Mathematics - Complex Number Question with Solution | TestHub

Let $z=\cos \theta +i\sin \theta$ $$\begin{array}{rl}\frac{z}{1-z^2}& =\frac{\cos \theta +i\sin \theta }{1-(\cos 2\theta +i\sin 2\theta )}\\ & =\frac{\cos \theta +i\sin \theta }{2{\sin }^2\theta -2i\sin \theta \cos \theta }\\ & =\frac{\cos \theta +i\sin \theta }{-2i\sin \theta (\cos \theta +i\sin \theta )}\\ & =\frac{i}{2\sin \theta }\end{array}$$ Hence, $\frac{z}{1-z^2}$ lies on the imaginary axis, i.e. $x=0$. ALITER Let $E=\frac{z}{1-z^2}=\frac{z}{z\bar{z}-z^2}=\frac{1}{\bar{z}-z}$ which is imaginary.