$\frac{az+ b}{cz+ d},\frac{αz+β}{γz+ δ}$有公共不动点的条件

hbghlyj

4. COMMON FIXED POINTS

$a,b,c,d\inC,ad-bc=1,$
$α,β,γ,δ\inC,αδ-βγ=1,$
则关于$z\inC\cup\{\infty\}$的方程$$z=\frac{az+ b}{cz+ d}=\frac{αz+β}{γz+ δ}$$有解的充要条件是
\[(\delta a -\gamma  b)(a \alpha +b \gamma )
+(\delta  c -\gamma  d) (a \beta +b \delta )
+(\alpha  b-\beta  a)(\alpha  c+\gamma  d)
+(\alpha  d -\beta  c)(\beta  c+d \delta )=
(a+d)^2-2\]

hbghlyj

验证：\[\tiny\text{Tr}\left[\left(
\begin{array}{cc}
a & b \\
c & d \\
\end{array}
\right).\left(
\begin{array}{cc}
\alpha  & \beta  \\
\gamma  & \delta  \\
\end{array}
\right).\left(
\begin{array}{cc}
a & b \\
c & d \\
\end{array}
\right).\left(\frac{\left(
\begin{array}{cc}
\alpha  & \beta  \\
\gamma  & \delta  \\
\end{array}
\right)}{\alpha  \delta -\beta  \gamma }\right)^{-1}\right]=(a \alpha +b \gamma ) (a \delta -b \gamma )+(\alpha  b-a \beta ) (\alpha  c+\gamma  d)+(a \beta +b \delta ) (c \delta -\gamma  d)+(\alpha  d-\beta  c) (\beta  c+d \delta )\]

1. FullSimplify[Tr[{{a,b},{c,d}}.{{\[Alpha],\[Beta]},{\[Gamma],\[Delta]}}.{{a,b},{c,d}}.Inverse[{{\[Alpha],\[Beta]},{\[Gamma],\[Delta]}}/(\[Alpha] \[Delta]-\[Beta] \[Gamma])]]==(\[Delta] a-\[Gamma] b) (a \[Alpha]+b \[Gamma])+(\[Delta] c-\[Gamma] d) (a \[Beta]+b \[Delta])+(\[Alpha] b-\[Beta] a) (\[Alpha] c+\[Gamma] d)+(\[Alpha] d-\[Beta] c) (\[Beta] c+d \[Delta])]

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\[\tiny
\frac{\text{Resultant}[(a z+b)-z (c z+d),(\beta +\alpha  z)-z (\delta +\gamma  z),z]}{\alpha  \delta -\beta  \gamma }+\text{Tr}\left[\left(
\begin{array}{cc}
a & b \\
c & d \\
\end{array}
\right).\left(
\begin{array}{cc}
\alpha  & \beta  \\
\gamma  & \delta  \\
\end{array}
\right).\left(
\begin{array}{cc}
a & b \\
c & d \\
\end{array}
\right).\left(
\begin{array}{cc}
\alpha  & \beta  \\
\gamma  & \delta  \\
\end{array}
\right)^{-1}\right]=(a+d)^2-2 (a d-b c)\]

1. FullSimplify[Resultant[(a z+b)-z(c z+d),(\[Alpha] z+\[Beta])-z(\[Gamma] z+\[Delta]),z]/(\[Alpha] \[Delta]-\[Beta] \[Gamma])+Tr[{{a,b},{c,d}}.{{\[Alpha],\[Beta]},{\[Gamma],\[Delta]}}.{{a,b},{c,d}}.Inverse[{{\[Alpha],\[Beta]},{\[Gamma],\[Delta]}}]]==(a+d)^2-2(a d -b c)]

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True

如何证明呢