A complex number $z_1$  is rotated in anticlockwise direction by an angle $\theta$  and we get $z_2$  and if the same complex number $z_1$  is rotated by an angle  $\theta$ in clockwise direction and we get  $z_3$, then

$z\text{'}=z{e}^{i\alpha }\left(1\right)$

$z\text{'}\text{'}=z{e}^{-i\alpha } \left(2\right)$

$\therefore z\text{'}z\text{'}\text{'}=z^2$

$\Rightarrow z\text{'},z,z\text{'}\text{'}$are in G.P
Also, ${\left(\frac{z\text{'}}{z}\right)}^2+{\left(\frac{z\text{'}\text{'}}{z}\right)}^2=2\cos 2\alpha$$\Rightarrow {\left(z\text{'}\right)}^2+{\left(z\text{'}\text{'}\right)}^2=2z^2\cos 2\alpha$$=2z^2\left(2{\cos }^2\alpha -1\right)$
$\Rightarrow {\left(z\text{'}\right)}^2+{\left(z\text{'}\text{'}\right)}^2+2z^2=4z^2{\cos }^2\alpha$

$\Rightarrow {\left(z\text{'}\right)}^2+{\left(z\text{'}\text{'}\right)}^2+2z\text{'}z\text{'}\text{'}=4z^2{\cos }^2\alpha$

$\Rightarrow {\left(z\text{'}+z\text{'}\text{'}\right)}^2=4z^2{\cos }^2\alpha$

$\Rightarrow z\text{'}+z\text{'}\text{'}=2z\cos \alpha$