{z}={3+2 {i} θ}{1-2 {i} θ},({i}={-1}) will be purely imaginary if θ=

Correct Answer is : nπ±π3n π ± π/3nπ±3π, where n∈Zn {Z}n∈Z

Correct Answer is : nπ±π3n π ± π/3nπ±3π, where n∈Zn {Z}n∈Z

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MHTCET 22 Apr 2025 Shift 2 Paper

Question : 115 of 150

Marks: +1, -0

\mathrm{z}=\frac{3+2 \mathrm{i} \sin \theta}{1-2 \mathrm{i} \sin \theta},(\mathrm{i}=\sqrt{-1})

\theta=

Solution:

Concept:

A complex number is purely imaginary when its real part is equal to zero.

Explanation:

Given

z=\frac{3+2\mathrm{i}\sin\theta}{1-2\mathrm{i}\sin\theta}

Multiply the numerator and denominator by the conjugate

1+2\mathrm{i}\sin\theta

z=\frac{(3+2\mathrm{i}\sin\theta)(1+2\mathrm{i}\sin\theta)}{(1-2\mathrm{i}\sin\theta)(1+2\mathrm{i}\sin\theta)}

Simplify the numerator:

(3+2\mathrm{i}\sin\theta)(1+2\mathrm{i}\sin\theta)=3-4\sin^2\theta+8\mathrm{i}\sin\theta

Simplify the denominator:

(1-2\mathrm{i}\sin\theta)(1+2\mathrm{i}\sin\theta)=1+4\sin^2\theta

Therefore,

z=\frac{3-4\sin^2\theta}{1+4\sin^2\theta}+\mathrm{i}\frac{8\sin\theta}{1+4\sin^2\theta}

For

z

to be purely imaginary, the real part must be zero:

\frac{3-4\sin^2\theta}{1+4\sin^2\theta}=0

So,

3-4\sin^2\theta=0

, giving

\sin^2\theta=\frac{3}{4}

Since

\frac{3}{4}=\sin^2\frac{\pi}{3}

\sin^2\theta=\sin^2\frac{\pi}{3}

Hence,

\theta=n\pi\pm\frac{\pi}{3}

, where

n\in\mathbb{Z}

Answer:

Option C:

n\pi\pm\frac{\pi}{3}

, where

n\in\mathbb{Z}

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