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Question

If z=x+iy such that |z+1|=|z1| and amp(z1z+1)=π4 then

A
x=2+1;y=0
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B
x=0;y=2+1
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C
x=0;y=21
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D
x=21;y=0
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Solution

The correct option is B x=0;y=2+1
Let z=x+iy

Given that |z+1|=|z1|
$\Rightarrow |(x+1)+iy|=|(x-1)+iy| $

(x+1)2+y2=(x1)2+y2

(x+1)2+y2=(x1)2+y2

x2+1+2x=x2+12x

x=0

Given that amp(z1z+1)=π4


amp(z1z+1)=amp(x+iy1x+iy+1)

But x=0

amp(z1z+1)=amp(1+iy1+iy)

amp(1+iy1+iy)=π4

amp(1+iy1+iy×1iy1iy)=π4

amp(1+iy+iy+y21+y2)=π4

amp(y21+2iy1+y2)=π4

tan1(2yy21)=π4

2yy21=1

y22y1=0

On solving we get,

y=1±2

Since θ lies in Q1

y=2+1 and x=0

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