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. $|z|=2$
• C. the $x$-axis
• D. the $y$-axis

Answer 1

Answer: (D)

Solution: $\mathrm{As}|z|=1$, we get $z\bar{z}=1$.
Let $w=\frac{z}{1-z^2}$, then
$w+\bar{w}=\frac{z}{1-z^2}+\frac{\bar{z}}{1-{\bar{z}}^2}=\frac{z}{1-z^2}+\frac{1/z}{1-(1/z{)}^2}$
$=\frac{z}{1-z^2}+\frac{z}{z^2-1}=0$
$\Rightarrow 2\mathrm{Re}(w)=0\Rightarrow$ $\mathrm{Re}(w)=0$
Thus, $w$ lies on the $y$-axis.
Alternate Solution
$w=\frac{z}{1-z^2}=\frac{z}{z\bar{z}-z^2}=\frac{1}{\bar{z}-z}$
$=\frac{1}{-2i\mathrm{Im}(z)}$
$\Rightarrow w$ is purely imaginary, that is, $w$ lies on the $y$-axis.