$P$ is a point on the Argand diagram. On the circle with $OP$ as diameter two points $Q$ and $R$ are taken such that $\angle POQ = \angle QOR = \theta$. If $O$ is the origin and $P$, $Q$, $R$ are represented by the complex numbers $Z_1$, $Z_2$, $Z_3$ respectively, show that $$Z_{2}^{2} \cos 2\theta = Z_1 \cdot Z_3 \cos^2 \theta$$
I tried it a lot but don't know how to start. Can anybody provide a hint?
Assume $z_1\ne z_3$ and take $\theta \in \left( 0,\frac{\pi}{4} \right)$ as anti-clockwise whereas $\theta \in \left( -\frac{\pi}{4},0 \right)$ as clockwise.
\begin{align*} z_2 &= \frac{z_1}{2}(1+e^{2i\theta}) \\ z_3 &= \frac{z_1}{2}(1+e^{4i\theta}) \\ z_2^2 \cos 2\theta &= \frac{z_1^2}{4}(1+e^{2i\theta})^2 \times \frac{e^{2i\theta}+e^{-2i\theta}}{2} \\ &= \frac{z_1^2}{4}(1+e^{2i\theta})^2 \times \frac{1+e^{4i\theta}}{2e^{2i\theta}} \\ &= \frac{z_1^2(1+e^{4i\theta})}{2} \left( \frac{e^{i\theta}+e^{-i\theta}}{2} \right)^2 \\ &= z_1 z_3 \cos^2 \theta \end{align*}
