Misc 15 - Find modulus of (1 + i)/(1 - i) - (1 - i)/(1 + i) - Miscellaneous

  1. Chapter 5 Class 11 Complex Numbers
  2. Serial order wise
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Misc 15 Find the modulus of (1 + ๐‘–)/(1 โˆ’ ๐‘–) โˆ’ (1 โˆ’ ๐‘–)/(1 + ๐‘–) First we solve (1 + ๐‘–)/(1 โˆ’ ๐‘–) โˆ’ (1 โˆ’ ๐‘–)/(1 + ๐‘–) (1 + ๐‘–)/(1 โˆ’ ๐‘–) โˆ’ (1 โˆ’ ๐‘–)/(1 + ๐‘–) = ((1 + ๐‘–) ร— (1 + ๐‘–) โˆ’(1 โˆ’ ๐‘–) ร— (1 โˆ’ ๐‘–))/((1โˆ’๐‘–) (1+ ๐‘–)) = ((1 + ๐‘–)2 โˆ’ ( 1 โˆ’ ๐‘– )2)/((1 โˆ’ ๐‘–) (1 + ๐‘–)) Using a2 โ€“ b2 = (a + b) (a โ€“ b) = ([( 1+ ๐‘– ) + ( 1 โˆ’ ๐‘– )] [( 1+ ๐‘– ) โˆ’ ( 1 โˆ’ ๐‘– )])/((1)2 โˆ’(๐‘–)2) = (( 1 + 1 + ๐‘– โˆ’ ๐‘– ) ( ๐‘– + ๐‘– + 1 โˆ’1 ))/( 1 โˆ’ ๐‘–2) = (( 2 + 0 ) ( 2๐‘– + 0 ))/(1 โˆ’ ๐‘–2) = ((2) (2๐‘–))/(1 โˆ’๐‘–2) = 4๐‘–/(1 โˆ’๐‘–2) Putting i2 = - 1 = 4๐‘–/( 1โˆ’ ( โˆ’ 1 ) ) = 4๐‘–/( 1 + 1) = ( 4๐‘–)/( 2) = 2๐‘– = 0 + 2๐‘– Thus , (1 + ๐‘–)/(1 โˆ’ ๐‘–) โˆ’ (1 โˆ’ ๐‘–)/(1 + ๐‘–) = 0 + 2๐‘– Thus , (1 + ๐‘–)/(1 โˆ’ ๐‘–) โˆ’ (1 โˆ’ ๐‘–)/(1 + ๐‘–) = 0 + 2๐‘– We need to find modulus of (1 + ๐‘–)/(1 โˆ’ ๐‘–) โˆ’ (1 โˆ’ ๐‘–)/(1 + ๐‘–) Complex number (0 + 2๐‘–) is of the form x + ๐‘–y Hence x = 0 and y = 2 Modulus of z = |z| |z| = โˆš(๐‘ฅ^2+๐‘ฆ2) = โˆš((0)2+(2)2) = โˆš(0+4) = โˆš4 = 2 Hence, Modulus of (1 + ๐‘–)/(1 โˆ’ ๐‘–) โˆ’ (1 โˆ’ ๐‘–)/(1 + ๐‘–) = 2

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