Question

The locus of z satisfying the inequality $lo{g}_{1/3}|z+1|>lo{g}_{1/3}|z-1|$ is

• A. R(z) < 0
• B. R(z) > 0
• C. I(z) < 0
• D. none of these

Answer 1

The correct option is A R(z) < 0

We know that ${\log }_{a}m>{\log }_{a}n\Rightarrow m>n$ or $m<n$, according as $a>1$ or $0<a<1$. Hence for $z=x+iy$.
${\log }_{\left(1/3\right)}|z+1|>{\log }_{\left(1/3\right)}|z-1|\Rightarrow |z+1|<|z-1|\{\because 0<\frac{1}{3}<1\}$
$\Rightarrow |x+iy+1|<|x+iy-1|$
$\Rightarrow (x+1{)}^2+y^2<(x-1{)}^2+y^2$
$\Rightarrow 4x<0\Rightarrow x<0$
$\Rightarrow R\left(z\right)<0$