Find the value of θ for which \(\rm z=\frac {3-2i\sinθ}{2+i\sinθ}\) is purely real.

Concept:

Let z = x + iy be a complex number, Where x is called the real part of a complex number or Re (z) and y is called the imaginary part of the complex number or Im (z)

Condition for purely real: Imaginary part equals zero.

Condition for purely imaginary: Real part equals zero.

Calculation:

\(\rm z=\frac {3-2i\sinθ}{2+i\sinθ}\)

Multiplying the numerator and denominator by 2 + i sin θ

\(\rm⇒ z=\frac{3-2isinθ}{2+isinθ}\times \frac{2-isinθ}{2-isinθ}\)

\(\rm =\frac{6-3isinθ-4isinθ+2i^2sin^2θ}{4-i^2sin^2θ}\)

\(\rm =\frac{6-7isinθ-2sin^2θ}{4+sin^2θ}\)       (∵ i² = -1)

\(\rm⇒ z=\frac{6+2sin^2θ}{4+sin^2θ}+\frac{-7isinθ}{4+sin^2θ}\)

The imaginary part of z 

\(\rm ⇒ Im(z)=\frac{-7\sin θ}{4+\sin^2θ}\)

For z to be purely real Im(z) = 0

\(\therefore \frac{-7\sinθ}{4+\sin^2θ}=0\)

⇒ sin θ = 0

So, θ = nπ, where n belongs to an integer