The locus of z satisfying the inequality $l o g_{1 / 3} | z + 1 | > l o g_{1 / 3} | z - 1 |$ is

R(z) < 0

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R(z) > 0

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I(z) < 0

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none of these

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Solution

The correct option is A R(z) < 0
We know that ${log}_{a} m > {log}_{a} n \Rightarrow m > n$ or $m < n$ , according as $a > 1$ or $0 < a < 1$ . Hence for $z = x + i y$ .
${log}_{(1 / 3)} | z + 1 | > {log}_{(1 / 3)} | z - 1 | \Rightarrow | z + 1 | < | z - 1 | {∵ 0 < \frac{1}{3} < 1}$
$\Rightarrow | x + i y + 1 | < | x + i y - 1 |$
$\Rightarrow (x + 1)^{2} + y^{2} < (x - 1)^{2} + y^{2}$
$\Rightarrow 4 x < 0 \Rightarrow x < 0$
$\Rightarrow R (z) < 0$

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Similar questions

The locus of z satisfying the inequality $l o g_{\frac{1}{3}}$ |z+1| > $l o g_{\frac{1}{3}}$ |z-1| is

l o g_{1 / 3} | z + 1 | > l o g_{1 / 3} | z - 1 |

z = x + i y

{log}_{1 / 2} | z - 1 | > {log}_{1 / 2} | z - i |

z

{log}_{1 / 2} (\frac{| z - 1 | + 4}{3 | z - 1 | - 2}) > 1 (

| z - 1 | \neq \frac{2}{3})

{log}_{1 / 3} | z + 1 | > {log}_{1 / 3} | z - 1 |

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