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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Transcript

Ex5.2, 3 Convert the given complex number in polar form: 1 – i Given 𝑧 = 1 – 𝑖 Let polar form be z = 𝑟 (cos⁡θ+𝑖 sin⁡θ ) From (1) and (2) 1 - 𝑖 = r (cos θ + 𝑖 sin θ) 1 – 𝑖 = r cos θ + 𝑖 r sin θ Comparing real part 1 = r cos θ Squaring both sides (1)2 =( 𝑟 cos⁡θ )^2 1 = 𝑟2 cos2θ Adding (3) and (4) 1 + 1 = 𝑟2 cos2 θ + 𝑟2 sin2 θ 1 + 1 = r2 cos2 θ + r2 sin2 θ 2 = r2 ( cos2 θ + sin2 θ ) 2 = r2 × 1 2 = r2 √2 = r r = √2 Now finding argument 1 – 𝑖 = r cos θ + 𝑖 r sin θ Comparing real part 1 = r cos θ Putting r =√2 1 = √2 cos θ 1/√2 = cos θ cos θ = 1/√2 Hence, cos θ = 1/√2 & sin θ = (− 1)/√2 Hence, cos θ = 1/√2 & sin θ = (− 1)/√2 Since, sin θ is negative and cos θ is positive, Hence, θ lies in IVth quadrant Argument = − 45° = − 45° × 𝜋/(180°) = (− 𝜋)/4 Hence, argument of 𝑧 = (− 𝜋)/4 Hence 𝑟 = √2 and θ = ( − 𝜋)/4 Polar form of 𝑧=𝑟 (cos⁡θ+sin⁡θ ) = √2 (cos(( − 𝜋)/4)+sin(( − 𝜋)/4))

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.