For $\mathrm{a},\mathrm{b},\mathrm{c},\mathrm{d} \in \mathrm{R},$ if ${\mathrm{z}}_1=\mathrm{a}+\mathrm{ib}, {\mathrm{z}}_2=\mathrm{c}+\mathrm{id}$ such that $\left|{\mathrm{z}}_1\right|=\left|{\mathrm{z}}_2\right|=1$and $\mathrm{Re}\left({\mathrm{z}}_1{\overline{\mathrm{z}}}_2\right)=0$, then the pair of complex numbers ${\mathrm{w}}_1=\mathrm{a}+\mathrm{ic}$ and ${\mathrm{w}}_2=\mathrm{b}+\mathrm{id}$ satisfy

$\left|z_1\right|=\left|z_2\right|=1\Rightarrow a^2+b^2=c^2+d^2=1$

$\mathrm{Re}\left(z_1{\overline{z}}_2\right)=0\Rightarrow \mathrm{Re}\left\{\right(a+ib\left)\right(c-id\left)\right\}=0\Rightarrow ac+bd=0$

$c^2+d^2=1\Rightarrow c^2+\frac{a^2{c}^2}{b^2}=1\Rightarrow b^2=c^2$

$c^2+d^2=1\Rightarrow c^2+\frac{a^2{c}^2}{b^2}=1\Rightarrow b^2=c^2$

$\begin{array}{r}\mathrm{Re}\left({\omega }_1{\overline{\omega }}_2\right)=\mathrm{Re}\left\{\right(a+ic\left)\right(b-id\left)\right\}\\ =ab+cd\end{array}$

$\begin{array}{r}\mathrm{Re}\left({\omega }_1{\overline{\omega }}_2\right)=ab+\left(-\frac{a{c}^2}{b}\right)\\ =\frac{a{b}^2-a{c}^2}{b}=0\end{array}$

$\left|{\omega }_1\right|=\sqrt{a^2+c^2}=\sqrt{a^2+b^2}=1$

$\left|{\omega }_2\right|=\sqrt{b^2+d^2}=\sqrt{a^2+b^2}=1$