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

Question

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

Accepted Answer

$\begin{gathered} z=\frac{1+2i\sin 	heta }{1-i\sin 	heta }	imes \frac{1+i\sin 	heta }{1+i\sin 	heta } \ \frac{1+i\sin 	heta +2i\sin 	heta -2{\sin }^{2}	heta }{1-{i}^{2}{\sin }^{2}	heta } \ =\frac{(1-2{\sin }^{2}	heta )+i(3\sin 	heta )}{1+{\sin }^{2}	heta } \end{gathered}$

$\because z$ is purely imaginary, so $Rez=0$

$\begin{gathered} \Rightarrow \frac{1-2{\sin }^{2}	heta }{1+{\sin }^{2}	heta }=0 \ \Rightarrow 2{\sin }^{2}	heta =1\Rightarrow {\sin }^{2}	heta =\frac{1}{2} \ \Rightarrow \sin 	heta =\pm \frac{1}{\sqrt{2}} \end{gathered}$

$\begin{gathered} 	herefore A=[\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},\frac{7\pi }{4}]\because 	heta \in (0,2\pi ) \ 	herefore =\frac{\pi +3\pi +5\pi +7\pi }{4}=\frac{16\pi }{4}=4\pi \end{gathered}$

Solutions · 1