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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?

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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 $ad-bc\ne0$.

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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 $$T(z)={iz+2\over z-2i}\ .$$

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