Let A=(-2,):3+2i1-2i is purely imaginary.

Question

Let $A=\left\{	heta\in\left(-\dfrac\pi2,\piight):\dfrac{3+2i\sin	heta}{1-2i\sin	heta}	ext{ is purely imaginary}ight\}$. Then the sum of the elements in $A$ is:

Accepted Answer

Given complex number,

$${{3 + 2i\sin \theta } \over {1 - 2i\sin \theta }}$$

$$ = {{\left( {3 + 2i\sin \theta } \right)\left( {1 + 2i\sin \theta } \right)} \over {1 + 4{{\sin }^2}\theta }}$$

$$ = {{3 + 6i\sin \theta  + 2i\sin \theta  - 4{{\sin }^2}\theta } \over {1 + 4{{\sin }^2}\theta }}$$

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

As complex number is purely imaginary, So real part of this complex number is zero.

$$ \therefore $$   $${{3 - 4{{\sin }^2}\theta } \over {1 + 4{{\sin }^2}\theta }}$$  = 0

$$ \Rightarrow $$   $$3 - 4{\sin ^2}\theta  = 0$$

$$ \Rightarrow $$   $$\sin \theta  =  \pm {{\sqrt 3 } \over 2}$$

as   $$\theta $$  $$ \in $$ $$\left( { - {\pi  \over 2},\pi } \right)$$

$$ \therefore $$   $$\theta $$ $=$ - $${\pi  \over 3},{\pi  \over 3},{{2\pi } \over 3}$$

$$ \therefore $$   Sum of those values of A is

$$ =  - {\pi  \over 3} + {\pi  \over 3} + {{2\pi } \over 3}$$

$$ = {{2\pi } \over 3}$$

Solutions · 1