Question

If $|z|=1$ and $z\ne \pm 1$, then all the values of $\frac{z}{1-z^2}$ lie on

• A. $a$ line not passing through the origin
• B. $|z|=\sqrt{2}$
• C. The X - axis
• D. The Y - axis

Answer 1

The correct option is D The Y - axis

Let $z=cos\theta +isin\theta \Rightarrow \frac{z}{1-z^2}=\frac{cos\theta +sin\theta }{1-(cos2\theta +isin2\theta )}=\frac{cos\theta +isin\theta }{2si{n}^2\theta -2isin\theta cos\theta }=\frac{cos\theta +isin\theta }{-2isin\theta (cos\theta +isin\theta )}=\frac{i}{2sin\theta }$
Hence, $\frac{z}{1-z^2}$ lies on the imaginary axis i.e. Y-axis.