Ex 5.2, 1 - Find modulus and argument of z = -1 - i root 3 - Ex 5.2

 

  1. Chapter 5 Class 11 Complex Numbers
  2. Serial order wise
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Ex 5.2, 1 Find the modulus and the argument of the complex number z = -1 - i√3 Method (1) for modulus z = − 1 − 𝑖 √3 Complex number z is of the form x + 𝑖y Here x = − 1 , y = −√3 Modulus of z = |z| = √(𝑥^2+𝑦^2 ) = √("(−1)2 + (− " √3 ")2" ) = √(1+3) = √4 = 2 Hence |𝑧| = 2 Modulus = 2 Method (2) for modulus z = − 1 − 𝑖√3 Let z = r (cos⁡θ + 𝑖 sin⁡θ) Here r is modulus, and θ is argument From (1) & (2) − 1 − 𝑖 √3 = r ( cos⁡θ + 𝑖 sin⁡θ) − 1 − 𝑖 √3 = r〖 cos〗⁡θ + 𝑖 r sin⁡θ Comparing Real parts −1 = 𝑟 cos⁡θ Squaring both sides ( -1 )2 = ( r〖 cos〗⁡θ)2 1 = r2〖 cos"2" 〗⁡θ Adding (3) & (4) 1 + 3 = 𝑟2 cos2⁡θ + r2 sin2⁡θ 4 = 𝑟2 cos2⁡θ + r2 sin2⁡θ 4 = 𝑟2 ( cos2⁡θ + sin2⁡θ ) 4 = 𝑟2 × 1 4 = 𝑟2 𝑟2 = 4 𝑟 = √4 𝑟 = 2 Hence, modulus = r = 2 Finding argument − 1 − 𝑖 √3 = r〖 cos〗⁡θ + 𝑖 r sin⁡θ Comparing real parts −1 = 𝑟 cos⁡θ Putting r = 2 −1 = 2 cos⁡θ −1 = 2cos⁡θ −1/2 = cos⁡θ cos⁡θ= −1/2 Hence, sin θ = −√3/2 & cos θ = −1/2 Since both the values of sin θ and cos θ are negative , Argument will be in IIIrd quadrant Argument = – (180° −60°) = – 120 ° = – 120 × 𝜋/180 = – 2 × 𝜋/3 = – 2𝜋/3

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