JEE Main · 2010 · 11 Apr 2010 · Offline · Complex Numbers

Reading $z$ as a point in the plane, $|z-1|=|z+1|=|z-i|$ asks for a point equidistant from the three points $1$, $-1$ and $i$, that is $(1,0)$, $(-1,0)$ and $(0,1)$. Three non-collinear points have exactly one equidistant point, their circumcentre. Here $|z-1|=|z+1|$ forces $x=0$, and $|z-1|=|z-i|$ forces $x=y$, so together $x=y=0$ and $z=0$. Checking, $|0-1|=|0+1|=|0-i|=1$, so $z=0$ satisfies all three and is the only such number; there is exactly one.