Question

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

• A. 1
• B. 2
• C. $\infty$

Answer 1

The correct option is A 1

Let z = x + iy
$|z-1|=|z+1|\Rightarrow (x-1{)}^2+y^2=(x+1{)}^2+y^2\Rightarrow Re\left(z\right)=0\Rightarrow x=0|z-1|=|z-i|\Rightarrow (x-1{)}^2+y^2=x^2+(y-1{)}^2\Rightarrow x=y|z+1|=|z-i|\Rightarrow (x+1{)}^2+y^2=x^2+(y-1{)}^2$
Only (0, 0) satisfies all conditions.