If |z|=1 and z neq pm 1, then all the points | vedclass

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(D) Let $z = e^{i 	heta}$, where $	heta \in \mathbb{R}$ and $	heta 
eq n\pi$ for $n \in \mathbb{Z}$ (since $z 
eq \pm 1$).Let $w = \frac{z}{1-z^2

Vedclass · Accepted Answer

the $y$-axis

Vedclass · Answer

(D) Let $z = e^{i 	heta}$, where $	heta \in \mathbb{R}$ and $	heta 
eq n\pi$ for $n \in \mathbb{Z}$ (since $z 
eq \pm 1$).Let $w = \frac{z}{1-z^2}$.Substituting $z = e^{i 	heta}$, we get:$w = \frac{e^{i 	heta}}{1 - e^{i 2 	heta}}$Divide numerator and denominator by $e^{i 	heta}$:$w = \frac{1}{e^{-i 	heta} - e^{i 	heta}}$Using the identity $e^{i 	heta} = \cos 	heta + i \sin 	heta$, we have $e^{-i 	heta} - e^{i 	heta} = -2i \sin 	heta$.Thus, $w = \frac{1}{-2i \sin 	heta} = \frac{i}{2 \sin 	heta}$.Since $w$ is of the form $0 + i \left( \frac{1}{2 \sin 	heta} ight)$, the real part is $0$.Therefore, the locus of $w$ is the $y$-axis.

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

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