• # question_answer The locus of $z$satisfying the inequality ${{\log }_{1/3}}|z+1|\,>$ ${{\log }_{1/3}}|z-1|$ is A) $R\,(z)<0$ B) $R\,(z)>0$ C) $I\,(z)<0$ D) None of these

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|$$<\,|z-1|$$\left\{ \because 0<\frac{1}{3}<1 \right\}$ Þ$|x+iy+1|<|x+iy-1|$ Þ  ${{(x+1)}^{2}}+{{y}^{2}}<{{(x-1)}^{2}}+{{y}^{2}}$ Þ $4x<0\,\Rightarrow x<0\,\,\Rightarrow \operatorname{Re}(z)<0$

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