If $\frac{3+2 i \sin \theta}{1-2 i \sin \theta}$ is a real number and $0 < \theta < 2 \pi$, then $\theta$ is…
If $\frac{3+2 i \sin \theta}{1-2 i \sin \theta}$ is a real number and $0 < \theta < 2 \pi$, then $\theta$ is equal to
- $\pi$
- $\frac{\pi}{6}$
- $\frac{\pi}{3}$
- $\frac{\pi}{2}$
Solution
We have
$
\begin{aligned}
\frac{(3+2 i \sin \theta)}{(1-2 i \sin \theta)} & =\frac{(3+2 i \sin \theta)(1+2 i \sin \theta)}{(1-2 i \sin \theta)(1-2 i \sin \theta)} \\
& =\frac{3-4 \sin ^2 \theta+8 i \sin \theta}{1+4 \sin ^2 \theta} \\
& =\frac{\left(3-4 \sin ^2 \theta\right)}{\left(1+4 \sin ^2 \theta\right)}+i\left(\frac{8 \sin \theta}{1+4 \sin ^2 \theta}\right)
\end{aligned}
$
For a real number,
$\begin{gathered}\frac{8 \sin \theta}{1+4 \sin ^2 \theta}=0 \\ \Rightarrow \quad \sin \theta=0 \Rightarrow \theta=\pi\end{gathered}$
Asked in: AP EAMCET 2002
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