Question

Let S be the set of (α,β),π<α,β<2π,for which the complex number
\(\frac{1-i\sinα}{1+2i\sinα}\) is purely imaginary and \(\frac{1+i\cosβ}{1-2i\cosβ}\) is purely real,
Let \(Zαβ = \sin2α+i\cos2β, (α,β) ∈ S\). Then
\(\sum_{(\alpha, \beta) \in S} \left(iZ_{\alpha\beta} + \frac{1}{iZ_{\alpha\beta}}\right)\)
is equal to

• A. 3
• B. 3i
• C. 1
• D. 2-i

Answer 1

The Correct Option is C

The correct answer is (C) :  1
\(∵ \frac{1-i\sinα}{1+2i\sinα}\)is purely imaginary
\(\therefore \frac{1-i\sinα}{1+2i\sinα} + \frac{1+\sinα}{1-2i\sinα} = 0\)
\(⇒ 1-2\sin^2α = 0\)
\(∴ α = \frac{5π}{4}, \frac{7π}{4}\)
and
\(\frac{1+i\cosβ}{1-2i\cosβ}\)
is purely real.
\(\frac{1-i\cosβ}{1-2i\cosβ} - \frac{1-i\cosβ}{1+2i\cosβ }= 0\)
\(⇒ cosβ = 0\)
\(∴ β = \frac{3π}{2}\)
\(∴ S = \left\{(\frac{5π}{4},\frac{3π}{2}),(\frac{7π}{4},\frac{3π}{2})\right\}\)
\(Z_{αβ}\) = 1-i and \(Z_{αβ}\) = -1-i
\(\therefore \sum_{(\alpha, \beta) \in S} (iZ_{\alpha\beta} + \frac{1}{i\overline{Z}_{\alpha\beta}})\)
\(= i(-2i) + \frac{1}{i}\left[\frac{1}{1 + i} + \frac{1}{-1 + i}\right]\)
\(= 2+\frac{1}{i}\frac{2i}{-2}\)
= 1