If z₁ = a + ib and z₂ = c + id are complex numbers such that |z₁| = |z₂| = 1 and $\mathrm{Re}\left({\mathrm{z}}_1{\overline{\mathrm{z}}}_2\right)=0$, then the pair of complex numbers ${\mathrm{\omega }}_1=\mathrm{a}+\mathrm{ic}\text{ and }{\mathrm{\omega }}_2=\mathrm{b}+\mathrm{id}$ satisfies

$\left|{\mathrm{z}}_1\right|=\left|{\mathrm{z}}_2\right|=1$

$\Rightarrow {\mathrm{a}}^2+{\mathrm{b}}^2={\mathrm{c}}^2+{\mathrm{d}}^2=1$          (1)

and $\mathrm{Re}\left({\mathrm{z}}_1{\overline{\mathrm{z}}}_2\right)=0$

$\begin{array}{l}\Rightarrow \mathrm{Re}\left\{\right(\mathrm{a}+\mathrm{ib}\left)\right(\mathrm{c}+\mathrm{id}\left)\right\}=0\\ \Rightarrow \mathrm{ac}+\mathrm{bd}=0 \left(2\right)\end{array}$

Now, from Eqs. (1) and (2), we get

$\begin{array}{l}{\mathrm{a}}^2+{\mathrm{b}}^2=1\\ \Rightarrow {\mathrm{a}}^2+\frac{{\mathrm{a}}^2{\mathrm{c}}^2}{{\mathrm{d}}^2}=1\\ \Rightarrow {\mathrm{a}}^2={\mathrm{d}}^2 \left(3\right)\end{array}$

Also, ${\mathrm{c}}^2+{\mathrm{d}}^2=1$

$\begin{array}{l}\Rightarrow {\mathrm{c}}^2+\frac{{\mathrm{a}}^2{\mathrm{c}}^2}{{\mathrm{b}}^2}=1\\ \Rightarrow {\mathrm{b}}^2={\mathrm{c}}^2 \left(4\right)\end{array}$

$\left|{\mathrm{\omega }}_1\right|=\sqrt{{\mathrm{a}}^2+{\mathrm{c}}^2}=\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}=1$           [From (1) and (4)]

and $\left|{\mathrm{\omega }}_2\right|=\sqrt{{\mathrm{b}}^2+{\mathrm{d}}^2}=\sqrt{{\mathrm{c}}^2+{\mathrm{d}}^2}=1$    [from (1) and (4)]

Further $\mathrm{Re}\left({\mathrm{\omega }}_1{\overline{\mathrm{\omega }}}_2\right)=\mathrm{Re}\left\{\right(\mathrm{a}+\mathrm{ic}\left)\right(\mathrm{b}-\mathrm{id}\left)\right\}$

                              = ab + cd

                             $=\mathrm{ab}-\frac{{\mathrm{ac}}^2}{\mathrm{b}}$          [From (2)]

                             $=\frac{{\mathrm{ab}}^2-{\mathrm{ac}}^2}{\mathrm{b}}=0$  [from(4)]

Also, $\mathrm{Im}\left({\mathrm{\omega }}_1{\overline{\mathrm{\omega }}}_2\right)=\mathrm{bc}-\mathrm{ad}$

                          $=\mathrm{bc}-\mathrm{a}\left(-\frac{\mathrm{ac}}{\mathrm{b}}\right)=\frac{\left({\mathrm{a}}^2+{\mathrm{b}}^2\right)\mathrm{c}}{\mathrm{b}}=\frac{\mathrm{c}}{\mathrm{b}}=\pm 1\ne 0$

$\therefore \left|{\mathrm{\omega }}_1\right|=1, \left|{\mathrm{\omega }}_2\right|=1\text{ and }\mathrm{Re}\left({\mathrm{\omega }}_1{\overline{\mathrm{\omega }}}_2\right)=0$