Test Index

Complex Numbers Part 1

Section: Mathematics

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Question : 66 of 100

Marks: +1, -0

Let A =

\{ 0 \in (-\pi/2 , \pi) : \frac{3+2i\sin\theta}{1-2i\sin\theta} \text{ is purely imaginary} \}

Then the sum of the elements in A is :

[9-Jan-2019 Shift 1]

$\frac{5\pi}{6}$
$\frac{2\pi}{3}$
$\frac{3\pi}{4}$
π

Solution:

Given z =

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

is purely img

so real part becomes zero

z =

\left(\frac{3+2i\sin\theta}{1-2i\sin\theta}\right)

×

\left(\frac{1-2i\sin\theta}{1+2i\sin\theta}\right)

z =

\frac{(3-4\sin^2\theta)+i(8\sin\theta)}{i+4\sin^2\theta}

Now Re (z) = 0

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

= 0

\sin^2

θ =

\frac{3}{4}

sin θ = ±

\frac{\sqrt{3}}{2}

⇒ θ =

-\frac{\pi}{3}, \frac{\pi}{3}, \frac{2\pi}{3}

Since ₹ ∊

\left(-\frac{\pi}{2}, \pi\right)

then sum of the elements in A is

-\frac{\pi}{3}+\frac{\pi}{3}+\frac{2\pi}{3}

=

\frac{2\pi}{3}

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