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Misc 12 - Let z1 = 2 - i, z2 = -2 + i. Find (i) Re (z1 z2) - Miscellaneous

  1. Chapter 5 Class 11 Complex Numbers
  2. Serial order wise
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Misc 12 Let z1 = 2 โ€“ i, z2 = -2 + i . Find Re ((๐‘ง_1 ๐‘ง_2)/(๐‘ง_1 )ย ฬ… ) We need to find Re ((๐‘ง_1 ๐‘ง_2)/(๐‘ง_1 )ย ฬ… ) i.e.Real part of ((๐‘ง_1 ๐‘ง_2)/(๐‘ง_1 )ย ฬ… ) Lets first calculate ((๐‘ง_1 ๐‘ง_2)/(๐‘ง_1 )ย ฬ… ) z1 = 2 โ€“ ๐‘– z2 = โˆ’2 + ๐‘– ("z1" )ย ฬ… = 2 + ๐‘– ((๐‘ง_1 ๐‘ง_2)/(๐‘ง_1 )ย ฬ… ) = ((2 โˆ’ ๐‘–) (โˆ’2 + ๐‘–))/(2 + ๐‘–) = (2(โˆ’2 + ๐‘–) โˆ’๐‘– (โˆ’2 + ๐‘–))/(2 + ๐‘–) = (2 ร— (โˆ’ 2) + 2 ร— ๐‘– + (โˆ’ ๐‘–) ร— (โˆ’ 2) + (โˆ’ ๐‘–) ร— ๐‘–)/(2 + ๐‘–) = (โˆ’4 + 2๐‘– + 2๐‘– โˆ’ ๐‘–2)/(2 + ๐‘–) Putting i2 = โˆ’ 1 = (โˆ’4 + 2๐‘– + 2๐‘– โˆ’ (โˆ’1))/(2 + ๐‘–) = (โˆ’4 + 2๐‘– + 2๐‘– + 1)/(2 + ๐‘–) = (โˆ’ 4 + 1 + 2๐‘– + 2๐‘– )/(2 + ๐‘–) = (โˆ’3 + 4๐‘–)/(2 + ๐‘–) Rationalizing = (โˆ’ 3 + 4๐‘–)/(2 + ๐‘–) ร— (2 โˆ’ ๐‘–)/(2 โˆ’ ๐‘–) = (( โˆ’3 + 4๐‘–) ( 2 โˆ’ ๐‘– ))/(( 2 + ๐‘– ) ( 2 โˆ’ ๐‘– )) = (โˆ’ 3 ( 2 โˆ’ ๐‘– ) + 4๐‘– ( 2 โˆ’ ๐‘– ))/(( 2 + ๐‘– ) ( 2 โˆ’ ๐‘– )) = (โˆ’ 3 ร— 2 + (โˆ’3) ร— (โˆ’๐‘–) + 4๐‘– ร— 2 + 4๐‘– ร— (โˆ’๐‘–))/(( 2 + ๐‘– ) ( 2 โˆ’ ๐‘– )) = (โˆ’ 6 + 3๐‘– + 8๐‘– โˆ’ 4๐‘–2)/(( 2 + ๐‘– ) ( 2 โˆ’ ๐‘– )) Using (a+b)(a-b) = a2 โ€“ b2 = (โˆ’ 6 + 3๐‘– + 8๐‘– โˆ’ 4๐‘–2)/(22 โˆ’ ๐‘–2) Putting i2 = - 1 = (โˆ’ 6 + 3๐‘– + 8๐‘– โˆ’ 4 ( โˆ’1))/(4 โˆ’ (โˆ’1) ) = (โˆ’ 6 + 3๐‘– + 8๐‘– + 4)/(4 + 1) = (โˆ’ 6 + 4 + 3๐‘– + 8๐‘– )/5 = (โˆ’ 2 +11๐‘–)/5 = (โˆ’ 2)/5 + ๐‘– 11/5 โ‡’ ((๐‘ง_1 ๐‘ง_2)/(๐‘ง_1 )ย ฬ… ) = (โˆ’ 2)/5 + ๐‘– 11/5 โ‡’ Re ((z1 z2)/("z1" )ย ฬ… ) = (โˆ’2)/5 Misc 12 Let ๐‘ง1 = 2 โ€“ ๐‘–, ๐‘ง2 = โˆ’ 2 + ๐‘– . Find (ii) Im (1/(๐‘ง_1 (๐‘ง_1 )ย ฬ… )) We need to find Im (1/(๐‘ง_1 (๐‘ง_1 )ย ฬ… )) i.e. imaginary part of (1/(๐‘ง_1 (๐‘ง_1 )ย ฬ… )) Lets first calculate (1/(๐‘ง_1 (๐‘ง_1 )ย ฬ… )) ๐‘ง1 = 2 โ€“ ๐‘– ("z1" )ย ฬ… = 2 + ๐‘– 1/(๐‘ง_1 (๐‘ง_1 )ย ฬ… ) = 1/(( 2 โˆ’ ๐‘– ) ( 2 + ๐‘– )) Using ( a + b ) ( a โ€“ b ) = a2 โ€“ b2 = 1/((2)2 โˆ’ (๐‘–)2) = 1/(4 โˆ’ ( โˆ’1) ) = 1/(4+1) = 1/5 = 1/5 + 0 = 1/5 + 0๐‘– So, imaginary part is 0 Hence, Im (1/(๐‘ง_1 (๐‘ง_1 )ย ฬ… )) = 0

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