Question

If $z=(4{\sin }^2\theta -1)+i({\cos }^2\theta +1)$ is purely imaginary number, then the number of value(s) of $\theta \in [0,2n\pi ]$ where $n\in I$ is

• A. $n$
• B. $2n$
• C. $4n$
• D. $0$

Answer 1

The correct option is C $4n$

Given : $z=(4{\sin }^2\theta -1)+i({\cos }^2\theta +1)$ is purely imaginary.
$\Rightarrow Re\left(z\right)=0\Rightarrow 4{\sin }^2\theta -1=0\Rightarrow \sin \theta =\pm \frac{1}{2}\Rightarrow \theta =\frac{\pi }{6},\frac{5\pi }{6},\frac{7\pi }{6},\frac{11\pi }{6}$
When $\theta \in [0,2\pi ]$
Four values of $\theta$ in $[0,2\pi ]$
$\therefore$ Total number of values in $[0,2n\pi ]$ is $4n.$