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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 – i Given 𝑧 = 1 – 𝑖 Let polar form be z = π‘Ÿ (cos⁑θ+𝑖 sin⁑θ ) From (1) and (2) 1 - 𝑖 = r (cos ΞΈ + 𝑖 sin ΞΈ) 1 – 𝑖 = r cos ΞΈ + 𝑖 r sin ΞΈ Comparing real part 1 = r cos ΞΈ Squaring both sides (1)2 =( π‘Ÿ cos⁑θ )^2 1 = π‘Ÿ2 cos2ΞΈ Adding (3) and (4) 1 + 1 = π‘Ÿ2 cos2 ΞΈ + π‘Ÿ2 sin2 ΞΈ 1 + 1 = r2 cos2 ΞΈ + r2 sin2 ΞΈ 2 = r2 ( cos2 ΞΈ + sin2 ΞΈ ) 2 = r2 Γ— 1 2 = r2 √2 = r r = √2 Now finding argument 1 – 𝑖 = r cos ΞΈ + 𝑖 r sin ΞΈ Comparing real part 1 = r cos ΞΈ Putting r =√2 1 = √2 cos ΞΈ 1/√2 = cos ΞΈ cos ΞΈ = 1/√2 Hence, cos ΞΈ = 1/√2 & sin ΞΈ = (βˆ’ 1)/√2 Hence, cos ΞΈ = 1/√2 & sin ΞΈ = (βˆ’ 1)/√2 Since, sin ΞΈ is negative and cos ΞΈ is positive, Hence, ΞΈ lies in IVth quadrant Argument = βˆ’ 45Β° = βˆ’ 45Β° Γ— πœ‹/(180Β°) = (βˆ’ πœ‹)/4 Hence, argument of 𝑧 = (βˆ’ πœ‹)/4 Hence π‘Ÿ = √2 and ΞΈ = ( βˆ’ πœ‹)/4 Polar form of 𝑧=π‘Ÿ (cos⁑θ+sin⁑θ ) = √2 (cos(( βˆ’ πœ‹)/4)+sin(( βˆ’ πœ‹)/4))

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