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Question

lf log12|z2|>log12|z| then

A
x>1
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B
x<1
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C
x<2
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D
x>2
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Solution

The correct option is A x>1
Letz=x+iy
log12|z2|>log12|z|
|z2|<|z|
|x2+iy|<|x+iy|
x24x+4+y2<x2+y2
4x+4<0
x>1.

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