# The derivative of a Möbius transformation at infinity

Can someone explain, plz, why if $L(z)$ is a Möbius transformation, $$L'(\infty)=-(ad-bc)/c^2?$$ I know about I need to make previous inverse transformation $z'=1/z$ so the derivative at infinity is actually the derivative at $0$ of $L(z')$.

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## 1 Answer

A Möbius transformation has the form $L(z)=\frac{az+b}{cz+d}$ for some $a,b,c,d\in\mathbb{C}$ with $ad-bc\neq 0$. Let $w=1/z$, so

$L(w)=\frac{aw+b}{cw+d}=\frac{bz+a}{dz+c}$.

By quotient rule,

$\frac{d}{dz}[L(w)]=\frac{b(dz+c)-d(bz+a)}{(dz+c)^2}=\frac{-(ad-bc)}{(dz+c)^2}$,

and for $z=0$, we get precisely the expression given.

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thx a lot for you help –  myko Apr 16 '12 at 13:38