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I have to draw the region on the complex plane defined by the following relation:

$|z-z_{1}|=|z-z_{2}|$.

After squaring both sides, we obtain the equality $(z-z_{1})\overline{(z-z_{1})} = (z-z_{2})\overline{(z-z_{2})}$. Then, letting $z = x + i y$, $z_{1} = x_{1} + iy_{1}$, and $z_{2} = x_{2} + i y_{2}$, multiplying out and collecting terms, we obtain the following expression:

$(x-x_{1})^{2} + (y-y_{1})^{2} = (x-x_{2})^{2}+(y-y_{2})^{2}$,

which, to me, looks like two unit circles set equal to each other, so I assume the region defined by the above relation is the area where the two circles intersect. Please let me know if I am correct.

However, I now have to sketch this. Is this going to require separate cases depending on which quadrant $(x_{1},y_{1})$ and $(x_{2},y_{2})$ are in, and then cases where they don't intersect at all? I'm going a bit crazy overthinking this. Please help me figure out what to sketch.

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  • $\begingroup$ Geometrically, this is the set of affixes of points equidistant from the points of affixes $z_1$ and $z_2$, that is... the line segment bisector of the segment $[z_1,z_2]$. Algebraically, you can develop the identity you arrived at, the terms $x^2$ and $y^2$ cancel out hence you are left with the equation of a line. $\endgroup$ – Did Jan 25 '16 at 21:30
  • $\begingroup$ @Did, and isn't that then the overlap of the two circles centered at $z_{1}$ and $z_{2}$, respectively? $\endgroup$ – ALannister Jan 25 '16 at 21:31
  • $\begingroup$ "and isn't that then the overlap of the two circles centered at z1 and z2, respectively?" Did you look at the link in my comment? $\endgroup$ – Did Jan 25 '16 at 21:32
  • $\begingroup$ @Did, yes and the animated graphic looks an awful lot like the intersection of two circles, if you keep going with the arcs all the way around. $\endgroup$ – ALannister Jan 25 '16 at 21:34
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Note that terms cancel and you end up with a straight line which is the perpendicular bisector of $z_1$ and $z_2$.

In simple language the equation $|z-z_1|=|z-z_2|$ says "$z$ moves in such a way that its distance to $z_1$ is equal to its distance to $z_2$".

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  • $\begingroup$ Multiplying everything out, I see what you mean about the $x^2$ and $y^2$ terms cancelling out. Eventually, you do get a line with equation $y = \frac{(2x_{1}-2x_{2})}{(2y_{2}-2y_{1}}x + \frac{(x_{2}^{2}-x_{1}^{2}+y_{2}^{2}-y_{1}^{2})}{2y_{2}-2y_{1}}$. Do I need to sketch this line exactly (no idea how to even begin), or just on the complex plane somewhere, WLOG perhaps in the first quadrant, sketch $z_{1}$ and $z_{2}$ and draw the perpendicular bisector? $\endgroup$ – ALannister Jan 25 '16 at 21:45
  • $\begingroup$ Just choose any two points $z_1$ and $z_2$ anywhere you like and draw the perpendicular bisector $\endgroup$ – David Quinn Jan 25 '16 at 21:49
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The equation of a circle is of the form

$$ (x - a)^2 + (y - b)^2 = r^2 $$

where $r$ is a constant radius equal to the distance from any point $(x,y)$ on the circle to the circle's center, $(a,b)$.

If you have $ (x - a)^2 + (y - b)^2 = $ something that is not a constant, generally you do not have a circle. You have a circle only if you can reduce the equation to the form $ (x - a)^2 + (y - b)^2 = r^2 $ where $r$ is a constant.

What you actually have here is a set of points that are each equidistant from $z_1$ and $z_2$. In terms of plane geometry, suppose you have two points $A$ and $B$ and a point $P$ that is equidistant from them, that is, $PA = PB$. What is the shape of the set of all such points $P$?

Here's another example: let $z_1 = i$ and $z_2 = -i$. Is it possible to find a $z$ such that $|z - i| = |z - (-i)| = 10$? Approximately where on the complex plane would $z$ be? What about $|z - i| = |z - (-i)| = 100$? Or $|z - i| = |z - (-i)| = 1000$?

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the points satisfying the constraint $$|z-z_{1}| = |z-z_{2}|$$ is on the perpendicular bisector of the line joining $z_1$ and $z_2.$ it is given by $$ z = \frac12(z_1+z_2) + \frac12i(z_1-z_2)k, \text{ $k$ any real number.}\tag 1$$

we can check the work by computing $$\begin{align}z-z_1 &= \frac12(z_2-z_1)+\frac12ik(z_1-z_2) = \frac12(z_1-z_2)(-1+ik)\\ z-z_2 &= \frac12(z_1-z_2)+\frac12ik(z_1-z_2) = \frac12(z_1-z_2)(1+ik)\\ \vert z-z_1\vert &=\vert z-z_2\vert = \frac12\vert z_1 - z_2 \vert(1+k^2)\end{align}$$

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  • $\begingroup$ that's the equation of the perpendicular bisector or the equation of the line joining $z_{1}$ and $z_{2}$? $\endgroup$ – ALannister Jan 25 '16 at 21:52
  • $\begingroup$ the equation $(1)$ represents the perpendicular bisector with $k$ a real parameter. $\endgroup$ – abel Jan 25 '16 at 21:53
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Hint:

$ |z-z_1|$ is the distance of $z$ from $z_1$ and $ |z-z_2|$ is the distance from $z_2$. Interpreting the complex numbers as point in a plane $|z-z_1|=|z-z_2| $ means that $z$ has the same distance from $z_1$ and from $z_2$. Do you know the locus of the points with such a property?

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You are going in the right track except that the equation doesn't give you the intersection of two circle.

It basically states that the region is the locus of all points that are equidistant from $z_1$ and $z_2$ which is a line perpendicular to and passes through the midpoint of the segment joining $z_1$ and $z_2$.

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