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← JEE Main 2022 · 27 Jul · Shift 2
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JEE Main 2022 · 27 Jul · Shift 2, Mathematics question 2

Let S be the set of all (α,β)(\alpha,\beta), π<α,β<2π\pi<\alpha,\beta<2\pi, for which the complex number 1−isin⁡α1+2isin⁡α\dfrac{1-i\sin\alpha}{1+2i\sin\alpha} is purely imaginary and 1+icos⁡β1−2icos⁡β\dfrac{1+i\cos\beta}{1-2i\cos\beta} is purely real. Let Zαβ=sin⁡2α+icos⁡2βZ_{\alpha\beta}=\sin2\alpha+i\cos2\beta, (α,β)∈S(\alpha,\beta)\in S. Then ∑(α,β)∈S(iZαβ+1iZˉαβ)\displaystyle\sum_{(\alpha,\beta)\in S}\left(iZ_{\alpha\beta}+\frac1{i\bar Z_{\alpha\beta}}\right) is equal to :

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Answer and worked solution

Official answer: C

First condition. (1−isin⁡α)(1−2isin⁡α)(1-i\sin\alpha)(1-2i\sin\alpha) has real part 1−2sin⁡2α=01-2\sin^2\alpha=0, so sin⁡α=−12\sin\alpha=-\frac1{\sqrt2}: α=5π4,7π4\alpha=\frac{5\pi}4,\frac{7\pi}4.

Second. (1+icos⁡β)(1+2icos⁡β)(1+i\cos\beta)(1+2i\cos\beta) has imaginary part 3cos⁡β=03\cos\beta=0, so β=3π2\beta=\frac{3\pi}2.

Values. Z=1−iZ=1-i or −1−i-1-i. Then iZ+1iZˉiZ+\frac1{i\bar Z} is 12+i2\frac12+\frac i2 and 12−i2\frac12-\frac i2, which add to 1.

(C)

Topics:Algebra of complex numbersTrigonometric equations
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