We have,
$z_1=x_1+i y_1$
$z_2=x_2+i y_2$
$z_3=x_1+\frac{i x_2}{2}$
$z_4=2 y_1+i y_2$
$\left|z_1\right|=1,\left|z_2\right|=2$ and $\operatorname{Re}\left(z_1 z_2\right)=0$
Let $z_1=e^{i \alpha}$ and $z_2=2 e^{i \beta}$
$z_1 z_2=2 e^{i(\alpha+\beta)}$
$z_1 z_2=2(\cos (\alpha+\beta)+i \sin (\alpha+\beta))$
$\operatorname{Re}\left(z_1+z_2\right)=2 \cos (\alpha+\beta)=0$
$\therefore \quad \alpha+\beta=\frac{\pi}{2}$
$\beta=\frac{\pi}{2}-\alpha$
$\therefore \quad z_1=\cos \alpha+i \sin \alpha$
$z_2=2(\cos \beta+i \sin \beta)$
$z_2=2(\sin \alpha+i \cos \alpha)$
$\therefore \quad z_3=\cos \alpha+i\left(\frac{2 \sin \alpha}{2}\right)$
$z_3=\cos \alpha+i \sin \alpha$
$\left|z_3\right|=1$
$z_4=2(\sin \alpha+i \cos \alpha)$
$\left|z_4\right|=2$
$z_3 z_4=2(\cos \alpha+i \sin \alpha)(\sin \alpha+i \cos \alpha)$
$z_3 z_4=2[(\cos \alpha \sin \alpha-\cos \alpha \sin \alpha)$ $\left.+i\left(\sin ^2 \alpha+\cos ^2 \alpha\right)\right]$
$z_3 z_4=2 i$
$\operatorname{Re}\left(z_3 z_4\right)=0$