# Is the bilinear transformation that maps the crescent-shaped region into the unique horizontal strip unique?

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?

• I don't know whether your second choice is good or whether you made some algebra mistake, but there are many bilinear transformations from any horizontal strip to itself (example: $z \mapsto z + 1$). Composing the author's solution with one of these will yield another solution, so the author's solution is not unique. (By the way, is "bilinear" really the word you want here? It looks as if "fractional linear" is the correct one.) – John Hughes Jun 28 '18 at 10:56
• @JohnHughes It is really written as "bilinear" in the textbook. And I agree with you. I think it is not unique either because of the different orientation that I have chosen. – Ivy Jun 28 '18 at 11:01

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.