How does Collinearity Justify This? I cannot make sense of one line in the given solution I am reading to this question:
Problem: Let $A_0,A_1,\cdots,A_6$ be a regular $7$-gon. Prove that $\displaystyle \frac1{A_0A_1}=\frac1{A_0A_2}+\frac1{A_0A_3}$.
Solution: Let $\varepsilon = e^{2i\pi/7}$. Take $a_k=\varepsilon^k$ to be the complex coordinates of $A_k$ where $k$ ranges from $0$ to $6$. Rotate $a_1$ (to $a_1^\prime$) and $a_2$ (to $a_2^\prime$) around $a_0$ by $2\pi/7$ and $\pi/7$ radians, respectively, so that they are collinear with $a_3$. It suffices, now, to show that:
\begin{equation}
\frac1{a_1^\prime-1}=\frac1{a_2^\prime-1}+\frac1{a_3-1}
\end{equation}
Why are we justified in writing the above, as opposed to:
\begin{equation}
\frac1{|a_1^\prime-1|}=\frac1{|a_2^\prime-1|}+\frac1{|a_3-1|} ?
\end{equation}
I suspect it has something to do with the fact all three lie on one line, but I am missing something obvious?

Since you're rotated $a_1$ and $a_2$ around $a_0=1$ to get $a'_1$ and $a'_2$, we have more than just collinearity of $(a'_1-1, a'_2-1,a_3-1)$: the line they share is a ray going out from the origin. So what you need is that
$$ \frac{1}{z_1} = \frac{1}{z_2} + \frac{1}{z_3} \iff \frac{1}{|z_1|} = \frac{1}{|z_2|} + \frac{1}{|z_3|} $$
when $z_1, z_2, z_3$ have the same argument.
Call the common argument $\theta$ so we have $z_n = e^{i\theta}|z_n|$. But then
$$ \frac{1}{z_n} = e^{-i\theta}\frac{1}{|z_n|}$$
so every term in in the left equation is simply a constant (that is, $e^{-\theta}$) times the corresponding term in the right equation.
