Let A=θ ∈ (0,2π ): 1+2i sinθ 1-i sinθ is purely imaginary Then the sum of the elements is in A is

Let

$$z=\frac{1+2i\mathrm{sin}\theta }{1-i\mathrm{sin}\theta }$$z=1−isinθ1+2isinθ​

$$\Rightarrow z=\frac{1+2i\mathrm{sin}\theta }{1-i\mathrm{sin}\theta }\times \frac{1+i\mathrm{sin}\theta }{1+i\mathrm{sin}\theta }$$⇒z=1−isinθ1+2isinθ​×1+isinθ1+isinθ​

$$\Rightarrow z=\frac{1-2{\mathrm{sin}}^{2}\theta +3i\mathrm{sin}\theta }{1+{\mathrm{sin}}^{2}\theta }$$⇒z=1+sin2θ1−2sin2θ+3isinθ​

$$\Rightarrow z=(\frac{1-2{\mathrm{sin}}^{2}\theta }{1+{\mathrm{sin}}^{2}\theta })+i(\frac{3\mathrm{sin}\theta }{1+{\mathrm{sin}}^{2}\theta })$$⇒z=(1+sin2θ1−2sin2θ​)+i(1+sin2θ3sinθ​)

Since, $z$z is a purely imaginary number, so real part must be zero, hence

$$\frac{1-2{\mathrm{sin}}^{2}\theta }{1+{\mathrm{sin}}^{2}\theta }=0$$1+sin2θ1−2sin2θ​=0

$$\Rightarrow 1-2{\mathrm{sin}}^{2}\theta =0$$⇒1−2sin2θ=0

$$\Rightarrow \mathrm{cos}2\theta =0$$⇒cos2θ=0

$\Rightarrow \theta =\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},\frac{7\pi }{4}$⇒θ=4π​,43π​,45π​,47π​, for $\theta \in (0,2\pi )$θ∈(0,2π)

$\therefore$∴Sum of all values $=\frac{16\pi }{4}=4\pi$=416π​=4π