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

Example 15 - Chapter 5 Class 11 Complex Numbers - Part 2
Example 15 - Chapter 5 Class 11 Complex Numbers - Part 3
Example 15 - Chapter 5 Class 11 Complex Numbers - Part 4

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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

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.