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I need to create a fractional linear map of the form $$ F(z) = \frac {az+b}{cz+d}, $$ where a,b,c, and d are complex numbers such that $ad-bc \neq 0$ and $F(0) = 1$, $F(1)=\infty$, and $F(\infty)=0.$ The only solution which I can think of is $$F(z) = \frac {1- \frac {z}{\infty}}{1-z}.$$ However, I'm not sure about this. I know that infinity is considered a "point" to some extent in complex analysis, but am I allowed to use it in this way to define this map? If not, how else could I solve this problem?

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  • $\begingroup$ Maybe the function is $F(z)=\frac 1 {1-z}$? $\endgroup$ Mar 11, 2016 at 7:19

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Well, your intuition is correct. However, since $\infty$ is not a complex number, you're not allowed to have it in the expression of your map. But "$\frac1{\infty}=0$". Check if $\frac1{1-z}$ fits what you want.

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  • $\begingroup$ Ah, you're right. I think I was somehow worried about the fact that it would be one divided by negative infinity. But of course, -0 is the same as 0...duh. Thanks! $\endgroup$ Mar 11, 2016 at 7:40
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Yes, the "naive" (or rather "straightforward"?) approach to the use of $\infty$ is fine and justified for these maps. That is, $F(z)=\infty$ precisely one point, namely either for the - finite - solution of $cz+d=0$ (if $c\ne 0$) or for $z=\infty$ if $c=0$ 8and consequently $a\ne 0$). Similarly $F(\infty)=\frac ac$ if $c\ne 0$ and $=\infty$ if $c=0$ (as follows by taking the limit as $|z|\to\infty$)

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