If $\text{Z}\text{=a}\text{+}\text{ib,}{\text{Z}}^{\text{'}}\text{=}\text{c}\text{+}\text{id}$  be two complex numbers such that $\left|{\displaystyle Z}\right|{\displaystyle =}\left|{{\displaystyle Z}}^{{\displaystyle \text{'}}}\right|{\displaystyle =}{\displaystyle 1}$  and $\text{Re}\left(\text{Z}\overline{{\text{Z}}^{\text{'}}}\right)\text{=}\text{0}$   then $\text{ω}\text{=}\text{a}\text{+}\text{ic}$  and ${\text{ω}}^{\text{'}}\text{=}\text{b}\text{+}\text{id}$  will be such that

Given $\left|z\right|=1\Rightarrow a^2+b^2=1\to 1$

            Given $\left|z^1\right|=1\Rightarrow c^2+d^2=1\to 2$

            $z\overline{z^1}=\left(a+ib\right)\left(c-id\right)$

            Given $Rc\left(z\overline{z^1}\right)=0\Rightarrow ac+bd=0$

                        $\Rightarrow b=-\frac{ac}{d}\to 3$

            From 1, 3,  $a^2+\frac{a^2{c}^2}{d^2}=1\Rightarrow a^2=d^2\left[\because u\sin g2\right]$

            $\left|w\right|=\sqrt{a^2+c^2}=\sqrt{d^2+c^2}=1\left[\because from2\right]$

            $\left|w^1\right|=\sqrt{b^2+d^2}=\sqrt{b^2+a^2}=1\left[\because from1\right]$

            $R\left|w\overline{w^1}\right|=ab+cd=-a\left(\frac{ac}{d}\right)+cd=0$

            $\left|w\overline{w^1}\right|=\left|w\right|\left|w^1\right|=1$