Question

if 3+2isinalpha/1-2isinalpha is a purely a real number then find the real of alpha ​

Answer 1

$\large\rm{ z = \dfrac{3 + 2i \ \sin \alpha}{1-2i \ \sin \alpha}}$

$\large\rm{ \ }$

$\large\rm{ = \dfrac{(3+2i \ \sin \alpha)(1+2i \ \sin \alpha)}{1+4 \ \sin^{2} \alpha}}$

$\large\rm{ \ }$

$\large\rm{ = \dfrac{3+6i \ \sin \alpha + 2i \ \sin \alpha -4 \ sin^{2} \alpha}{ 1 + 4 \ \sin^{2} \alpha}}$

$\large\rm{ \ }$

$\large\rm{ = \dfrac{ 3-4 \ \sin^{2} \alpha + 8i \ \sin \alpha}{ 1 + 4 \ \sin^{2} \alpha}}$

$\large\rm{ \ }$

z is purely imaginary when $\small\rm{ (3-4 \sin^{2} \alpha) = 0}$

$\large\rm { \ }$

$\large\rm { \implies \sin \alpha = \pm \dfrac{\sqrt{3}}{2}}$

$\large\rm{ \ }$

$\large\boxed{\rm{ \therefore \alpha = n \pi + \frac{\pi}{3}}}$