Question

If $|z|=1$ and $\omega =\frac{z-1}{z+1}$ (where $z\ne -1$), then Re($\omega$) is

• 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 1

Answer: (A)

0

If $\left|z\right|=1$
Let $z=\cos \theta +i\sin \theta$
$\frac{-\sin \frac{\theta }{2}}{\cos \frac{\theta }{2}}\left(\frac{\sin \frac{\theta }{2}-i\cos \frac{\theta }{2}}{\cos \frac{\theta }{2}+i\sin \frac{\theta }{2}}\right)$
$\Rightarrow w=\frac{-\sin \frac{\theta }{2}}{\cos \frac{\theta }{2}}\left(\frac{\sin \frac{\theta }{2}-i\cos \frac{\theta }{2}}{\cos \frac{\theta }{2}+i\sin \frac{\theta }{2}}\right)=\frac{i\sin \frac{\theta }{2}}{\cos \frac{\theta }{2}}\left(\frac{i\sin \frac{\theta }{2}+\cos \frac{\theta }{2}}{\cos \frac{\theta }{2}+i\sin \frac{\theta }{2}}\right)$
$\Rightarrow w=i\tan \frac{\theta }{2}$
$\therefore Re\left(w\right)=0$
Ans: A