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Meinme

  • 3 years ago

The locus of Z satisfying the inequality given below is

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  1. Meinme
    • 3 years ago
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    \[\log_{1/3} \left| z+1 \right|>\log_{1/3}\left| z-1 \right| \]

  2. Meinme
    • 3 years ago
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    options are a) R(z) < 0 b) R(z) > 0 c) I(z) < 0 d) None of these Please provide step wise procedure with answer

  3. vf321
    • 3 years ago
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    \[\log_{1/3}|z+1|> \log_{1/3}|z-1|\]Since \(e^x\) is monotonic increasing over the real domain \((\frac{1}{3})^x\) is monotonic decreasing over all reals. Thus, \[(1/3)^{\log_{1/3}|z+1|} < (1/3)^{\log_{1/3}|z-1|}\]\[|z+1| < |z -1|\]For a complex number \(z = a + b i\), the only way the magnitude can be decreased by increasing the real component is if \(a\) is negative, because then the \((a-1)^2\) term in the magnitude \(\sqrt{(a-1)^2+b^2}\) increases, as opposed to if it was \(a+1\). Thus, the answer is (a)

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