Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

If $z$ and $w$ are two complex numbers such that $|zw|=1$

and $arg(z)-arg(w)=\frac{\pi}{2},$ then :
Option 1)
Option 2)
$z\bar w=\frac{-1+i}{\sqrt2}$
Option 3)
$\bar z w=-i$
Option 4)
$z\bar w=\frac{1-i}{\sqrt2}$

Answers (1)

best_answer

$|zw|=1$ and $arg(z)-arg(w)=\frac{\pi}{2},$

Let $|z|=r$ => $z=re^{i\theta}$

$|\omega |=\frac{1}{r}$ $=> \omega =\frac{1}{r}e^{i\phi }$

$arg(z)-arg(w)=\frac{\pi}{2}$

$\theta -\phi =\frac{\pi}{2}$

$\theta =\frac{\pi}{2}+\phi$

$z\bar{\omega }=re^{i\theta }.\frac{1}{r}e^{-i\phi }$

$=re^{i(\theta -\phi)}$

$=re^{i(\frac{\pi}{2}+\phi -\phi)}$

$=re^{i(\frac{\pi}{2})}$

$=cos(\frac{\pi}{2})+isin(\frac{\pi}{2})$

$=0+i.1$

$=i$

Option 1)

Option 2)

$z\bar w=\frac{-1+i}{\sqrt2}$

Option 3)
$\bar z w=-i$
Option 4)

$z\bar w=\frac{1-i}{\sqrt2}$

Posted by

Aadil

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2026 application correction begins on jeemain.nta.nic.in; edit details by December 2

anam-khan

JEE Mains 2026 LIVE: NTA to conduct session 1 from January 21 to 30; exam pattern, syllabus

anam-khan

JEE Main 2026 correction window opens tomorrow; NTA allows exam city change

Latest Question