Solving Absolute-Value Equations of Complex Numbers I just started doing A-Level Further Maths and I am unsure of how to solve this question: $|z-3|= |z-1|$
I understand that with $|z| = 1$ you would get a circle of a radius of 1 on the origin on the argand diagram however I'm not sure what you would do for the question above.
(Sorry if there are any formatting mistakes or errors, new to this)
 A: Hint. Let $z=x+iy$ then your equation is equivalent to
$$(x-3)^2+y^2=|z-3|^2=|z-1|^2=(x-1)^2+y^2.$$
After the algebraic solution. Are you able to solve the problem also from the geometric point of view?
Note that you are looking for the points whose distances from 1 and 3 are the same. That is the line segment bisector of the real segment $[1,3]$ whose equation is $x=2$ or $\mbox{Re}(z)=2$.
P.S. In general $|z-u|= |z-v|$ is the line segment bisector of the segment of extreme point $u$ and $v$. The equation of this line can be found by solving
$$(x-u_x)^2+(y-u_y)^2=(x-v_x)^2+(y-v_y)^2$$
where $u=u_x+iu_y$ and  $v=v_x+iv_y$. Finally we get
$$(v_x-u_x) x+(v_y-u_y)y=\frac{|v|^2-|u|^2}{2}.$$
A: $\left| {z - c} \right| = r$ is a circle of radius r with center at $c+0i$, so
$\left| {z - 1} \right| = r = \left| {z - 3} \right|$ clearly is the intersection of ..., all which stay on a ...
A: $|z-3|=|z-1|$ is the locus of points $z$ in the complex plane that are equidistant from $3$ and $1$. It is geometrically obvious that this is the vertical line $\Re(z) = 2$.
For a calculation leading to the same result, square both sides of the equality and use $|z|^2 = z \bar{z}$:
$$(z-3)(\bar z - 3) = (z - 1)(\bar z - 1)$$
$$z \bar z -3(z + \bar z) + 9 = z \bar z -(z + \bar z) + 1$$
$$ 2(z+\bar z) = 8$$
$$ z+\bar z = 4$$
Since $z+\bar z = 2 \;\Re(z)$ the latter simplifies to $\Re(z) = 2$.
