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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