If two Möbius maps are equivalent, then they relate by constant

Question

Let $\mu_1=\frac{a_1z+b_1}{c_1z+d_1}$ and $\mu_2=\frac{a_2z+b_2}{c_2z+d_2}$. Then, if $\mu_1=\mu_2$, there exists $0\ne\kappa\in\mathbb{C}$ such that $a_2=\kappa a_1$, $b_2=\kappa b_1$, $c_2=\kappa c_1$, $d_2=\kappa d_1$.

My approach:

Since $\frac{\gamma}{\gamma} \frac{a_1z+b_1}{c_1z+d_1}=\frac{a_1z+b_1}{c_1z+d_1} $ We can express the matrix $\gamma A$ (or $A$) corresponding to $\mu_1$ as $\begin{bmatrix} \gamma a_1 & \gamma b_1 \\ \gamma c_1 & \gamma d_1 \end{bmatrix}$ or $\begin{bmatrix} a_1 & b_1 \\ c_1 & d_1 \end{bmatrix}$, respectively.

The matrix corresponding to $\mu_2$ is

$\begin{bmatrix} a_2 & b_2 \\ c_2 & d_2 \end{bmatrix}$

$A$ and $B$ are related as follows:

(*)-> $\frac{1}{\det(A)}A=\frac{1}{\det(B)}B $, or $\frac{\det(B)}{\det(A)}A=B$. Let $\kappa:=\frac{\det(B)}{\det(A)}$, then we have the needed result.

Please let me know if I'm not missing anything. I'm not sure if this proof is correct since I think I need to justify the relation (*)->.

Answer 1

For $\mu_1=\dfrac{az+b}{cz+d}$ and $\mu_2=\dfrac{\alpha z+\beta }{\gamma z+\delta }$, $\mu_1=\mu_2$ iff $$\frac{az+b}{cz+d}=\frac{\alpha z+\beta }{\gamma z+\delta }$$ $$(az+b)(\gamma z+\delta )=(cz+d)(\alpha z+\beta )$$ $$(a\gamma -c\alpha )z^2+(a\delta +b\gamma -c\beta -\alpha d)z+(b\delta -d\beta )=0$$ this valid for all $z$ so \begin{cases} a\gamma -c\alpha =0\\ b\delta -d\beta =0\\ a\delta +b\gamma -c\beta -\alpha d=0 \end{cases} then for $p\in\mathbb{C}$ and $q\in\mathbb{C}$ $$\frac{a}{\alpha }=\frac{c}{\gamma }=p$$ $$\frac{b}{\beta }=\frac{d}{\delta }=q$$ othe equality shows $$a\delta +b\gamma -c\beta -\alpha d=\alpha p\delta +\beta q\gamma -\gamma p\beta -\delta q\alpha =(\alpha \delta -\beta \gamma )(p-q)=0$$ shows $p=q=\kappa\in\mathbb{C}$ such that $$\frac{a}{\alpha }=\frac{c}{\gamma }=\frac{b}{\beta }=\frac{d}{\delta }=\kappa$$