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

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Aicharya · Accepted Answer

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 $z=1$

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

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