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anonymous
 3 years ago
The locus of Z satisfying the inequality given below is
anonymous
 3 years ago
The locus of Z satisfying the inequality given below is

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\log_{1/3} \left z+1 \right>\log_{1/3}\left z1 \right \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0options are a) R(z) < 0 b) R(z) > 0 c) I(z) < 0 d) None of these Please provide step wise procedure with answer

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\log_{1/3}z+1> \log_{1/3}z1\]Since \(e^x\) is monotonic increasing over the real domain \((\frac{1}{3})^x\) is monotonic decreasing over all reals. Thus, \[(1/3)^{\log_{1/3}z+1} < (1/3)^{\log_{1/3}z1}\]\[z+1 < z 1\]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 \((a1)^2\) term in the magnitude \(\sqrt{(a1)^2+b^2}\) increases, as opposed to if it was \(a+1\). Thus, the answer is (a)
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