Fractional Linear Transformation

Question

Suppose that $a, b, c, d\in\mathbb{C}$ and $ad-bc=1$. Let $T$ be the fractional linear transformation $$z\mapsto\frac{az+b}{cz+d}.$$ Show that if $a=i, b=-i, c=1, d=i$, then the corresponding fraction linear transformation maps the upper half plane onto the unit disc.

I tried to find the image of $$w=\frac{iz-i}{z+i}$$ From here, $$z=\frac{-i-wi}{w-i}$$ Hence \begin{align} x+iy&=z \\ &= \frac{-i-wi}{w-i} \\ &=\frac{-i-i(u+iv)}{u+i(v-1)} \\ &=\frac{(-i-iu+v)(u-iv+i)}{u^2+(v-1)^2} \\ &= \frac{1-v+u}{u^2+(v-1)^2}+i\frac{-u-u^2-v^2+v}{u^2+(v-1)^2} \\ \end{align} Thus we want the image of $y>0$, so $$\frac{-u-u^2-v^2+v}{u^2+(v-1)^2}>0$$ But this does not yield the unit circle. Where have I gone wrong in my method?

Answer 1

Hint: you have to prove two things

First: if $\Im(z)>0$ then $|w|\leq 1$

(and if you try $z=-2+i$ you will see this is not true, so your problem is incorrect).

Second: if $|w|\leq 1$ then $\Im(z)>0$

Answer 2

If you change to $a=1, b=-i, c=1, d=i$, then the image of $z\in \mathbb R$ is $\{w\ | \ |w|=1\}$, since $z+i$ is the conjugate of $z-i$. But $i$ goes to $0$, and infinity goes to 1, so by continuity the upper half plane goes to the interior of the circle, and the lower half plane to the exterior. Since the function is a bijection of $\mathbb C \cup \{\infty\}$ to itself, you get the result.