Let [A= (0,2 ): (1+2 i )/(1-i ).] is purely imaginary [].

Question

Let $$A=\left\{\theta \in(0,2 \pi): \frac{1+2 i \sin \theta}{1-i \sin \theta}\right.$$ is purely imaginary $$\}$$. Then the sum of the elements in $$\mathrm{A}$$ is :

Accepted Answer

$$
\begin{aligned}
& \text { Here, } z=\frac{1+2 i \sin \theta}{1-i \sin \theta} \times \frac{1+i \sin \theta}{1+i \sin \theta} \\
& \frac{1+i \sin \theta+2 i \sin \theta-2 \sin ^2 \theta}{1-i^2 \sin ^2 \theta} \\
& =\frac{\left(1-2 \sin ^2 \theta\right)+i(3 \sin \theta)}{1+\sin ^2 \theta}
\end{aligned}
$$

$\because z$ is purely imaginary, so $\operatorname{Re} z=0$

$$
\begin{aligned}
& \Rightarrow \frac{1-2 \sin ^2 \theta}{1+\sin ^2 \theta}=0 \\
& \Rightarrow 2 \sin ^2 \theta=1 \Rightarrow \sin ^2 \theta=\frac{1}{2} \\
& \Rightarrow \sin \theta= \pm \frac{1}{\sqrt{2}}
\end{aligned}
$$

$$
\begin{aligned}
& \therefore A=\left[\frac{\pi}{4}, \frac{3 \pi}{4}, \frac{5 \pi}{4}, \frac{7 \pi}{4}\right] \because \theta \in(0,2 \pi)\\
& \therefore \text { Sum }=\frac{\pi+3 \pi+5 \pi+7 \pi}{4}=\frac{16 \pi}{4}=4 \pi
\end{aligned}
$$

Solutions · 1