If $$|z|\,=1$$ and $$\omega =\frac{z-1}{z+1}$$ (where $$z\ne -1)$$, then $$\operatorname{Re}(\omega )$$ is [IIT Screening 2003]

Question

If $$|z|\,=1$$ and $$\omega =\frac{z-1}{z+1}$$ (where $$z\ne -1)$$, then $$\operatorname{Re}(\omega )$$ is  [IIT Screening 2003]

TuteeHUB · Accepted Answer

$$-\frac{1}{|z+1{{|}^{2}}}$$

TuteeHUB · Answer

$$0$$

TuteeHUB · Answer

$$\left| \frac{z}{z+1} \right|\,.\frac{1}{|z+1{{|}^{2}}}$$

TuteeHUB · Answer

$$\frac{\sqrt{2}}{|z+1{{|}^{2}}}$$

1.	If \[\|z\|\,=1\] and \[\omega =\frac{z-1}{z+1}\] (where \[z\ne -1)\], then \[\operatorname{Re}(\omega )\] is [IIT Screening 2003]
A.	\[0\]
B.	\[-\frac{1}{\|z+1{{\|}^{2}}}\]
C.	\[\left\| \frac{z}{z+1} \right\|\,.\frac{1}{\|z+1{{\|}^{2}}}\]
D.	\[\frac{\sqrt{2}}{\|z+1{{\|}^{2}}}\]
Answer» B. \[-\frac{1}{\|z+1{{\|}^{2}}}\]

If \[|z|\,=1\] and \[\omega =\frac{z-1}{z+1}\] (where \[z\ne -1)\], then \[\operatorname{Re}(\omega )\] is [IIT Screening 2003]