Find z: $|\frac{z-1}{z-i}|=1$ and $|\frac{z-3i}{z+i}|=1$ Give a complex number z:
$|\frac{z-1}{z-i}|=1$ and $|\frac{z-3i}{z+i}|=1$
Find z ?
Could someone help me through this problem ?

${\left| {z - 1} \right|^2} = {\left| {z - i} \right|^2}$ and ${\left| {z - 3i} \right|^2} = {\left| {z + i} \right|^2}$, for $z = x + iy$ this gives you ${\left( {x - 1} \right)^2} + {y^2} = {x^2} + {\left( {y - 1} \right)^2}$ and ${x^2} + {\left( {y - 3} \right)^2} = {x^2} + {\left( {y + 1} \right)^2}$, since ${\left| {z - a - ib} \right|^2} = {\left( {x - a} \right)^2} + {\left( {y - b} \right)^2}$.
After simplifying, you get 2 linear equations with 2 unknowns
$\left\{ \begin{gathered}
  2x + 1 = 2y + 1 \\
   - 6x + 9 = 2y + 1 \\ 
\end{gathered}  \right.$
, which you can solve.
