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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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Transcript

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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.