Misc 1 - Evaluate (i18 + (1/i)25)3 - Chapter 5 Class 11 - Miscellaneou - Miscellaneous

part 2 - Misc 1 - Miscellaneous - Serial order wise - Chapter 4 Class 11 Complex Numbers
part 3 - Misc 1 - Miscellaneous - Serial order wise - Chapter 4 Class 11 Complex Numbers part 4 - Misc 1 - Miscellaneous - Serial order wise - Chapter 4 Class 11 Complex Numbers

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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

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