If z_1 = a + ib and z_2 = c + id are co | vedclass

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Question

Since $\left|z_{1}ight|=\left|z_{2}ight|=1,$ we have

$z_{1}=\cos 	heta_{1}+i \sin 	heta_{1}, z_{2}=\cos 	heta_{2}+i \sin 	heta_{2}$

wher

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Since $\left|z_{1}ight|=\left|z_{2}ight|=1,$ we have

$z_{1}=\cos 	heta_{1}+i \sin 	heta_{1}, z_{2}=\cos 	heta_{2}+i \sin 	heta_{2}$

where $	heta_{1}=\arg \left(z_{1}ight)$ and $	heta_{2}=\arg \left(z_{2}ight)$

Also, $z_{1}=a+ ib$ and $z_{2}=c+i d$

Therefore $a=\cos 	heta_{1}, b=\sin 	heta_{1}, c=\cos 	heta_{2}$

and $d=\sin 	heta_{2}$

Also, $\quad \mathrm{R}\left(\mathrm{z}_{1} \overline{\mathrm{z}}_{2}ight)=0$

$\Rightarrow \quad \mathrm{R}\left[\left(\cos 	heta_{1}+i \sin 	heta_{1}ight)\left(\cos 	heta_{2}-i \sin 	heta_{2}ight)ight]=0$

$\Rightarrow \quad \mathrm{R}\left[\left(\cos \left(	heta_{1}-	heta_{2}ight)+i \sin \left(	heta_{1}-	heta_{2}ight)ight]=0ight.$

$\Rightarrow \cos \left(	heta_{1}-	heta_{2}ight)=0 \quad \Rightarrow \quad 	heta_{1}-	heta_{2}=\frac{\pi}{2}$

$\Rightarrow 	heta_{1}=	heta_{2}+\frac{\pi}{2}$

Now, $\omega_{1}=a+i c=\cos 	heta_{1}+i \cos 	heta_{2}=\cos 	heta_{1}+i \sin 	heta_{1}$

$\Rightarrow\left|\omega_{1}ight|=1 .$ Similarly, $\left|\omega_{2}ight|=1$

Next $\omega_{1} \bar{\omega}_{2}=\left(\cos 	heta_{1}+i \sin 	heta_{1}ight)\left(\cos 	heta_{2}-i \sin 	heta_{2}ight)$

${=\cos \left(	heta_{1}-	heta_{2}ight)+i \sin \left(	heta_{1}-	heta_{2}ight)} $

${=\cos \pi / 2+i \sin \pi / 2=0+i}$

$	herefore \quad \mathrm{R}\left(\omega_{1} \bar{\omega}_{2}ight)=0$.

If $z_1 = a + ib$ and $z_2 = c + id$ are complex numbers such that $| z_1 | = | z_2 |=1$ and $R({z_1}\overline {{z_2}} ) = 0$, then the pair of complex numbers $w_1 = a + ic$ and $w_2 = b + id$ satisfies

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