Prove there exist a complex number that satisfies the given conditions

Question

Prove that there exists a complex number $z=a+ib$ on the complex plane such that there are no two complex numbers $z_1=x_1+i y_1$ and $z_2=x_2+i y_2$ such that $\left|z_1-z\right|=\left|z_2-z\right|$ where $a, b \in \mathbb{R}$ and $x_1, x_2, y_1, y_2 \in \mathbb{Z}$

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I assumed the contrary, hence I tried using distance formula to get $\left(x_1-a\right)^2+\left(y_1-b\right)^2=\left(x_2-a\right)^2+\left(y_2-b\right)^2 \implies \left(x_1-x\right)\left(x_1+x_2-2 a\right)=\left(y_2-y\right)\left(y_2+y_1-2 b\right)$. Here, $x_1-x, x_1+x_2, y_2-y, y_2+y_1$ are all integers while $2a,2b$ aren't necessarily integers. My aim was to look for some contradiction, but I can't progress from here.

Answer 1

For two points on the real or complex plane, the set of points that are equidistant from both of them lies on the perpendicular bisector of the line segment joining the two points.

For the $\mathbb{Z}\times\mathbb{Z}$ grid in $\mathbb{R}\times\mathbb{R}$ or $\mathbb{C}$, there will be a perpendicular bisector for any two points $(x_1,y_1), (x_2,y_2)$. The set of perpendicular bisectors for two points of $\mathbb{Z}\times\mathbb{Z}$ will be of cardinality $|\mathbb Z^4|=\aleph_0$ and therefore, it's union can't cover all of $\mathbb{R}\times\mathbb{R}$ (lines are of measure zero). So, there is a point that satisfies said requirement.

Answer 2

Disclaimer: this answers the original problem, but perhaps not OP's question. But it's too long for a comment.

Claim: The complex number $\ z = \pi + i\frac{1}{\pi}\ $ satisfies the conditions.

Proof: Let $\ z = \pi + i\frac{1}{\pi}\ $ and suppose there are two complex numbers $z_1=x_1+i y_1$ and $z_2=x_2+i y_2$ such that $\left|z_1-z\right|=\left|z_2-z\right|$ where $x_1, x_2, y_1, y_2 \in \mathbb{Z}.$ Then we have:

$$ (\pi - x_1 )^2 + \left(\frac{1}{\pi} - y_1 \right)^2 = (\pi - x_2 )^2 + \left(\frac{1}{\pi} - y_2 \right)^2. $$

Expanding, simplifying and rearranging, we get:

$$ 2(x_2-x_1)\pi + (y_2-y_1)\frac{2}{\pi} + ({x_1}^2 + {y_1}^2 - {x_2}^2 - {y_2}^2) = 0 $$

Multiplying through by $\ \pi,\ $ we see that $\ u=\pi\ $ satisfies the quadratic equation with rational coefficients:

$$ 2(x_2-x_1)u^2 + ({x_1}^2 + {y_1}^2 - {x_2}^2 - {y_2}^2)u + 2(y_2-y_1) = 0,$$

contradicting the transcendence of $\ \pi.$ So our assumption that two such numbers $\ z_1,\ z_2\ $ exist is false. This completes the proof.