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The number of complex numbers $z$ such that $|z-1|=|z+1|=|z-i|$ equals
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Verified Answer
The correct answer is:
1
1
Let $z=x+i y$
$$
\begin{array}{lll}
|z-1|=|z+1| & \Rightarrow \operatorname{Re} z=0 & \Rightarrow x=0 \\
|z-1|=|z-i| & \Rightarrow x=y & \\
|z+1|=|z-i| & \Rightarrow y=-x &
\end{array}
$$
Only $(0,0)$ will satisfy all conditions.
$\Rightarrow$ Number of complex number $z=1$
$$
\begin{array}{lll}
|z-1|=|z+1| & \Rightarrow \operatorname{Re} z=0 & \Rightarrow x=0 \\
|z-1|=|z-i| & \Rightarrow x=y & \\
|z+1|=|z-i| & \Rightarrow y=-x &
\end{array}
$$
Only $(0,0)$ will satisfy all conditions.
$\Rightarrow$ Number of complex number $z=1$
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