For $a, b, c, d \in R$, if $z_1=a+i b, z_2=c+i d$ are such that $\left|z_1\right|=\left|z_2\right|=1$ and…
For $a, b, c, d \in R$, if $z_1=a+i b, z_2=c+i d$ are such that $\left|z_1\right|=\left|z_2\right|=1$ and $\operatorname{Re}\left(z_1 \bar{z}_2\right)=0$, then the pair of complex numbers $w_1=a+i c$ and $w_2=b+i d$ satisfy
$\operatorname{Re}\left(w_1 \bar{w}_2\right)=0$
$\operatorname{Re}\left(w_1 \bar{w}_2\right)=1$
$\left|w_1\right| \neq\left|w_2\right|$
$\left|w_1\right|=\left|w_2\right|=0$
Solution
As it is given, $\left|z_1\right|=\left|z_2\right|=1$, so let
$
\begin{aligned}
& z_1=a+i b=\cos \alpha+i \sin \alpha \\
& \text { and } z_2=c+i d=\cos \beta+i \sin \beta \\
& \text { Now, } \operatorname{Re}\left(z_1 \bar{z}_2\right)=a c+b d=\cos \alpha \cos \beta+\sin \alpha \sin \beta
\end{aligned}
$
$=\cos (\alpha-\beta)=0 \quad$ [given]
So, $\alpha-\beta=\frac{\pi}{2}$ or $-\frac{\pi}{2}$
Now, for the pair of complex numbers
$
\begin{aligned}
& \quad w_1=a+i c=\cos \alpha+i \cos \beta \\
& \text { and } w_2=b+i d=\sin \alpha+i \sin \beta \\
& \text { The } \operatorname{Re}\left(w_1 \bar{w}_2\right)=a b+c d=\cos \alpha \sin \alpha+\cos \beta \sin \beta \\
& =\frac{1}{2}[\sin 2 \alpha+\sin 2 \beta] \\
& =\frac{1}{2} \times 2 \sin (\alpha+\beta) \cos (\alpha-\beta)=0 \quad\left[\because \alpha-\beta=\frac{\pi}{2} \text { or }-\frac{\pi}{2}\right]
\end{aligned}
$
Hence, option (a) is correct