Find the bilinear transformation which maps $z=(1, i, -1)$ respectively to $w=(i, 0, -i)$. Also, find the image of $|z|\leq 1$ under the transformation
My try: Here, $w_1=i$, $w_2=0$, $w_3=-i$, $z_1=1$, $z_2=i$, $z_3=-1$. According to the formula, $$\begin{align}\\ &{\begin{aligned}\\ \frac{(w-w_1)(w_2-w_3)}{(w-w_3)(w_2-w_1)}&=\frac{(z-z_1)(z_2-z_3)}{(z-z_3)(z_2-z_1)}\\ \end{aligned}\\}\\ &\implies\frac{(w-i)(i)}{(w+i)(-i)}=\frac{(z-1)(i+1)}{(z+1)(i-1)}\\ &\implies-\frac{w-i}{w+i}=\frac{(z-1)(2i)}{(z+1)(-2)}\\ &\implies\frac{w-i}{w+i}=\frac{iz-i}{z+1}....(i)\\ &\implies-\frac wi=\frac{z(i+1)+(1-i)}{z(i-1)-(1+i)}\\ &\implies w=\frac{z(1-i)-(i+1)}{z(1+i)+(i-1)}\\ \end{align}\\ $$ Now, from eq.(i) we get, $$\begin{align}\\ &\frac{z-1}{z+1}=\frac{w-i}{iw-1}\\ &{\begin{aligned}\\ \implies\require{cancel}\frac{z-\cancel{1}+z+\cancel{1}}{\cancel{z}-1-\cancel{z}-1}&=\frac{w(i+1)-(i+1)}{w(1-i)+(1-i)}\\ \end{aligned}\\}\\ &\implies-\frac z2=\frac{w-1}{w+1}.\frac{\cancel{2}i}{\cancel{2}}\\ &\implies z=\frac{2i-2iw}{w+1}\\ \end{align}\\ $$ As, $|z|\leq 1$ $$\begin{align}\\ &\therefore\left|\frac{2i-2iw}{w+1}\right|\leq 1\\ &\implies\left|\frac{2i-2iu+v}{(u+1)+iv}\right|\leq 1\\ &\implies 4(1-u)^2+v^2\leq(u+1)^2+v^2\\ &\implies 4-8u+4u^2\leq u^2+2u+1\\ &\implies 3u^2-10u+3\leq 0\\ &\implies(3u-1)(u-3)\leq 0\\ \end{align}\\ $$ But, in my book it is written that the transformation maps to the right half of w-plane. Which is correct?