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Ex 5.2, 2 - Find modulus, argument of z = - root 3 + i - Ex 5.2

Ex 5.2, 2 - Chapter 5 Class 11 Complex Numbers - Part 2
Ex 5.2, 2 - Chapter 5 Class 11 Complex Numbers - Part 3 Ex 5.2, 2 - Chapter 5 Class 11 Complex Numbers - Part 4 Ex 5.2, 2 - Chapter 5 Class 11 Complex Numbers - Part 5 Ex 5.2, 2 - Chapter 5 Class 11 Complex Numbers - Part 6

 

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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 (cos⁡θ + 𝑖 sin⁡θ) Here r is modulus, and θ is argument From (1) & (2) − √3 + 𝑖 = r (cos⁡θ+𝑖 sin⁡θ ) − √3 + 𝑖 = r〖 cos〗⁡θ + 𝑖 r sin⁡θ Comparing Real parts √3 = r cos⁡θ Squaring both sides (√3)^2 = (𝑟 cos⁡θ)2 3 = 𝑟2 〖 cos"2" 〗⁡θ Adding (3) & (4) 3 + 1 = r2 cos2⁡θ + r2 sin2⁡θ 4 = 𝑟2 cos2⁡θ + r2 sin2⁡θ 4 = 𝑟2 ( cos2⁡θ + sin2⁡θ ) 4 = 𝑟2 × 1 4 = 𝑟2 √4 = 𝑟 r = 2 Hence, Modulus = 2 Finding argument − √3 + 𝑖 = r〖 cos〗⁡θ + 𝑖 r sin⁡θ Comparing real part −√3 = r cos⁡θ Putting r = 2 −√3 = 2cos⁡θ − √3/2 = cos⁡θ cos⁡θ = − √3/2 Hence, sin⁡θ = 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

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.