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Ex 5.2

Ex 5.2, 1

Ex 5.2, 2 Important You are here

Ex 5.2, 3 Important Deleted for CBSE Board 2023 Exams

Ex 5.2, 4 Deleted for CBSE Board 2023 Exams

Ex 5.2, 5 Important Deleted for CBSE Board 2023 Exams

Ex 5.2, 6 Deleted for CBSE Board 2023 Exams

Ex 5.2, 7 Important Deleted for CBSE Board 2023 Exams

Ex 5.2, 8 Deleted for CBSE Board 2023 Exams

Last updated at Sept. 3, 2021 by Teachoo

Ex5.2, 2 Find the modulus and the argument of the complex number 𝑧 = − √3 + 𝑖 Method (1) To calculate modulus of z z = - √3 + 𝑖 Complex number z is of the form x + 𝑖y Where x = - √3 and y = 1 Modulus of z = |z| = √(𝑥^2+𝑦^2 ) = √(( − √3 )2+( 1 )2 ) = √(3+1) = √4 = 2 Hence |z| = 2 Modulus of z = 2 Method (2) to calculate Modulus of z Given z = − √3 + 𝑖 Let z = r (cosθ + 𝑖 sinθ) Here r is modulus, and θ is argument From (1) & (2) − √3 + 𝑖 = r (cosθ+𝑖 sinθ ) − √3 + 𝑖 = r〖 cos〗θ + 𝑖 r sinθ Comparing Real parts √3 = r cosθ Squaring both sides (√3)^2 = (𝑟 cosθ)2 3 = 𝑟2 〖 cos"2" 〗θ Adding (3) & (4) 3 + 1 = r2 cos2θ + r2 sin2θ 4 = 𝑟2 cos2θ + r2 sin2θ 4 = 𝑟2 ( cos2θ + sin2θ ) 4 = 𝑟2 × 1 4 = 𝑟2 √4 = 𝑟 r = 2 Hence, Modulus = 2 Finding argument − √3 + 𝑖 = r〖 cos〗θ + 𝑖 r sinθ Comparing real part −√3 = r cosθ Putting r = 2 −√3 = 2cosθ − √3/2 = cosθ cosθ = − √3/2 Hence, sinθ = 1/2 & cos θ = −√3/2 Since sin θ is positive and cos θ is negative , Argument will be in IInd quadrant Argument = 180° − 30° = 150° = 150 × 𝜋/180 = (5 𝜋)/3