Question

Find real $\theta$ Such that $\sf\dfrac{3+2iSin\theta}{1-2iSin\theta}$ is
A) real B) Purely imaginary

Answer 1

Topic :-

Complex Number

Given :-

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

To Find :-

Real $\theta$ such that given fraction is :-

• Real
• Purely Imaginary

Concept Used :-

A complex number, z = a + ib is real when b = 0 and purely imaginary if a = 0.

General solution of trigonometric equation :-

$\mathtt{If\:\sin\theta=\sin \alpha, \:then\:\theta=n\pi +(-1)^n\alpha,\:n\:\epsilon\:Integer}$

$\mathtt{If\:\sin^2\theta=\sin^2 \alpha, \:then\:\theta=n\pi \pm\alpha,\:n\:\epsilon\:Integer}$

Solution :-

For Real number

For number to be real, coefficient of 'i' should be Zero.

$\mathtt{2\sin\theta=0}$

$\mathtt{\sin\theta=\sin0}$

$\mathtt{\theta=n\pi +(-1)^n(0)}$

$\mathtt{\theta=n\pi}$

For purely imaginary number

For number to be purely imaginary, real part should vanish.

$\mathtt{\dfrac{3+2i\sin \theta}{1-2i\sin\theta}\times \dfrac{1+2i\sin \theta}{1+2i\sin\theta} }$

$\mathtt{\dfrac{3+6i\sin\theta+2i\sin\theta+4i^2\sin^2\theta}{1-4i^2\sin^2\theta}}$

As we know, i² = -1,

$\mathtt{\dfrac{(3-4\sin^2\theta)+i(8\sin\theta)}{1+4\sin^2\theta}}$

Real part should vanish. So,

$\mathtt{\dfrac{3-4\sin^2\theta}{1+4\sin^2\theta}=0}$

$\mathtt{3-4\sin^2\theta=0}$ as $\theta$ is real.

$\mathtt{\sin^2\theta=\dfrac{3}{4}=\left (\dfrac{\sqrt3}{2} \right)^2}$

$\mathtt{\sin^2\theta=\left (\sin\dfrac{\pi}{3} \right)^2}$

$\mathtt{\theta=n\pi \pm\dfrac{\pi}{3}}$

Answer :-

So, values of $\theta$ for :-

$\bullet\:\:\mathtt{Real\:is\:\:\theta=n\pi}$

$\bullet\:\: \mathtt{Purely\:Imaginary\:is\:\:\theta=n\pi \pm\dfrac{\pi}{3}}$