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Question

The locus of z satisfying the inequality |z+2i2z+i|<1, where z=x+iy, is

A
x2+y2<1
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B
x2y2<1
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C
x2+y2>1
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D
2x2+3y2<1
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Solution

The correct option is B x2+y2>1
Given inequality:

|z+2i2z+i|<1 and z=x+iy

Hence,

|x+iy+2i||2(x+iy)+i|<1

|x+(2+y)i||2x+(2y+1)i|<1

|x+(2+y)i|<|2x+(2y+1)i|

x2+(2+y)2<(2x)2+(2y+1)2

Squaring both sides:

x2+4+y2+4y<4x2+4y2+4y+1

3x2+3y2>3

x2+y2>1 is required locus.

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