Complex Numbers And Quadratic Equations question 80
Question: If $ |z|=1 $ and $ \omega =\frac{z-1}{z+1} $ (where $ z\ne -1) $ , then $ Re(\omega ) $ is [IIT Screening 2003]
Options:
A) $ 0 $
B) $ -\frac{1}{|z+1{{|}^{2}}} $
C) $ | \frac{z}{z+1} |.\frac{1}{|z+1{{|}^{2}}} $
D) $ \frac{\sqrt{2}}{|z+1{{|}^{2}}} $
Show Answer
Answer:
Correct Answer: A
Solution:
$ |z|=1\Rightarrow |x+iy|=1\Rightarrow x^{2}+y^{2}=1 $ $ \omega =\frac{z-1}{z+1}=\frac{(x-1)+iy}{(x+1)+iy}\times \frac{(x+1)-iy}{(x+1)-iy} $ $ =\frac{(x^{2}+y^{2}-1)}{{{(x+1)}^{2}}+y^{2}}+\frac{2iy}{{{(x+1)}^{2}}+y^{2}}=\frac{2iy}{{{(x+1)}^{2}}+y^{2}} $ $ (\because x^{2}+y^{2}=1) $ \ $ Re(\omega )=0 $ .
BETA
