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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)

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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 $$(v_x-u_x) x+(v_y-u_y)y=\frac{|v|^2-|u|^2}{2}.$$

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  • $\begingroup$ I need to show a set of solutions so the answer would have to be in a geometric point of view $\endgroup$ – Max Sep 9 '16 at 22:09
  • $\begingroup$ @SylentNyte Note the $y^2$'s would cancel out, so any value of $y$ plus the correct values of $x$ will satisfy the problem. Its a straight line along $x=\dots$ $\endgroup$ – Simply Beautiful Art Sep 9 '16 at 22:11
  • $\begingroup$ @SimpleArt, got it, x = 2, thank you! $\endgroup$ – Max Sep 9 '16 at 22:14
  • $\begingroup$ @Sylent Nyte Line segment bisector is the key-word. $\endgroup$ – Robert Z Sep 9 '16 at 22:15
  • $\begingroup$ @RobertZ Meh, no need to confuse him. I think $z=2+iy$ is fine. $\endgroup$ – Simply Beautiful Art Sep 9 '16 at 22:17
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$\left| {z - c} \right| = r$ 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 ...

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$|z-3|=|z-1|$ 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 $\Re(z) = 2$.

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