Complex Numbers And Quadratic Equations question 48
Question: If $ \frac{z-i}{z+i}(z\ne -i) $ is a purely imaginary number, then $ z.\bar{z} $ is equal to
Options:
A) 0
B) 1
C) 2
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
Here $ \frac{z-i}{z+i}=\frac{x+i(y-1)}{x+i(y+1)}.\frac{x-i(y+1)}{x-i(y+1)} $ $ =\frac{(x^{2}+y^{2}-1)+i(-2x)}{x^{2}+{{(y+1)}^{2}}} $ As $ \frac{z-i}{z+i} $ is purely imaginary, we get $ x^{2}+y^{2}-1=0 $ Þ $ x^{2}+y^{2}=1 $ Þ $ z\overline{z}=1 $ .
BETA
