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Complex Numbers question

2023 · 8 Apr · Shift 2 · Q22

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Complex Numbers
2023 · 8 Apr · Shift 2 · Q22

Complex Numbers question

2023 · 8 Apr · Shift 2 · Q22

JEE MainMathematicsComplex NumbersMCQ+4 / −1

Let

A=\left\{\theta \in(0,2 \pi): \frac{1+2 i \sin \theta}{1-i \sin \theta}\right.

is purely imaginary

\}

. Then the sum of the elements in

\mathrm{A}

is :

A
$3 \pi$
B
$\pi$
C
$2 \pi$
D
$4 \pi$

View written solutionFree

Correct answer: D

Let $z=\frac{1+2i\sin\theta}{1-i\sin\theta}.$ We need $z$ to be purely imaginary, so its real part must be zero.
Rationalize the denominator: $z=\frac{1+2i\sin\theta}{1-i\sin\theta}\cdot\frac{1+i\sin\theta}{1+i\sin\theta}.$ Thus, $z=\frac{(1+2i\sin\theta)(1+i\sin\theta)}{1+\sin^2\theta}.$
Expand the numerator:

=1+i\sin\theta+2i\sin\theta+2i^2\sin^2\theta.$$ Since $i^2=-1$, $$=1+3i\sin\theta-2\sin^2\theta.$$ So, $$z=\frac{1-2\sin^2\theta}{1+\sin^2\theta}+i\frac{3\sin\theta}{1+\sin^2\theta}.$$ 4. For $z$ to be purely imaginary, $$\Re(z)=\frac{1-2\sin^2\theta}{1+\sin^2\theta}=0.$$ Since $1+\sin^2\theta>0$, we get $$1-2\sin^2\theta=0$$ $$\Rightarrow \sin^2\theta=\frac12.$$ Hence, $$\sin\theta=\pm\frac{1}{\sqrt2}.$$ 5. In the interval $\theta\in(0,2\pi)$, the solutions are $$\theta=\frac{\pi}{4},\;\frac{3\pi}{4},\;\frac{5\pi}{4},\;\frac{7\pi}{4}.$$ So, $$A=\left\{\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}\right\}.$$ 6. Sum of the elements: $$\frac{\pi}{4}+\frac{3\pi}{4}+\frac{5\pi}{4}+\frac{7\pi}{4} =\frac{16\pi}{4}=4\pi.$$ 7. Therefore, the correct option is $$\boxed{D: 4\pi}. $$

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