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