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Ex 5.2, 3 - Convert in polar form: 1 - i - Chapter 5 Class 11 - Polar representation

Ex 5.2, 3 - Chapter 5 Class 11 Complex Numbers - Part 2
Ex 5.2, 3 - Chapter 5 Class 11 Complex Numbers - Part 3
Ex 5.2, 3 - Chapter 5 Class 11 Complex Numbers - Part 4
Ex 5.2, 3 - Chapter 5 Class 11 Complex Numbers - Part 5


Transcript

Ex5.2, 3 Convert the given complex number in polar form: 1 โ€“ i Given ๐‘ง = 1 โ€“ ๐‘– Let polar form be z = ๐‘Ÿ (cosโกฮธ+๐‘– sinโกฮธ ) From (1) and (2) 1 - ๐‘– = r (cos ฮธ + ๐‘– sin ฮธ) 1 โ€“ ๐‘– = r cos ฮธ + ๐‘– r sin ฮธ Comparing real part 1 = r cos ฮธ Squaring both sides (1)2 =( ๐‘Ÿ cosโกฮธ )^2 1 = ๐‘Ÿ2 cos2ฮธ Adding (3) and (4) 1 + 1 = ๐‘Ÿ2 cos2 ฮธ + ๐‘Ÿ2 sin2 ฮธ 1 + 1 = r2 cos2 ฮธ + r2 sin2 ฮธ 2 = r2 ( cos2 ฮธ + sin2 ฮธ ) 2 = r2 ร— 1 2 = r2 โˆš2 = r r = โˆš2 Now finding argument 1 โ€“ ๐‘– = r cos ฮธ + ๐‘– r sin ฮธ Comparing real part 1 = r cos ฮธ Putting r =โˆš2 1 = โˆš2 cos ฮธ 1/โˆš2 = cos ฮธ cos ฮธ = 1/โˆš2 Hence, cos ฮธ = 1/โˆš2 & sin ฮธ = (โˆ’ 1)/โˆš2 Hence, cos ฮธ = 1/โˆš2 & sin ฮธ = (โˆ’ 1)/โˆš2 Since, sin ฮธ is negative and cos ฮธ is positive, Hence, ฮธ lies in IVth quadrant Argument = โˆ’ 45ยฐ = โˆ’ 45ยฐ ร— ๐œ‹/(180ยฐ) = (โˆ’ ๐œ‹)/4 Hence, argument of ๐‘ง = (โˆ’ ๐œ‹)/4 Hence ๐‘Ÿ = โˆš2 and ฮธ = ( โˆ’ ๐œ‹)/4 Polar form of ๐‘ง=๐‘Ÿ (cosโกฮธ+sinโกฮธ ) = โˆš2 (cos(( โˆ’ ๐œ‹)/4)+sin(( โˆ’ ๐œ‹)/4))

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.