If $\frac{az+b}{cz+d}=\frac{ez+f}{gz+h}$, then, $a = z_0 e, b = z_0 f, c = z_0 g, d = z_0 h$ for some $z_0 \in \mathbb{C}-\{0\}$. How to prove the following proposition elegantly?
My proof is very bad.

Let $a,b,c,d,e,f,g,h \in \mathbb{C}$ and $ad-bc\ne 0, eh-fg\ne 0$.
  If
  $$\frac{az+b}{cz+d}=\frac{ez+f}{gz+h}$$
  for all $z \in \mathbb{C} \cup \{\infty\}$,
  then, $a = z_0 e, b = z_0 f, c = z_0 g, d = z_0 h$ for some $z_0 \in \mathbb{C}-\{0\}$.  

Proof:  
Assume that $a=0\ne e$.
Then,  
$$0\ne \frac{b}{c\frac{-f}{e}+d}=\frac{e \frac{-f}{e}+f}{g \frac{-f}{e} +h}=0.$$ 
This is a contradiction.  
So, if $a=0$ then, $e =0$.
Similarly, if $e=0$ then, $a=0$.  


*

*Suppose $a=e=0$.
Then, $b, c, f, g \notin \{0\}$.
Assume $d =0\ne h$.
Then, $\infty = \frac{b}{c 0}=\frac{f}{g 0 + h} \in \mathbb{C}$.
This is a contradiction.
So, if $d=0$, then $h=0$.
Similarly, if $h=0$, then $d=0$.
a. Suppose $h=d=0$.
Then, $\frac{b}{c} = \frac{b}{c 1} = \frac{f}{g 1} = \frac{f}{g}$.
So, in this case, $a =0 = \frac{b}{f} 0=\frac{b}{f} e, b = \frac{b}{f} f, c = \frac{b}{f} g, d=0 = \frac{b}{f} 0=\frac{b}{f} h$.
b. Suppose $h, d \notin \{0\}$.
Then $\infty=\frac{b}{c \frac{-d}{c}+d}=\frac{f}{g \frac{-d}{c}+h}$.
So, $g \frac{-d}{c}+h=0, -g d + c h = 0, \frac{c}{g} = \frac{d}{h} =: z_0.$
$\frac{b}{c z + d} = \frac{b}{z_0 g z + z_0 h} = \frac{\frac{b}{z_0}}{g z + h} = \frac{f}{g z+h}$ for all $z \in \mathbb{C} \cup \{\infty\}$.
So, $\frac{b}{z_0} = f, b = z_0 f$.
So, in this case, $a = 0 = z_0 0 = z_0 e,b = z_0 f, c = z_0 g, d = z_0 h$.

*Suppose $a, e \notin \{0\}$.
Assume $c=0 \ne g$.
Then $$\mathbb{C} \ni \frac{a \frac{-h}{g} + b}{d} = \frac{e \frac{-h}{g} + f}{g \frac{-h}{g} + h} = \infty.$$
This is a contradiction.
So, if $c=0$, then $g=0$.
Similarly, if $g = 0$, then $c=0$.
a. Suppose $c=g=0$.
Assume $b=0 \ne f$.
Then, $0 = \frac{a 0}{d} = \frac{e 0 + f}{h} = \frac{f}{h} \ne 0$.
This is a contradiction.
So, if $b=0$, then $f=0$.
Similarly, if $f=0$, then $b=0$.  


*

*Suppose $b=f=0$.
Then, $\frac{a}{d} = \frac{a 1}{d} = \frac{e 1}{h} = \frac{e}{h}$.
So, $\frac{a}{e}=\frac{d}{h} =: z_0$.
So, in this case, $a = z_0 e, b = 0 = z_0 0 = z_0 f, c = 0 = z_0 0 = z_0 g, d = z_0 h$.  

*Suppose $b, f \notin \{0\}$.
$$0=\frac{a \frac{-b}{a}+b}{d} = \frac{e \frac{-b}{a}+f}{h}, -eb+af=0, \frac{a}{e}=\frac{b}{f}=:z_0.$$
$$\frac{a z +b}{d} = \frac{z_0 e z + z_0 f}{d} = \frac{e z + f}{\frac{d}{z_0}} = \frac{e z + f}{h}$$
for all $z \in \mathbb{C} \cup \{\infty\}$.
So, $\frac{d}{z_0}=h$.
So, in this case, $a = z_0 e, b=z_0 f, c =0=z_0 0 = z_0 g, d = z_0 h$.
$\cdots$

Let $$P = 
\begin{pmatrix}
a & b \\
c & d
\end{pmatrix}\,\,\,\text{ and }\,\,\,Q = 
\begin{pmatrix}
e & f \\
g & h
\end{pmatrix}
$$
The first two conditions mean that $P$ and $Q$ are both invertible.
Now, let $$u_z=\begin{pmatrix}z\\1
\end{pmatrix}$$
Then the third condition means that there exists $\lambda_z\in\mathbb C-\{0\}$ such that
$$Pu_z=\lambda_z Qu_z$$
In other words $Q^{-1}Pu_z=\lambda_z u_z$ for all $z\in\mathbb C\cup \{\infty\}$. 
It's easy to verify that it implies that any vector of $\mathbb C^2$ is an eigenvector of $Q^{-1}P$ and therefore there must exist $z_0\in\mathbb C-\{0\}$ such that $$Q^{-1}P=z_0 I$$
where $I$ is the identity matrix. This yields the desired result.
