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Question

The locus of z satisfying the inequality log1/3|z+1|>log1/3|z1| is

A
R (z) < 0
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B
R (z) > 0
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C
I (z) < 0
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D
None of these
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Solution

The correct option is A R (z) < 0
We know that logam>logan m > n or m < n, according as a > 1 or 0 < a < 1. Hence for z = x + iy
log1/3|z+1|>log(1/3)|z1| { 0<13<1}
|x+iy+1|<|x+iy1|
(x+1)2+y2<(x1)2+y2
4x<0x<0Re(z)<0

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