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Transcript

Misc 1 Evaluate: (๐‘–^18+(1/i)^25 )^3 (๐‘–^18+(1/๐‘–)^25 )^3 = (๐‘–^18+ 1/(๐’Š)^๐Ÿ๐Ÿ“ )^3 = (๐‘–^18+ 1/(๐’Š ร— ๐’Š^๐Ÿ๐Ÿ’ ))^3 = ((๐’Š^๐Ÿ )^๐Ÿ—+1/(๐‘– ร— (๐’Š^๐Ÿ )^๐Ÿ๐Ÿ ))^3 Putting i2 = โˆ’๐Ÿ = ((โˆ’๐Ÿ)^๐Ÿ—+1/ใ€–๐‘– ร— (โˆ’๐Ÿ)ใ€—^12 )^3 = (โˆ’๐Ÿ+1/(๐‘– ร— ๐Ÿ))^3 = (โˆ’1+1/๐‘–)^3 Removing ๐’Š from the denominator = (โˆ’1+1/๐‘–ร—๐’Š/๐’Š)^3 = (โˆ’1+๐‘–/๐’Š^๐Ÿ )^3 = (โˆ’1+๐‘–/((โˆ’๐Ÿ)))^3 = (โˆ’๐Ÿ โ€“ ๐’Š )๐Ÿ‘ = (โˆ’1(1+ ๐‘– ))3 = (โˆ’๐Ÿ)๐Ÿ‘ (๐Ÿ + ๐’Š )๐Ÿ‘ = (โˆ’1)(1 + ๐‘– )3 = โˆ’(๐Ÿ + ๐’Š )๐Ÿ‘ Using (a + b) 3 = a3 + b3 + 3ab(a + b) = โˆ’(13 + ๐‘–3 + 3 ร— 1 ร— ๐‘– (1 + ๐‘–)) = โˆ’(1 + ๐’Š๐Ÿ‘ +3๐‘– (1 + ๐‘–)) = โˆ’(1 + ๐’Š๐Ÿ ร— ๐’Š +3๐‘– (1 + ๐‘–)) Putting i2 = โˆ’๐Ÿ = โˆ’(1 +(โˆ’๐Ÿ) ร— ๐‘– +3๐‘– (1 + ๐‘–)) = โˆ’(1 โˆ’๐‘– +3๐‘– (1 + ๐‘–)) = โˆ’(1 โˆ’๐‘– +3๐‘–+3๐‘– ร— ๐‘–) = โˆ’(1+2๐‘–+3๐’Š^๐Ÿ ) Putting i2 = โˆ’๐Ÿ = โˆ’(1+2๐‘–+3 ร— โˆ’๐Ÿ) = โˆ’(1+2๐‘–โˆ’3) = โˆ’(2๐‘–โˆ’2) = โˆ’2๐‘–+2 = ๐Ÿโˆ’๐Ÿ๐’Š

  1. Chapter 4 Class 11 Complex Numbers
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About the Author

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo