Ex 5.2, 7 - Convert root 3 + i in polar form - Chapter 5 CBSE - Ex 5.2

  1. Chapter 5 Class 11 Complex Numbers
  2. Serial order wise
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Ex 5.2, 7 Convert the given complex number in polar form: √3 + i. Given z = √3 + i Let polar form be z = r (cos⁡θ + i sin⁡θ) From (1) & (2) √3 + i = r (cos⁡θ + i sin⁡θ) √3 + 𝑖 = r〖 cos〗⁡θ + 𝑖 r sin⁡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 Hence, sin⁡θ = 1/2 & cos θ = √3/2 Since sin θ and cos θ both are positive, Argument will be in Ist quadrant Argument = 30° = 30 × 𝜋/180 = 𝜋/6 Hence θ = 𝜋/6 and r = 2 Polar form of z = r ( cos θ + 𝑖 sin θ ) = 2 ( cos 𝜋/6 + 𝑖 sin 𝜋/6)

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