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 (*)->.