The problem in question is this one:
Let $z_1$,$z_2$ $\in$ $\mathbb{C}$; $\rvert{z_1}\rvert$= $\rvert{z_2}\rvert$ and $\alpha$ $\in$ $\mathbb{R}$. Prove that:
$\rvert \alpha z_1+z_2\rvert$=$\rvert z_1+\alpha z_2\rvert$
My approach was that since $\rvert{z_1}\rvert$= $\rvert{z_2}\rvert$ then their coefficients must be the same,so I could just rearrange the coefficients to prove the statement but then I realized that is wrong because there is an infinite amount of real numbers that could satisfy both sides of the equality $\rvert{z_1}\rvert$= $\rvert{z_2}\rvert$ so the coefficients need not to be the same. When I looked at the answer I could understand pretty much everything except for the penultimate step. Here is the given answer.
\begin{align} \rvert{z_1}\rvert= \rvert{z_2}\rvert=r;r \in (0,\infty); \rvert \alpha z_1+z_2\rvert=\rvert z_1+\alpha z_2\rvert \Rightarrow \rvert \alpha z_1+z_2\rvert^2=\rvert z_1+\alpha z_2\rvert^2 \Rightarrow (\alpha z_1+z_2) (\overline{ \alpha z_1+z_2})= (z_1+\alpha z_2) (\overline{ z_1+\alpha z_2})\Rightarrow (\alpha z_1+z_2) ( \alpha \overline{z_1}+\overline{z_2})=( z_1+\alpha z_2)( \overline{z_1}+\alpha \overline{z_2})\Rightarrow \alpha^2 z_1 \overline z_1 + \alpha z_1 \overline z_2+\alpha z_2 \overline z_1+z_2 \overline z_2= z_1\overline{z_1}+\alpha z_1 \overline{z_2}+\alpha z_2\overline{z_1}+\alpha^2z_2\overline{z_2}\Rightarrow \alpha^2 r^2+\alpha^2=\alpha^2 + \alpha^2 r^2\end{align}
My question is how does the penultimate step is implied by the last, specifically regarding the squares.
The final expression in that proof is wrong ($\alpha^2 r^2 + \alpha^2$ should be $\alpha^2 r^2 + r^2$), also all implications are in the wrong direction. I would write it as follows:
For $|z_1| = |z_2| = r > 0$ and $\alpha \in \Bbb R$ is $$ \begin{align} |\alpha z_1 + z_2|^2 &= (\alpha z_1 + z_2) \cdot \overline{(\alpha z_1 + z_2)} \\ &= (\alpha z_1 + z_2) \cdot (\alpha \overline{z_1} + \overline{z_2}) \\ &= \alpha^2 z_1 \overline{z_1} + z_2 \overline{z_2} + \alpha(z_1 \overline{z_2} + z_2 \overline{z_1}) \\ &= \alpha^2 r^2 + r^2 + \alpha(z_1 \overline{z_2} + z_2 \overline{z_1}) \, . \end{align} $$ The same holds with $z_1$ and $z_2$ exchanged: $$ | z_1 + \alpha z_2|^2 = \alpha^2 r^2 + r^2 + \alpha(z_2 \overline{z_1}+z_1 \overline{z_2}) \, . $$ The right-hand sides of both equations are the same, it follows that $$ |\alpha z_1 + z_2|^2 = | z_1 + \alpha z_2|^2 \implies |\alpha z_1 + z_2| = | z_1 + \alpha z_2|\, . $$
An alternative proof. This idea is to rotate the configuration so that the rotated points $z_1', z_2'$ are symmetric with respect to the real axis.
Let $s$ be a square root of $\overline{z_2}/z_1$. Then $|s| = 1$ and $$ s z_1 = \frac{\overline{z_2}}{s} = \overline{s z_2} $$ and therefore $$ \begin{align} |\alpha z_1 + z_2| &= |\alpha s z_1 + s z_2| \\ &= |\overline{\alpha s z_1 + s z_2}| \\ &= |\alpha s z_2 + s z_1| \\ &= |\alpha z_2 + z_1| \, . \end{align} $$