Complex numbers $z_1$, $z_2$, $z_3$ and $z_4$ lie on a circle, show $w=\frac{(z_1-z_2)(z_3-z_4)}{(z_4-z_1)(z_2-z_3)}$ is real Can anyone help with this:

I have tried breaking it down like this:

and trying to play with the angles like this:

I feel there has to be a more elegant way to do this, could anyone see one?
 A: Let $w_1=\frac{z_1-c}{z_4-c}$, $w_2=\frac{z_2-c}{z_1-c}$, $w_3=\frac{z_3-c}{z_2-c}$, $w_4=\frac{z_4-c}{z_3-c}$, with $c$ the center of the circle. Then, $w=\frac{ (w_2-1)  (w_4-1) }{ (\bar w_1-1)  (\bar w_3-1)}$  and
$$|w_1|= |w_2| = |w_3| = |w_4|=1$$
$$w_1w_2w_3w_4= \bar w_1 \bar w_2 \bar w_3 \bar w_4=1$$
It is straightforward to verify that
$$ (w_1-1)  (w_2-1)  (w_3-1)  (w_4-1) =( \bar w_1-1)  (\bar w_2-1)  (\bar w_3-1)  (\bar w_4-1)\tag 1$$
where $ w_1= \bar w_2 \bar w_3 \bar w_4$, $ w_1w_2 =\bar w_3 \bar w_4$, ... etc. are recognized. Rearrange (1) to get
$$ \frac{ (w_2-1)  (w_4-1) }{ (\bar w_1-1)  (\bar w_3-1)}
=\frac{ (\bar w_2-1)  (\bar w_4-1)}  {(w_1-1)  (w_3-1) }
$$
which is just $w=\bar w $. Thus, $w$ is real.
A: Suppose the origin of the circle is at $z_0=a+ib$. Then $z_i=r_i+z_0$. Where $r_i=r{e^{\theta_1+\theta_2+...+\theta_i}}$, $r$ is the radius of the circle. 
($\theta_1$ is the angle between x-axis and $z_1$. And $\theta_i$ is the angle between $z_i$ and $z_{i+1}$)
So, none of the expression like $z_i-z_j$ would involve $z_0$.
So, $z_1-z_2=re^{i\theta_1}(1-e^{i\theta_2})$
$z_3-z_4=re^{i(\theta_1+\theta_2+\theta_3)}(1-e^{i\theta_4})$
$z_4-z_1=(-re^{i\theta_1})(1-e^{i(\theta_2+\theta_3+\theta_4)})$
$z_2-z_3=(-re^{i(\theta_1+\theta_2+\theta_3)})(1-e^{-i\theta_3})$
So, $\frac{(z_1-z_2)(z_3-z_4)}{(z_4-z_1)(z_2-z_3)}$
$=\frac{(1-e^{i\theta_2})(1-e^{i\theta_4})}{(1-e^{i(\theta_2+\theta_3+\theta_4)})(1-e^{-i\theta_3})}$
$=\frac{1}{\zeta}(1+e^{i(\theta_2+\theta_4)}-(e^{i\theta_2}+e^{i\theta_4}))(1+e^{-i(\theta_2+\theta_4)}-(e^{-i(\theta_2+\theta_3+\theta_4)}+e^{i\theta_3}))$
$=\frac{1}{\zeta}(1-((e^{i\theta_3}+(e^{-i\theta_3})+(e^{i\theta_4}+e^{-i\theta_4})+(e^{i\theta_2}+e^{-i\theta_2}))+((e^{i(\theta_2+\theta_3)}+e^{-i(\theta_2+\theta_3)})+(e^{i(\theta_3+\theta_4)}+e^{-i(\theta_3+\theta_4)})+(e^{i(\theta_4+\theta_2)}+e^{-i(\theta_4+\theta_2)}))-(e^{i(\theta_2+\theta_3+\theta_4)}+e^{-i(\theta_2+\theta_3+\theta_4)}))$
(Where $\zeta=|(1-e^{i(\theta_2+\theta_3+\theta_4)})(1-e^{-i\theta_3})|$) 
Or, more compactly this is equal to 
$=\frac{1}{\zeta}(1-(\lambda_2+\lambda_3+\lambda_4)+(\lambda_{2,3}+\lambda_{3,4}+\lambda_{4,2})-(\lambda_{2,3,4}))$.
This is surely a real number as $e^{i\alpha}+e^{-i\alpha}$ is always real for $\alpha \in \mathbb R$.
A: HINT   
Note that  $$\arg(w)=\arg\left(\frac{z_1-z_2}{z_4-z_1}\cdot\frac{z_3-z_4}{z_2-z_3}\right)=\arg\left(\frac{z_1-z_2}{z_1-z_4}\right)+\arg\left(\frac{z_3-z_4}{z_3-z_2}\right).$$
If the points $z_1, z_2, z_3, z_4$ lie on a circle (in that order), then $z_1$ and $z_3$ lie on opposite arcs with respect to the chord with endpoints $z_2$ and $z_4.$
Depending on the orientation (clockwise or counterclockwise) of the points, the two arguments sum in $\pi$ or $-\pi.$
A: If $|z|=|w|=1$, then
$$
\begin{align}
\arg(z-w)
&=\arg\left(e^{i\arg(z)}-e^{i\arg(w)}\right)\tag1\\[9pt]
&=\arg\left(e^{i(\arg(z)-\arg(w))/2}-e^{-i(\arg(z)-\arg(w))/2}\right)+\arg\left(e^{i(\arg(z)+\arg(w))/2}\right)\tag2\\[3pt]
&=\frac\pi2-\pi[\arg(z)\lt\arg(w)]+\frac{\arg(z)+\arg(w)}2\tag3\\
&\equiv\frac{\pi+\arg(z)+\arg(w)}2\pmod\pi\tag4
\end{align}
$$
Explanation:
$(1)$: $|z|=1\implies z=e^{i\arg(z)}$
$(2)$: $\arg(zw)\equiv\arg(z)+\arg(w)\pmod{2\pi}$
$(3)$: $e^{i(\arg(z)-\arg(w))/2}-e^{-i(\arg(z)-\arg(w))/2}=2i\sin\left(\frac{\arg(z)-\arg(w)}2\right)$
$(4)$: remove the multiples of $\pi$
Applying $(4)$ yields
$$
\begin{align}
\arg\left(\frac{\color{#C00}{(z_1-z_2)}\color{#090}{(z_3-z_4)}}{\color{#00F}{(z_1-z_4)}\color{#C80}{(z_2-z_3)}}\right)
&\equiv\color{#C00}{\frac{\pi+\arg(z_1)+\arg(z_2)}2}+\color{#090}{\frac{\pi+\arg(z_3)+\arg(z_4)}2}\\
&-\color{#00F}{\frac{\pi+\arg(z_1)+\arg(z_4)}2}-\color{#C80}{\frac{\pi+\arg(z_2)+\arg(z_3)}2}\tag5\\[6pt]
&\equiv0\pmod\pi\tag6
\end{align}
$$
Equivalence $(6)$ says that
$$
\frac{(z_1-z_2)(z_3-z_4)}{(z_1-z_4)(z_2-z_3)}\in\mathbb{R}\tag7
$$
