What is angle θ such that z is purely imaginary ?

where n is an integer

Concept:

If z = x + iy is purely imaginary, then x = 0

Calculation:

Given, 

z = \(\frac{1+\text{i}\sin θ}{1−\text{i}\sin θ}\) where i = \(\sqrt{−1}\)

Multiply numerator and denominator by (1 + i sin θ), 

\(⇒ z =\frac{1+\text{i}\sin θ}{1−\text{i}\sin θ} \times \frac{1+\text{i}\sin θ}{1+\text{i}\sin θ} \),

\(⇒ z =\frac{(1+\text{i}\sin θ)^2}{1^2−(\text{i}\sin θ)^2} \)

\(⇒ z =\frac{(1+2{i}\sin θ - \sin^2\theta )}{1+\sin^2 θ} \)

\(⇒ z =\frac{(1 - \sin^2\theta )}{1+\sin^2 θ} +{i}\frac{2\sin θ}{1+\sin^2 θ}\)

\(⇒ z = \frac{ \cos^2\theta }{1+\sin^2 θ} +{i}\frac{2\sin θ}{1+\sin^2 θ}\)

Since z is purely imaginary,

⇒ \(\frac{\cos^2 θ}{1+\sin^2 θ} = 0\)

⇒ cos²θ = 0

⇒ θ =  \(\frac{(2n+1)\pi}{2}\)

∴ The correct answer is option (2).