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The number of complex numbers $z$ such that $|z-1|=|z+1|=|z-i|$ is ?

$\begin{array}{1 1}(A) \;0 \\(B)\;1 \\(C)\;2 \\(D)\;3 \end{array}$

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  • $|x+iy|=\sqrt {x^2+y^2}$
Let $z=x+iy$
$|z-1|=|x+iy-1|=\sqrt {(x-1)^2+y^2}$
$|z+1|=|x+iy+1|=\sqrt {(x+1)^2+y^2}$
$|z-i|=|x+iy-i|=\sqrt {x^2+(y-1)^2}$
$Given\:|z-1|=|z+1|=|z-i|$
$\Rightarrow\:\sqrt {(x-1)^2+y^2}=\sqrt {(x+1)^2+y^2}=\sqrt {x^2+(y-1)^2}$
$\Rightarrow\:x-1=\pm(x+1)$
$\Rightarrow\:x=0$
and
$(x+1)^2+y^2=x^2+(y-1)^2$
Substituting $x=0$, we get
$1+y^2=(y-1)^2=1+y^2-2y$
$\Rightarrow y=0$
$\Rightarrow z=0$
$\therefore $ there is only one complex number

 

answered Jul 3, 2013 by rvidyagovindarajan_1
 

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