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I need to show that $\left | z-z_{1} \right |=\left | z-z_{2} \right | $ for some $z_{1},z_{2} \in \mathbb{C}$ forms a line equation. I tried to expend the expression like that:

$|z|^2-2Re(z\bar{z_{1}})+|z_1|^2=|z|^2-2Re(z\bar{z_{2}})+|z_2|^2$

$-2Re(z\bar{z_{1}})+2Re(z\bar{z_{2}})=|z_2|^2-|z_1|^2$

But I'm not sure what thats means. Do I need to get an experession of $Im(z)$ vs $Re(z)$?

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  • $\begingroup$ What is the locus of points equidistant from two given points? $\endgroup$ – saulspatz Apr 13 '18 at 7:15
  • $\begingroup$ Its perpendicular bisector of segment joining $z_1$ and $z_2$ @saulspatz $\endgroup$ – jeea Apr 13 '18 at 7:18
  • $\begingroup$ Geometrically I understand why it is a line, but is it possible to get a line equation out of the initial expression? Something like $y=mx+b $ maybe? $\endgroup$ – bp7070 Apr 13 '18 at 7:30
  • $\begingroup$ And isn't that the answer to this question? $\endgroup$ – saulspatz Apr 13 '18 at 7:30
  • $\begingroup$ You have to get a relation between $\Re z$ and $\Im z$ as you said, but you know what the line has to be. If you get an equation like $y=ax+b$ you can immediately translate it to $\Re z$ and $\Im z$ $\endgroup$ – saulspatz Apr 13 '18 at 7:36
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Let $z=x+iy$, $z_1=x_1+iy_1$ and $z_2=x_2+iy_2$.

\begin{align*} |z-z_1|&=|z-z_2|\\ \sqrt{(x-x_1)^2+(y-y_1)^2}&=\sqrt{(x-x_2)^2+(y-y_2)^2}\\ x^2-2x_1x+x_1^2+y^2-2y_1y+y_1^2&=x^2-2x_2x+x_2^2+y^2-2y_2y+y_2^2\\ 2(x_2-x_1)x+2(y_2-y_1)y+(x_1^2+y_1^2-x_2^2-y_2^2)&=0 \end{align*}

which represents a straight line.

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Note, that $$z=x+iy,\; w=u+iw \Rightarrow Re(\bar z w) = xu+yv$$ So, let $z_0 = x_0 + iy_0,\; z_1 = x_1 + i y_1$: $$\left | z-z_{1} \right |^2=\left | z-z_{2} \right |^2 \Leftrightarrow |z|^2+|z_0|^2-2Re(\bar z z_0) = |z|^2+|z_1|^2-2Re(\bar z z_1)$$ $$|z_0|^2 - |z_1|^2 = 2Re(\bar z z_0)-2Re(\bar z z_1) = 2Re(\bar z(z_0-z_1)) = 2x(x_0-x_1)+2y(y_0-y_1)$$ This is obviously an equation of a line.

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$$\left | z-z_{1} \right |^2=\left | z-z_{2} \right |^2 \iff $$

$$ (x-x_1 )^2 + (y-y_1)^2 =(x-x_2 )^2 + (y-y_2)^2 $$

Note that the second degree terms in x and y cancel from both sides and we are left with a linear equation.

Thus the graph is a straight line which is the perpendicular bisector of the segment joining $z_1$ and $z_2.$

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