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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

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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 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.