First slide

Geometry of complex numbers

Question

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

Moderate

A

a line not passing through the origin.
B

$|z| = 2$
C

the $x -$ axis
D

the $y -$ axis

Solution

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

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