Question
The number of complex numbers $$z$$ such that $$\left| {z - 1} \right| = \left| {z + 1} \right| = \left| {z - i} \right|$$ equals
A.
$$1$$
B.
$$2$$
C.
$$\infty $$
D.
$$0$$
Answer :
$$1$$
Solution :
$$\eqalign{
& {\text{Let }}z = x + iy \cr
& \left| {z - 1} \right| = \left| {z + 1} \right|{\left( {x - 1} \right)^2} + {y^2} = {\left( {x + 1} \right)^2} + {y^2} \cr
& \Rightarrow \,\,{\text{Re}}\,z = 0\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,x = 0 \cr
& \left| {z - 1} \right| = \left| {z - i} \right|{\left( {x - 1} \right)^2} + {y^2} = {x^2} + {\left( {y - 1} \right)^2} \cr
& \Rightarrow \,\,x = y \cr
& \left| {z + 1} \right| = \left| {z - i} \right|{\left( {x + 1} \right)^2} + {y^2} = {x^2} + {\left( {y - 1} \right)^2} \cr} $$
Only $$(0, 0)$$ will satisfy all conditions.
⇒ Number of complex number $$z = 1$$