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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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Ex 5.2, 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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