Ex 5.2, 4 - Convert in polar form: -1 + i - Complex number - Ex 5.2

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

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Question 4 Convert the given complex number in polar form: – 1 + i Given 𝑧 = βˆ’1+ 𝑖 Let polar form be 〖𝑧 = π‘Ÿ (cos〗⁑θ+𝑖 sin⁑θ) From (1) & (2) βˆ’ 1+ 𝑖 = r ( cos⁑θ + 𝑖 sin⁑θ) βˆ’ 1+ 𝑖 = rγ€– cos〗⁑θ + 𝑖 r sin⁑θ Adding ( 3 ) and ( 4 ) 1 + 1 = π‘Ÿ2 cos2 ΞΈ+ π‘Ÿ2 sin2ΞΈ 2 = π‘Ÿ2 ( cos2 ΞΈ+ sin2 ΞΈ) 2 = π‘Ÿ2 Γ— 1 2 = π‘Ÿ2 √2 = π‘Ÿ π‘Ÿ = √2 Finding argument βˆ’ 1+ 𝑖 = rγ€– cos〗⁑θ + 𝑖 r sin⁑θ Hence, sin ΞΈ = 1/√2 & cos ΞΈ = (βˆ’ 1)/√2 Hence, sin ΞΈ = 1/√2 & cos ΞΈ = (βˆ’ 1)/√2 Here, sin ΞΈ is positive and cos ΞΈ is negative, Hence, ΞΈ lies in IInd quadrant Argument = 180Β° – 45Β° = 135Β° = 135Β° Γ— πœ‹/180o = ( 3 πœ‹)/4 So argument of z = ( 3 πœ‹)/4 Hence π‘Ÿ = √2 and ΞΈ = 3πœ‹/4 Polar form of z = r (cos ΞΈ + sin ΞΈ) = √2 (cos (( 3 πœ‹)/4)+ 𝑖 sin(( 3 πœ‹)/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 14 years. He provides courses for Maths, Science and Computer Science at Teachoo