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

Ā 

Remove Ads

Transcript

Misc 1 Evaluate: (š‘–^18+(1/i)^25 )^3 (š‘–^18+(1/š‘–)^25 )^3 = (š‘–^18+ 1/(š’Š)^šŸšŸ“ )^3 = (š‘–^18+ 1/(š’Š Ɨ š’Š^šŸšŸ’ ))^3 = ((š’Š^šŸ )^šŸ—+1/(š‘– Ɨ (š’Š^šŸ )^šŸšŸ ))^3 Putting i2 = āˆ’šŸ = ((āˆ’šŸ)^šŸ—+1/ć€–š‘– Ɨ (āˆ’šŸ)怗^12 )^3 = (āˆ’šŸ+1/(š‘– Ɨ šŸ))^3 = (āˆ’1+1/š‘–)^3 Removing š’Š from the denominator = (āˆ’1+1/š‘–Ć—š’Š/š’Š)^3 = (āˆ’1+š‘–/š’Š^šŸ )^3 = (āˆ’1+š‘–/((āˆ’šŸ)))^3 = (āˆ’šŸ – š’Š )šŸ‘ = (āˆ’1(1+ š‘– ))3 = (āˆ’šŸ)šŸ‘ (šŸ + š’Š )šŸ‘ = (āˆ’1)(1 + š‘– )3 = āˆ’(šŸ + š’Š )šŸ‘ Using (a + b) 3 = a3 + b3 + 3ab(a + b) = āˆ’(13 + š‘–3 + 3 Ɨ 1 Ɨ š‘– (1 + š‘–)) = āˆ’(1 + š’ŠšŸ‘ +3š‘– (1 + š‘–)) = āˆ’(1 + š’ŠšŸ Ɨ š’Š +3š‘– (1 + š‘–)) Putting i2 = āˆ’šŸ = āˆ’(1 +(āˆ’šŸ) Ɨ š‘– +3š‘– (1 + š‘–)) = āˆ’(1 āˆ’š‘– +3š‘– (1 + š‘–)) = āˆ’(1 āˆ’š‘– +3š‘–+3š‘– Ɨ š‘–) = āˆ’(1+2š‘–+3š’Š^šŸ ) Putting i2 = āˆ’šŸ = āˆ’(1+2š‘–+3 Ɨ āˆ’šŸ) = āˆ’(1+2š‘–āˆ’3) = āˆ’(2š‘–āˆ’2) = āˆ’2š‘–+2 = šŸāˆ’šŸš’Š

Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh is an IIT Kanpur graduate and has been teaching for 16+ years. At Teachoo, he breaks down Maths, Science and Computer Science into simple steps so students understand concepts deeply and score with confidence.

Many students prefer Teachoo Black for a smooth, ad-free learning experience.