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

If |z| = 1 and z ≠ ± 1, then all the values of z/(1-z²) lie on $|z| = 1,\quad z \neq \pm 1$ $\text{so } z \text{ lies on the unit circle, } z \neq 1,\ -1$ $\dfrac{z}{1-z^{2}}$ $\text{Find the locus these values satisfy}$ $|z| = 1 \Rightarrow \bar{z} = \dfrac{1}{z}$ $w = \dfrac{z}{1-z^{2}}$ $w = \dfrac{z}{1-z^{2}}$ $\bar{w} = \dfrac{\bar{z}}{1-\bar{z}^{2}}$ $\bar{w} = \dfrac{1/z}{1-1/z^{2}}$ $\bar{w} = \dfrac{1/z}{\dfrac{z^{2}-1}{z^{2}}}$ $= \dfrac{1}{z}\cdot\dfrac{z^{2}}{z^{2}-1}$ $\bar{w} = \dfrac{z}{z^{2}-1}$ $\bar{w} = \dfrac{z}{z^{2}-1} = -\dfrac{z}{1-z^{2}}$ $\bar{w} = -w$ $\bar{w} = -w \Rightarrow 2\,\text{Re}(w) = 0$ $w = 0 + iy \quad (\text{purely imaginary})$ $\text{Trace of } w: \text{ the y-axis}$ $z = e^{i\theta}, \quad \theta \neq 0,\ \pi$ $w = \dfrac{e^{i\theta}}{1-e^{2i\theta}}$ $1-e^{2i\theta} = e^{i\theta}\!\left(e^{-i\theta}-e^{i\theta}\right)$ $w = \dfrac{e^{i\theta}}{e^{i\theta}\!\left(e^{-i\theta}-e^{i\theta}\right)} = \dfrac{1}{e^{-i\theta}-e^{i\theta}}$ $e^{i\theta}-e^{-i\theta} = 2i\sin\theta$ $w = \dfrac{1}{-2i\sin\theta}$ $w = \dfrac{i}{2\sin\theta}$ $\sin\theta \neq 0 \Rightarrow w \neq 0,\ \text{but } w \text{ stays on the y-axis}$ $\text{Answer: (4) the y-axis}$