The locus of $\mathcal{z}$ satisfying the inequality $\log_{1/3}{\mathcal{|z} + 1|} \log_{1/3}{\mathcal{|z -}1|}$ is

Question

The locus of $\mathcal{z}$ satisfying the inequality $\log_{1/3}{\mathcal{|z} + 1|} > \log_{1/3}{\mathcal{|z -}1|}$ is

Dr. Abubacker · Accepted Answer

(a)
We know that, if $\log_{a}{m > \log_{a}n}$
$\Rightarrow m > n$ or $m < n$ according as $a > 1$ or $0 < a < 1$
$\therefore\ \log_{\left( \frac{1}{3} \right)}{\left| \mathcal{z} + 1 \right| > \log_{\left( \frac{1}{3} \right)}{\mathcal{|z -}1|}}$
$\Rightarrow \ \ \left| \mathcal{\ z} + 1 \right| < \left| \mathcal{z -}1 \right|\ \ \ \ \ \ \ \ \ \left( \because 0 < \frac{1}{3} < 1 \right)$
Let $\mathcal{\ z =}x + iy$
$|x + iy + 1| < |x + iy - 1|$
$\Rightarrow \ \ (x + 1)^{2} + y^{2} < (x - 1)^{2} + y^{2}$
$\Rightarrow \ 4x < 0\ \Rightarrow x < 0\ \Rightarrow \ Re\left( \mathcal{z} \right) < 0$

Q. The locus of $\mathcal{z}$ satisfying the inequality $\log_{1/3}{\mathcal{|z} + 1|} > \log_{1/3}{\mathcal{|z -}1|}$ is

Explanation

Reviewed by Dr. Abubacker

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