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The number of complex number $z$ such that $|z-1|=|z+1|=|z-i|$ equals

1) 1
2) 2
3) $\infty$
4) 0
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(1)

Explanation

Let $z=x+i y$ $|z-1|=|z+1| \Rightarrow \operatorname{Re} z=0 \Rightarrow x=0$

$|z-1|=|z-i| \Rightarrow x=y$

$|z+1|=|z-i| \Rightarrow y=-x$

Only $(0,0)$ will satisfy all conditions. $\Rightarrow$ Number of complex number $z=1$

by Diamond (74,866 points)

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