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
Made by
Davneet Singh
Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo
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