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)
On
$\left| {z - c} \right| = r$ 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 ...
On
$|z-3|=|z-1|$ 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 $\Re(z) = 2$.
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 $$(v_x-u_x) x+(v_y-u_y)y=\frac{|v|^2-|u|^2}{2}.$$