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Question

The locus of $$z$$ such that $$ \left| \dfrac { 1 + iz}{z+i} \right| = 1 $$ 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, $$ \left| \dfrac { 1 + iz}{z+i} \right| = 1 $$
Put, $$ z = x + iy $$
Therefore, $$  \dfrac {| 1 + i(x+iy)  |}{|x+iy+i|} = 1 $$       .....$$ \left( \therefore \left| \dfrac {z_1}{z_2} \right| = \dfrac { | z_1 | }{ | z_2 | } \right) $$
$$ \Rightarrow | (1-x) + ix| = | x + i (1+y) | $$
$$ \Rightarrow \sqrt { (1-y^2) + x^2 } = \sqrt{ x^2 + (1+x)^2 } $$
$$ \Rightarrow 1^2 + y^2 - 2y + x^2 = x^2 + 1^2 + y^2 + 2y $$
$$ \Rightarrow  4y=0$$
$$ \Rightarrow y = 0 $$ 

Mathematics

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