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Ex 5.2, 6 - Convert in polar form: -3 | Chapter 5 Class 11 - Polar representation

Ex 5.2, 6 - Chapter 5 Class 11 Complex Numbers - Part 2
Ex 5.2, 6 - Chapter 5 Class 11 Complex Numbers - Part 3 Ex 5.2, 6 - Chapter 5 Class 11 Complex Numbers - Part 4

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Ex 5.2, 6 Convert the given complex number in polar form: –3 Given z = – 3 z = – 3 + 0i Let polar form be z = r (cos⁡θ + i sin⁡θ) From (1) & (2) − 3+ 0i = r ( cos⁡θ + i sin⁡θ) − 3+ 0𝑖 = r〖 cos〗⁡θ + 𝑖 r sin⁡θ Adding (3) and (4) 9 + 0 = r2 cos2 θ + r2 sin2 θ 9 = r2 (cos2 θ + sin2 θ) 9 = r2 × 1 9 = r2 √9 = r 3 = r r = 3 Finding argument − 3+ 0𝑖 = r〖 cos〗⁡θ + 𝑖 r sin⁡θ Comparing real part –3 = r cos⁡θ Putting r = 3 – 3 = (3)cos θ (− 3)/3 = cos θ – 1 = cos θ cos θ = – 1 Hence, sin⁡θ = 0 & cos θ = –1 Hence, sin⁡θ = 0 & cos θ = –1 Since sin θ is zero And cos θ is negative θ lies in the llnd Quadrant Argument = 180° - 0° = 180° = 180° × 180"°" /180"°" = 180° × 𝜋/(180°) = π Hence r = 3 and θ = π Polar form of z = r (cos θ + sin θ) = 3 ( cos π + i sin π )

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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.