MHB Mobius transformation satisfying certain properties

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A Möbius transformation \( F \) maps \( 0 \) to \( \infty \) and \( \infty \) to \( 0 \) if and only if it can be expressed as \( F(z) = dz^{-1} \) for some \( d \in \mathbb{C} \). The transformation takes the form \( F(z) = \frac{pz+q}{rz+s} \), and for \( F(0) \) to equal \( \infty \), \( s \) must be zero. This simplifies \( F(z) \) to \( \frac{pz+q}{rz} \). To ensure \( F(\infty) = 0 \), it follows that \( p \) must also be zero, leading to the conclusion that \( F(z) = \frac{q}{rz} \). The discussion emphasizes the necessity of understanding the conditions under which these mappings occur.
kalish1
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I'm having some trouble showing that a Mobius transformation $F$ maps $0$ to $\infty$ and $\infty$ to $0$ iff $F(z)=dz^{-1}$ for some $d \in \mathbb{C}.$ Mainly with the "only if" part. Do I need to use pictures?

This is Exercise $23$ in Section $3.3$ of Conway's *Functions of One Complex Variable*.

This question has been crossposted here: complex analysis - Mobius transformation satisfying certain properties - Mathematics Stack Exchange
 
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kalish said:
I'm having some trouble showing that a Mobius transformation $F$ maps $0$ to $\infty$ and $\infty$ to $0$ iff $F(z)=dz^{-1}$ for some $d \in \mathbb{C}.$ Mainly with the "only if" part.
The Möbius transformation has the form $F(z) = \dfrac{pz+q}{rz+s}$. If $s\ne0$ then $f(0) = \dfrac qs$, which is not equal to $\infty$. Therefore we must have $s=0$, and $F(z) = \dfrac{pz+q}{rz} = \dfrac pr + \dfrac q{rz}$. For this to take $\infty$ to $0$ we must have $p=0$. Therefore $F(z) = \dfrac{q}{rz}$, which is of the form that you want.
 

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