\circ f^{-1}_{-i,0,2+i}$$

In a concrete way, it amounts to compute the matrix product :

$$\begin{pmatrix} \ \ \ 8i & 0\\ \ \ 2i&(-2i-4)\\ \end{pmatrix}\begin{pmatrix} \ \ \ -i(2+i)&i(2+i)\\ \ \ -i &-(2+i)\\ \end{pmatrix}^{-1}$$

being understood that these matrices are defined up to a multiplicative factor. For example the previous product can be written equivalently :

$$\begin{pmatrix} \ \ \ 8i & 0\\ \ \ 2i&-2(2+i)\\ \end{pmatrix}\begin{pmatrix} \ \ \ -i&i\\ \ \ -\dfrac{i}{2+i}&-1\\ \end{pmatrix}^{-1}$$

or, equivalently again :

$$\begin{pmatrix} \ \ \ 4i & 0\\ \ \ i&-(2+i)\\ \end{pmatrix}\begin{pmatrix} \ -1&-i\\ \ \ \dfrac{i}{2+i}&-i\\ \end{pmatrix}$$

or (simplification by $$i$$):

$$\begin{pmatrix} \ \ \ 4i & 0\\ \ \ i&-(2+i)\\ \end{pmatrix}\begin{pmatrix} \ i&1\\ \ \ \dfrac{1}{2+i}&-1\\ \end{pmatrix}$$

whence the looked for matrix :

$$\begin{pmatrix} 2& 2i\\ 1&-1\\ \end{pmatrix} \ \ \text{associated with} \ \ w=\dfrac{2z+2i}{z-1}.$$

Edit : what I have explained on your example could be the consequence of plugging the values of $$a, b, c, a', b', c'$$ (where $$f(a)=a',f(b)=b',f(c)=c'$$ into the following product, giving an explicit formula :

$$\begin{pmatrix} \ \ \ c'(a'-b')&a'(b'-c')\\ \ (a'-b')&(b'-c')\end{pmatrix}\begin{pmatrix} \ \ \ (b-c)&a(c-b)\\ \ (b-a)&c(a-b)\end{pmatrix}$$

Theorem: The linear fractional transformation $$w = T(z)$$ maps three distinct points $$z_1, z_2, z_3$$ uniquely into three distinct points $$w_1, w_2, w_3$$. The map is determined by the equation $$\frac{(w − w_1)(w_2 − w_3)}{(w − w_3)(w_2 − w_1)}=\frac{(z − z_1)(z_2 − z_3)}{(z − z_3)(z_2 − z_1)}$$

Here $$z_1=-i, z_2=0, z_3=2+i$$ and $$w_1=0, w_2=-2i, w_3=4$$

So, $$\frac{(w − w_1)(w_2 − w_3)}{(w − w_3)(w_2 − w_1)}=\frac{(z − z_1)(z_2 − z_3)}{(z − z_3)(z_2 − z_1)}$$ $$\implies \frac{(w − 0)(-2i − 4)}{(w − 4)(-2i − 0)}=\frac{(z +i)(0 − 2-i)}{(z − 2-i)(0 +i)}$$ $$\implies \frac{w (i +2)}{i(w − 4)}=\frac{-(z +i)(2+i)}{i(z − 2-i)}$$ $$\implies \frac{w}{(w − 4)}=\frac{-(z +i)}{(z − 2-i)}$$ $$\implies \frac{w+(w+4)}{w-(w-4)}=\frac{-(z +i)+(z − 2-i)}{-(z +i)-(z − 2-i)}$$ $$\implies \frac{2w-4}{4}=\frac{i+1}{z − 1}$$$$\implies w = 2 \frac{z+i}{z-1}$$

$$-i, 0, 2+i$$ from $$z$$ plane on to the points $$0, -2i, 4$$

$$f(z) = {az+b\over cz+d}$$

$$f(0) =-2i$$ so $$b=-2id$$

$$f(-i) = 0$$ so $$b=ia \implies a=-2d$$ So $$f(z) = {-2dz-2id\over cz+d}$$

Finally, since $$f(2+i)=4$$ we get ...