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 ?
 A: Rewrite the equations as
$$|z-1|=|z-i|, \ |z-3i|=|z--i|$$
Remember that the meaning of $|a-b|$ is the distance between the points $a$ and $b$ in the complex plane.
So consider what the locus of all solutions is for each equation, and your final solution is the intersection of those loci.
In other words, think of this geometrically.
A: For the first one, you have all points equidistant between $i$ and $1$.  Draw a picture and what do you see?  For the second one, you have all points equidistant between $3i$ and $i$.  
A: ${\left| {z - 1} \right|^2} = {\left| {z - i} \right|^2}$ 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.
A: HINT:
In such cases, write $z=x+iy$ where $x,y$ are real
Now, $|z-a-ib|=\sqrt{(x-a)^2+(y-b)^2}$
Finally square both sides of the given relation 
