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Misc 1 - Evaluate (i18 + (1/i)25)3 - Chapter 5 Class 11 - Power of i(odd and even)

  1. Chapter 5 Class 11 Complex Numbers
  2. Serial order wise
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Misc 1 Evaluate: (๐‘–^18+(1/i)^25 )^3 (i^18+(1/i)^25 )^3 = (๐‘–^18+ 1/(๐‘–)^25 )^3 = (๐‘–^18+ 1/((๐‘–) (๐‘–)^24 ))^3 = ((๐‘–^2 )^9+1/((๐‘–) (๐‘–^2 )^12 ))^3 = ((โˆ’1)^9+1/((โˆ’1)^12 ๐‘–))^3 = (โˆ’1+1/(1)๐‘–)^3 = (โˆ’1+1/๐‘–)^3 = (โˆ’1+1/๐‘–ร—๐‘–/๐‘–)^3 = (โˆ’1+๐‘–/๐‘–^2 )^3 = (โˆ’1+๐‘–/((โˆ’1)))^3 = (โ€“ 1 โ€“ ๐‘– )3 = (โˆ’1)3 (1 + ๐‘– )3 = (โˆ’1)(1 + ๐‘– )3 = โˆ’(1 + ๐‘– )3 Using (a + b) 3 =a3 + b3 + 3ab(a + b) = โˆ’(13 + ๐‘–3 + 3.1 .๐‘– (1 + ๐‘–)) = โˆ’(1 + ๐‘–2(๐‘–) + 3๐‘– + 3๐‘–2) = โˆ’(1 + (โˆ’1)๐‘– + 3๐‘– + 3(โˆ’1)) = โˆ’(1 โˆ’ ๐‘– + 3๐‘– โˆ’ 3) = โˆ’(1โˆ’ 3โˆ’ ๐‘– + 3๐‘–) = โˆ’[โˆ’2 + 2๐‘–] = 2 โ€“ 2๐‘–

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.
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