Question

IF \[\color{red}{|z|=1.}\]&\[\color{red}{z \ne \pm 1.}\]then all the values of \[\LARGE{\frac{z}{1-z^2}.}\]lie on ?

Answer 1

z = a complex number.

Answer 2

Its answer is imaginary axis.

Answer 3

bt hw to gt it?
I don't know:(

Answer 4

\[z=re^{i \theta} \\ r=1 \\ z=e^{i \theta}\\ \frac{z}{1-z^2}=\frac{e^{i \theta}}{1-e^{2i \theta}}\]

Answer 5

\[\frac{e^{i \theta }}{1-e^{2i \theta}}\]\[\implies \frac{e^{i \theta}}{(1+e^{i \theta })(1-e^{i \theta})}.\]\[\implies ?\]

Answer 6

sorry i didn't gt:(

Answer 7

\[e^{i \theta}-e^{-i \theta}=2i \sin \theta\]

Answer 8

no no
that nt
actually i didn't gt how come u gt the following: -\[e^{i \theta}-e^{- i \theta}\]

Answer 9

oh sorry Multiplying Fraction by e^(-i theta)
\[\frac{e^{i \theta}}{1-e^{2i \theta}}=\frac{e^{i \theta}e^{-i \theta}}{e^{-i \theta}-e^{-i \theta}e^{2i \theta}}\]

Answer 10

so i get \[- \frac{1}{2\space i \space im (z)}.\]now wt?

Answer 11

well thats your answer
\[-\frac{1}{2i \sin \theta}=-\frac{i}{2i^2 \sin \theta}=\frac{i}{2\sin \theta}=ib \\ b \in \mathbb{R} \\ \because \\ -1 \le \sin \theta \le1 \\ then \\ -\infty \le\frac{1}{\sin \theta} \le \infty \]

Answer 12

so (z) can lie on any point in whole imaginary axis, can't it?

Answer 13

not z but z/1-z^2

Answer 14

oops sorry my bad.
k btw thanx a lot
i gt it:)

Answer 15

ur welcome dear