Let A = \left\{ \theta \in \left (-\frac{\pi}{2}, \pi \right ): \frac{3+2 i\sin\theta}{1-2i\sin\theta} \textup{\;is\;purely\;imaginary}\right \} . Then the sum of the elements in A is:

  • Option 1)

    \frac{5\pi}{6}

  • Option 2)

    \pi

  • Option 3)

    \frac{3\pi}{4}

  • Option 4)

    \frac{2\pi}{3}

Answers (1)
A admin

 

Purely Imaginary Complex Number -

z=x+iy, \boldsymbol{x=0}, y\epsilon R

& i2=-1

- wherein

Real part of z = Re (z) = x & Imaginary part of z = Im (z) = y

from the concept if z is purely img, then real part of z is = 0

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

factorize Z

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

Z=\left ( \frac{3-4\sin ^{2}\theta }{1+4\sin ^{2}\theta } \right )+i\left ( \frac{8\sin \theta }{1+4\sin ^{2}\theta } \right )

Re\left ( z \right )=0

So,

\frac{3-4\sin ^{2}\left ( \theta \right )}{1+4\sin ^{2}\left ( \theta \right )}=0\Rightarrow \sin ^{2}\theta =\frac{3}{4}

\Rightarrow \sin \left ( \theta \right )=\pm \frac{\sqrt{3}}{2}\Rightarrow \theta =-\frac{\pi }{3},\frac{\pi }{3},\frac{2\pi }{3}

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

The sum of the element in A is 

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

 


Option 1)

\frac{5\pi}{6}

Option 2)

\pi

Option 3)

\frac{3\pi}{4}

Option 4)

\frac{2\pi}{3}

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