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Example 15 - (3 + 2i sin⁡)/(1 - 2isin) is purely real - Examples

  1. Chapter 5 Class 11 Complex Numbers
  2. Serial order wise
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Example, 15 Find real θ such that (3 + 2i sin⁡θ)/(1 − 2isin θ) is purely real Since (3 + 2i sin⁡θ)/(1 − 2isin θ) is purely real We need to first solve (3 + 2i sin⁡θ)/(1 − 2isin θ) and then take imaginary part as 0 (3 + 2i sin⁡θ)/(1 − 2isin θ) Rationalizing = (3 + 2i sin⁡θ)/(1 − 2isin θ) × (1 + 2isin θ)/(1 + 2isin θ) = ((3 + 2i sin⁡θ ) ( 1 + 2i sin⁡θ) )/(1 − 2i sin θ)(1 + 2i sin θ) = (3(1 + 2i sin⁡〖θ) + 2𝑖 sin⁡θ (1 + 2i sin θ") " 〗)/( 1 − 2i sin θ)(1 + 2i sin θ) = (3 + 6i sin⁡〖θ + 2𝑖 sin⁡θ + (2i sin θ)2" " 〗)/(1 − 2i sin θ)(1+ 2i sin θ) = (3 + 8i sin⁡〖θ + 4i2 sin2 θ" " 〗)/(1 − 2i sin θ)(1 + 2i sin θ) Using ( a – b ) ( a + b ) = a2 – b2 = (3 + 8i sin⁡〖θ + 4i2 sin2 θ" " 〗)/(12 −(2i sin θ)2) = (3 + 8i sin⁡〖θ + 4i2 sin2 θ" " 〗)/(1 − 4i2 sin2 θ) Putting i2 = − 1 = (3 + 8i sin⁡〖θ + 4(−1) sin2 θ" " 〗)/(1 − 4 (−1) sin2 θ) = (3 + 8i sin⁡〖θ − 4 sin2 θ" " 〗)/(1 + 4 sin2 θ) = (3 + 8i sin⁡〖θ − 4 sin2 θ" " 〗)/(1 + 4 sin2 θ) = (3 − 4 sin2 θ + 8i sin⁡〖θ 〗)/(1 + 4 sin2 θ) = (3 − 4 sin2 θ )/(1 + 4 sin2 θ) + 𝑖 ( 8 sin⁡〖θ 〗)/(1 + 4 sin2 θ) Hence, (3 + 2i sin⁡θ)/(1 − 2isin θ) = (3 − 4 sin2 θ )/(1 + 4 sin2 θ) + 𝑖 ( 8 sin⁡〖θ 〗)/(1 + 4 sin2 θ) Since (3 + 2i sin⁡θ)/(1 − 2isin θ) is purely real given Hence imaginary part of is equal to 0 i.e. ( 8 sin⁡〖θ 〗)/(1 + 4 sin2 θ) = 0 8 sin⁡θ= 0 ×(1 + 4 sin2⁡θ ) 8 sin θ = 0 sin θ = 0/8 sin⁡θ = 0 sin⁡θ = sin 0 Since sin θ = sin⁡𝑦 Then θ = n𝜋 ± y , where n ∈ Z Putting y = 0 θ = n𝜋 ± 0 θ = n𝜋 where n ∈ Z Hence for θ = n𝜋 ,where n ∈ Z (3 + 2𝑖 sin⁡𝜃)/(1 − 2𝑖 sin 𝜃) is purely real

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.
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