Let S be the set of all (α, β), π<α, β<2 π, for which the complex number (1-i sin α)/(1+2 i sin α) is purely imaginary and (1+i cos β)/(1-2 i cos β) is purely real. Let Z_(α β)=sin 2 α+i cos 2 β,(α, β) ∈ S. Then Σ_((α, β) ∈ S)(i Z_(α β)+1/(i Z̄_(α β))) is equal to :

Solution

Maths sir · 4 steps

Rationalise both fractions, read off α and β, and the sum collapses to two small complex numbers.

1. Condition on alpha. Put s = sin α and multiply top and bottom by 1 - 2is. This gives ((1 - 2s²) - 3is)/(1 + 4s²). For it to be purely imaginary, the real part must vanish, so s² = 1/2. Since π < α < 2π, sin α = -1/√2, so α = 5π/4 or 7π/4.
2. Condition on beta. Put c = cos β and multiply top and bottom by 1 + 2ic. This gives ((1 - 2c²) + 3ic)/(1 + 4c²). For it to be purely real, 3c = 0, so cos β = 0 and β = 3π/2. Then cos 2β = cos 3π = -1.
3. Build each Z. For α = 5π/4, sin 2α = sin 5π/2 = 1. For α = 7π/4, sin 2α = sin 7π/2 = -1. So S has two pairs, which give Z = 1 - i and Z = -1 - i.

   Z	iZ	iZ̄	1/iZ̄	iZ + 1/iZ̄
   1 - i	1 + i	-1 + i	(-1 - i)/2	1/2 + i/2
   -1 - i	1 - i	-1 - i	(-1 + i)/2	1/2 - i/2

4. So. Adding the two terms, the imaginary parts cancel and 1/2 + 1/2 = 1. The answer is (c).

The idea to keep. For a fraction of complex numbers, multiply by the conjugate of the denominator first. Purely imaginary means the real part is 0, and purely real means the imaginary part is 0. Then use the interval to choose the sign.

Answer from the NTA final answer key. Solution by PrepWiser.