Check sibling questions

Misc 17 - Chapter 5 Class 11 If are different complex numbers - Proof- Using modulus conjugate property (Pg 102)

Misc 17 - Chapter 5 Class 11 Complex Numbers - Part 2
Misc 17 - Chapter 5 Class 11 Complex Numbers - Part 3

This video is only available for Teachoo black users

Get live Maths 1-on-1 Classs - Class 6 to 12


Transcript

Misc 17 If and are different complex numbers with | | = 1, then find |( " " )/(1 )| . We know that |z|2 = (z) ( ) |( )/(1 )|^2=(( )/(1 )) ((( )/(1 )) ) = (( )/(1 )) (( ) /(1 ) ) = (( )/(1 )) (( )/(1 )) = (( )/(1 )) (( )/(1 )) = (( )( ))/((1 )(1 )) = ( ( ) ( ))/(1 (1 ) (1 ) ) = ( + )/(1 + ) = (| |^2 + | |^2)/(1 +| |^2 | |^2 ) Given that | |=1 So, | |^2=1 = (1 + | |^2)/(1 + | |^2 1) = (1 + | |^2)/(1 + | |^2 ) = 1 Hence |( )/(1 )|^2 = 1 |( )/(1 )| = 1 |( )/(1 )| =1

Ask a doubt
Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.