Question

If $\left|z\right|=1$ and $z\ne 1$, then all the values of $z/(1-z^2)$ lie on

• A. a line not passing through the origin
• B. $\left|z\right|=\sqrt{2}$
• C. the x-axis
• D. the y-axis

Answer 1

The correct option is D the y-axis

Given:

$|z|=1$

Let $z=x+iy$

$\therefore \sqrt{x^2+y^2}=1$

$\Rightarrow x^2+y^2=1$

$\therefore \frac{z}{1-z^2}$

$=\frac{x+iy}{1-(x^2-y^2+2ixy)}$

$=\frac{x+iy}{1-x^2+y^2-2ixy}$

$=\frac{x+iy}{2y^2-2ixy}$

$=\frac{x+iy}{2iy(-iy-x)}$

$=\frac{1}{-2iy}$

$=\frac{i}{2y}$ $\Rightarrow$ Imaginary number