Last updated at May 29, 2018 by Teachoo
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Example, 15 Find real θ such that (3 + 2i sinθ)/(1 − 2isin θ) is purely real Since (3 + 2i sinθ)/(1 − 2isin θ) is purely real We need to first solve (3 + 2i sinθ)/(1 − 2isin θ) and then take imaginary part as 0 (3 + 2i sinθ)/(1 − 2isin θ) Rationalizing = (3 + 2i sinθ)/(1 − 2isin θ) × (1 + 2isin θ)/(1 + 2isin θ) = ((3 + 2i sinθ ) ( 1 + 2i sinθ) )/(1 − 2i sin θ)(1 + 2i sin θ) = (3(1 + 2i sin〖θ) + 2𝑖 sinθ (1 + 2i sin θ") " 〗)/( 1 − 2i sin θ)(1 + 2i sin θ) = (3 + 6i sin〖θ + 2𝑖 sinθ + (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 sinθ)/(1 − 2isin θ) = (3 − 4 sin2 θ )/(1 + 4 sin2 θ) + 𝑖 ( 8 sin〖θ 〗)/(1 + 4 sin2 θ) Since (3 + 2i sinθ)/(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 sinθ= 0 ×(1 + 4 sin2θ ) 8 sin θ = 0 sin θ = 0/8 sinθ = 0 sinθ = sin 0 Since sin θ = sin𝑦 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 sin𝜃)/(1 − 2𝑖 sin 𝜃) is purely real
Examples
Example 2 (i)
Example 2 (ii) Important
Example 3
Example 4
Example 5 Important
Example 6 (i)
Example 6 (ii) Important
Example 7 Deleted for CBSE Board 2022 Exams
Example 8 Important Deleted for CBSE Board 2022 Exams
Example 9
Example 10
Example 11 Important
Example 12
Example 13 (i) Important Deleted for CBSE Board 2022 Exams
Example 13 (ii) Deleted for CBSE Board 2022 Exams
Example 14 Important
Example 15 You are here
Example 16 Important Deleted for CBSE Board 2022 Exams
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