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Misc 3 - Reduce (1/1 + 4i βˆ’ 2/1 + 𝑖) (3 βˆ’ 4i/5 + i) - Miscellaneous

Misc 3 - Chapter 5 Class 11 Complex Numbers - Part 2
Misc 3 - Chapter 5 Class 11 Complex Numbers - Part 3

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Misc 3 Reduce (1/(1 + 4 ) 2/(1 + )) ((3 4 )/(5 + )) to the standard form. We have to solve and then make it in Standard form a + ib (1/(1 + 4 ) 2/(1 + )) ((3 4 )/(5 + )) = (((1 + ) 2(1 4 ))/(1 4 )(1 + ) ) ((3 4 )/(5 + )) = ((1 + 2 + 8 )/(1 + 4 4 2)) ((3 4 )/(5 + )) = ((1 + 2 + 8 )/(1 + 4 4 ( 1))) ((3 4 )/(5 + )) = ((1 + 2 + 8 )/(1 + 4 + 4)) ((3 4 )/(5 + )) = (((1 2) + + 8 )/(1 +4 + 4 )) ((3 4 )/(5 + )) = (( 1 + 9 )/(5 3 )) ((3 4 )/(5 + )) = (( 1 + 9 ) (3 4 ))/(5 3 )( 5 + ) = ( 1(3 4 ) + 9 (3 4 ))/(5 ( 5 + ) 3 ( 5 + ) ) = ( 3 + 4 +27 36 2)/(25 + 5 15 3 2) = ( 3 + 4 +27 36 1)/(25 + 5 15 3 1) = ( 3 + 4 +27 + 36)/(25 + 5 15 + 3) = ( 3 + 36+ 4 +27 )/(25 + 3 + 5 15 ) = (33 +31 )/(28 10 ) = (33 +31 )/(2(14 5 ) ) = 1/2 (33 +31 )/((14 5 ) ) Rationalizing = 1/2 (33 + 31 )/((14 5 ) ) ((14 + 5i))/((14 + 5i)) = 1/2 (33(14 + 5i) + 31 (14 + 5i))/((14 5 )(14 + 5i)) = 1/2 (462 + 165 + 434 + 155 2)/(142 (5 )2) = 1/2 (462 + 165 + 434 + 155 2)/(196 25 2) " = " 1/2 (462 + 165 + 434 + 155( 1))/(196 25( 1)) " = " 1/2 (462 + 165 + 434 155)/(196 + 25) " = " 1/2 (462 155 + 165 + 434 )/221 " = " (307 + 599 )/2(221) " = " (307 + 599 )/442 = 307/442 + 599/442 This is the required standard form.

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.