The number of complex numbers z such that |z | vedclass

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(A) Let $z = x + iy$. The given condition is $|z - 1| = |z + 1| = |z - i|$.From $|z - 1| = |z + 1|$,we have $|x - 1 + iy| = |x + 1 + iy|$,which implie

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(A) Let $z = x + iy$. The given condition is $|z - 1| = |z + 1| = |z - i|$.From $|z - 1| = |z + 1|$,we have $|x - 1 + iy| = |x + 1 + iy|$,which implies $(x - 1)^2 + y^2 = (x + 1)^2 + y^2$.Expanding this,$x^2 - 2x + 1 + y^2 = x^2 + 2x + 1 + y^2$,which simplifies to $4x = 0$,so $x = 0$.Now,from $|z + 1| = |z - i|$,we have $|x + 1 + iy| = |x + i(y - 1)|$,which implies $(x + 1)^2 + y^2 = x^2 + (y - 1)^2$.Substituting $x = 0$,we get $(0 + 1)^2 + y^2 = 0^2 + (y - 1)^2$.$1 + y^2 = y^2 - 2y + 1$.This simplifies to $-2y = 0$,so $y = 0$.Thus,the only complex number satisfying the condition is $z = 0 + 0i = 0$.Therefore,there is only $1$ such complex number.

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

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