Complex Numbers And Quadratic Equations question 134
Question: The locus of $ z $ satisfying the inequality $ {\log_{1/3}}|z+1|> $ $ {\log_{1/3}}|z-1| $ is
Options:
A) $ R(z)<0 $
B) $ R(z)>0 $
C) $ I(z)<0 $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
We know that $ {\log_{a}}m>{\log_{a}}n $
Þ $ m>n $ or $ m<n $ , according as $ a>1 $ or $ 0<a<1 $ . Hence for $ z=x+iy $ $ {\log_{(1/3)}}|z+1|>{\log_{(1/3)}}|z-1|\Rightarrow |z+1| $ $ <|z-1| $ $ { \because 0<\frac{1}{3}<1 } $
Þ $ |x+iy+1|<|x+iy-1| $
Þ $ {{(x+1)}^{2}}+y^{2}<{{(x-1)}^{2}}+y^{2} $
Þ $ 4x<0\Rightarrow x<0\Rightarrow Re(z)<0 $
BETA
