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Question

The locus of z such that ∣∣∣1+izz+i∣∣∣=1 is :

A
yx=0
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B
y+x=0
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C
y=0
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D
xy=1
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E
x=0
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Solution

The correct option is D y=0
Given, 1+izz+i=1
Put, z=x+iy
Therefore, |1+i(x+iy)||x+iy+i|=1 .....(z1z2=|z1||z2|)
|(1x)+ix|=|x+i(1+y)|
(1y2)+x2=x2+(1+x)2
12+y22y+x2=x2+12+y2+2y
4y=0
y=0

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