If $|z|=1$ and $z\neq\pm1$, then all the values of $\dfrac{z}{1-z^2}$ lie on
Step-by-Step Solution
Key Concept: Writing $z=e^{i\theta}$ and using $e^{i\theta}-e^{-i\theta}=2i\sin\theta$ collapses the expression to $i/(2\sin\theta)$, immediately showing it is purely imaginary.
**Step 1: Parametrise z on the unit circle**
Let $z=e^{i\theta}$, $\theta\neq0,\pi$.
**Step 2: Simplify the expression**
$\dfrac{z}{1-z^2}=\dfrac{e^{i\theta}}{1-e^{2i\theta}}$. Multiply numerator and denominator by $e^{-i\theta}$: $\dfrac{1}{e^{-i\theta}-e^{i\theta}}=\dfrac{1}{-2i\sin\theta}=\dfrac{i}{2\sin\theta}$.
**Step 3: Identify the locus**
$\dfrac{i}{2\sin\theta}$ is purely imaginary for all $\theta\neq0,\pi$. So all values lie on the $Y$-axis.
Correct Answer: 4