# Finding the Möbius transformation when $z= \infty$

The following is available:

$$T(2i) = \infty$$

$$T(0) = -i$$

$$T(\infty) = i$$

So I've got:

$$\frac{a(2i)+b}{c(2i)+d} = \infty \Rightarrow d=-2ic$$

$$\frac{b}{d}=i \Rightarrow b = -2c$$

$$\frac{a \cdot \infty -2c}{c \cdot \infty + -2ic} = i$$

how do I continue when infinity is the argument?

Note that$$\lim_{z\to\infty}\frac{az+b}{cz+d}=\lim_{w\to0}\frac{a/w+b}{c/w+d}=\lim_{w\to0}\frac{bw+a}{dw+c}=\frac{a}{c}$$provided $$c\ne0$$. If $$c=0\ne a$$, the limit is $$\lim_{z\to\infty}\frac{az+b}{d}=\infty$$. We don't need to consider the case $$a=c=0$$, because Möbius transformations satisfy $$ad-bc\ne0$$.
We begin with the Ansatz $$T(z)={az+b\over cz+d}\ ,$$ noting that the coefficients are only determined up to a common $$\ne0$$ factor, and that there are certain exception rules concerning $$\infty$$.
Since $$T(2i)=\infty$$ we conclude that $$c\cdot2i+d=0$$, hence $$d=-2i c$$. We are now at $$T(z)={az+b\over c(z-2i)}\ .$$ This shows that $$c\ne0$$, and that we may as well assume $$c=1$$. We are now at $$T(z)={az+b\over z-2i}\ .$$ It follows that $$a=\lim_{z\to\infty}{az+b\over z-2i}=T(\infty)=i\ ,$$ so that we arrive at $$T(z)={iz+b\over z-2i}\ .$$ The condition $$T(0)=-i$$ then leads to $$b=2$$, so that we finally have $$T(z)={iz+2\over z-2i}\ .$$