4-1.Complex numbers
normal

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

A

$|w_1 |=1$

B

$|w_2 |=1$

C

$R({w_1}\overline {{w_2}} ) = 0$

D

All the above

Solution

Since $\left|z_{1}\right|=\left|z_{2}\right|=1,$ we have

$z_{1}=\cos \theta_{1}+i \sin \theta_{1}, z_{2}=\cos \theta_{2}+i \sin \theta_{2}$

where $\theta_{1}=\arg \left(z_{1}\right)$ and $\theta_{2}=\arg \left(z_{2}\right)$

Also, $z_{1}=a+ ib$ and $z_{2}=c+i d$

Therefore $a=\cos \theta_{1}, b=\sin \theta_{1}, c=\cos \theta_{2}$

and $d=\sin \theta_{2}$

Also, $\quad \mathrm{R}\left(\mathrm{z}_{1} \overline{\mathrm{z}}_{2}\right)=0$

$\Rightarrow \quad \mathrm{R}\left[\left(\cos \theta_{1}+i \sin \theta_{1}\right)\left(\cos \theta_{2}-i \sin \theta_{2}\right)\right]=0$

$\Rightarrow \quad \mathrm{R}\left[\left(\cos \left(\theta_{1}-\theta_{2}\right)+i \sin \left(\theta_{1}-\theta_{2}\right)\right]=0\right.$

$\Rightarrow \cos \left(\theta_{1}-\theta_{2}\right)=0 \quad \Rightarrow \quad \theta_{1}-\theta_{2}=\frac{\pi}{2}$

$\Rightarrow \theta_{1}=\theta_{2}+\frac{\pi}{2}$

Now, $\omega_{1}=a+i c=\cos \theta_{1}+i \cos \theta_{2}=\cos \theta_{1}+i \sin \theta_{1}$

$\Rightarrow\left|\omega_{1}\right|=1 .$ Similarly, $\left|\omega_{2}\right|=1$

Next $\omega_{1} \bar{\omega}_{2}=\left(\cos \theta_{1}+i \sin \theta_{1}\right)\left(\cos \theta_{2}-i \sin \theta_{2}\right)$

${=\cos \left(\theta_{1}-\theta_{2}\right)+i \sin \left(\theta_{1}-\theta_{2}\right)} $

${=\cos \pi / 2+i \sin \pi / 2=0+i}$

$\therefore \quad \mathrm{R}\left(\omega_{1} \bar{\omega}_{2}\right)=0$.

Standard 11
Mathematics

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