I want to find the transformation that maps $z_1 = 0,\,z_2 = 1,\,z_3 = \infty$ to $w_1 = 1,\,w_2 = 1+i,\,w_3 = i$. Under this mapping what is the image of the line $\text{Im}z = \text{Re}z$, the real axis and the imaginary axis.
So far I found that:
$$1 = \frac{b}{d},\quad 1+i = \frac{a+b}{c+d},\quad i = \frac{a}{c}\Rightarrow b=d,\,a = ic,\,id = -c.$$
this then gave me the transformation:
$$z\rightarrow w = \frac{z+1}{1-iz}$$
What do I do next to answer the question?
$$w=\frac{z+1}{1-iz} \Rightarrow z=\frac{w-1}{iw+1}$$ Now : $$x+iy=z=\frac{u-1+iv}{1-v+iu}.\frac{(1-v)-iu}{(1-v)-iu}=\frac{(u-1)(1-v)+uv}{(1-v)^2+u^2}+i\frac{v(1-v)+u(1-u)}{(1-v)^2+u^2}$$ The image of the line $Imz=Rez$ : $$x=y\Rightarrow u-1-uv+v+uv=v-v^2+u-u^2 \Rightarrow |w|=1$$