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4-1.Complex numbers
medium
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
Solution
(a)We know that ${\log _a}m > {\log _a}n$
==> $m > n$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|$
$\left\{ {\because 0 < \frac{1}{3} < 1} \right\}$
==>$|x + iy + 1| < |x + iy – 1|$
==> ${(x + 1)^2} + {y^2} < {(x – 1)^2} + {y^2}$
==> $4x < 0\, \Rightarrow x < 0\,\, \Rightarrow {\mathop{\rm Re}\nolimits} (z) < 0$
Standard 11
Mathematics
