Question 1074905
{{{z=a+bi}}}
So,
{{{abs(z-1)=sqrt((a-1)^2+b^2)}}}
{{{abs(z-3)=sqrt((a-3)^2+b^2)}}}
{{{abs(z-i)=sqrt(a^2+(b-1)^2)}}}
.
.
{{{(a-1)^2+b^2=(a-3)^2+b^2}}}
{{{a^2-2a+1+b^2=a^2-6a+9+b^2}}}
{{{-2a+1=-6a+9}}}
{{{4a=8}}}
{{{a=2}}}
and
{{{(a-1)^2+b^2=a^2+(b-1)^2}}}
{{{a^2-2a+1+b^2=a^2+b^2-2b+1}}}
{{{-2a+1=-2b+1}}}
{{{b=a}}}
{{{b=2}}}
.
.
.
{{{z=2+2i}}}