Describe the set of points z satisfying Re((z-z_1)/(z-z_2))=0

Describe the set of points $z$ satisfying $\ \text{Re}\left(\dfrac{z-z_1}{z-z_2}\right)=0$.

Step 1 · Translate "real part zero" into an argument condition

$\text{Re}(w)=0$ (with $w\neq0$) means $w$ is purely imaginary, so $\arg w=\pm\dfrac\pi2$. Applied to $w=\dfrac{z-z_1}{z-z_2}$:

$$ \arg(z-z_1)-\arg(z-z_2)=\pm\frac\pi2 $$

i.e. the vector from $z_2$ to $z$ and the vector from $z_1$ to $z$ are perpendicular — the angle $\angle z_1zz_2=90^\circ$.

Step 2 · Apply Thales' theorem

The locus of points $z$ from which a fixed segment $z_1z_2$ subtends a right angle is exactly the circle having $z_1z_2$ as a diameter (Thales' theorem, and its converse).

Step 3 · Write the circle's equation

$$ \boxed{\left|z-\frac{z_1+z_2}2\right|=\frac{|z_2-z_1|}2} $$

with center $\dfrac{z_1+z_2}2$ (the midpoint of $z_1,z_2$) and radius $\dfrac{|z_2-z_1|}2$.

Step 4 · Exclude the endpoints

At $z=z_1$ the quotient is $0$ (real, not purely imaginary — excluded by the strict "purely imaginary" reading, though $\text{Re}(0)=0$ technically holds only if $0$ counts; the geometrically meaningful locus excludes it since the "angle" is undefined there); at $z=z_2$ the quotient is undefined. Both $z_1$ and $z_2$ are excluded from the locus.

Verification (algebraic cross-check)

Setting $w+\bar w=0$ and expanding $(z-z_1)(\bar z-\bar z_2)+(z-z_2)(\bar z-\bar z_1)=0$ (the condition $w+\bar w=0$ after clearing denominators) leads, after collecting terms, to $2|z|^2-z(\bar z_1+\bar z_2)-\bar z(z_1+z_2)+z_1\bar z_2+z_2\bar z_1=0$, which is exactly $\left|z-\frac{z_1+z_2}2\right|^2=\frac{|z_2-z_1|^2}4$ after completing the square — matching Step 3.