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Misc 17 - Chapter 5 Class 11 Complex Numbers (Term 1)

Last updated at March 16, 2023 by

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

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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

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