Recently, I am learning Complex Analysis. An example of the text book wants us to find the bilinear transformation that maps the crescent-shaped region that lies inside the disk$|z-2|<2$ and outside the disk$|z-1|=1$onto a horizontal strip.Then, the author chooses $z_1=4, z_2=2+2i, z_3=0$ and the image values $w_1=0, w_2=1, w_3=\infty$.Then, he got the equation $w=\frac{-iz+4z}{z}$.
My question is when I choose $z_1=2$, $z_2=1+i$, $z_3=0$ which lies on the small circle, and maps them into $w_1=2$, $w_2=1$, $w_3=-\infty$ (to preserve the orientation), the result is not the same, it is $w=-\frac{(-i-2)z+2i}{z}$ . And the horizontal strip is totally different from the one obtained by the author. Is this correct?
Let $f(z)=\dfrac{z-4}{zi} = \dfrac{4-z}{z}i$. Then $f(4)=0$, $f(2+2i)=1$ and $f(0)=\infty$. So we calculate the image of the boundary $|z-2|=2$ and $|z-1|=1$. $$ f(0) = \infty \qquad f(2+2i)=1 \qquad f(4)= 0 $$ So $|z-2|=2$ goes to real axis. $$ f(0)=\infty \qquad f(2) = i \qquad f(1+i) = 2+i $$ So $|z-1|=1$ goes to the line $r(t)=t + i$. Note that this line and real axis are parallel and form an strip.