Question

If $z=\frac{1+2i\cos \theta }{4+3i\sin \theta }$, where $\theta \in (0,\frac{\pi }{2})$, then which of the following is/are INCORRECT?

• A. there exist one value of $\theta$ for which $Re\left(z\right)=0$
• B. there exist $2$ values of $\theta$ for which $Re\left(z\right)=0$
• C. there exist one value of $\theta$ for which $Im\left(z\right)=0$
• D. there exist $2$ values of $\theta$ for which $Im\left(z\right)=0$

Answer 1

Answer: (A), (B), (D)

there exist $2$ values of $\theta$ for which $Im\left(z\right)=0$

Given : $z=\frac{1+2i\cos \theta }{4+3i\sin \theta }$
Now,
$z=\frac{1+2i\cos \theta }{4+3i\sin \theta }\times \frac{4-3i\sin \theta }{4-3i\sin \theta }\Rightarrow z=\frac{(4+6\sin \theta \cos \theta )+i(8\cos \theta -3\sin \theta )}{16+9{\sin }^2\theta }$

When $Re\left(z\right)=0$, then
$\frac{4+6\sin \theta \cos \theta }{16+9{\sin }^2\theta }=0\Rightarrow 4+3\sin 2\theta =0\Rightarrow \sin 2\theta =-\frac{4}{3}$
Therefore $Re\left(z\right)=0$ not possible for any $\theta \in (0,\frac{\pi }{2})$

When $Im\left(z\right)=0$, then
$\frac{8\cos \theta -3\sin \theta }{16+9{\sin }^2\theta }=0\Rightarrow 8\cos \theta -3\sin \theta =0\Rightarrow \tan \theta =\frac{8}{3}$
Therefore there exist $1$ value of $\theta \in (0,\frac{\pi }{2})$