Describe the set of points z satisfying Im((z-z_1)/(z-z_2))=0
Describe the set of points $z$ satisfying $\ \text{Im}\left(\dfrac{z-z_1}{z-z_2}\right)=0$.
Explanation of Method
A quotient has zero imaginary part exactly when it is real, i.e. its argument is $0$ or $\pi$. Geometrically this means the vectors from $z$ to $z_1$ and from $z$ to $z_2$ are parallel (same or opposite direction) — which happens exactly when $z$, $z_1$, $z_2$ are collinear.
Step 1 · Translate "imaginary part zero" into an argument condition
$\text{Im}(w)=0$ means $w$ is real, so $\arg w\in\{0,\pi\}$. Applied to $w=\dfrac{z-z_1}{z-z_2}$:
$$ \arg(z-z_1)-\arg(z-z_2)\in\{0,\pi\} $$
i.e. the vectors $z-z_1$ and $z-z_2$ point in the same or exactly opposite direction — they are parallel.
Step 2 · Parallel vectors through a common point force collinearity
Since $z-z_1$ and $z-z_2$ both emanate conceptually from $z$ and are parallel, the three points $z,z_1,z_2$ lie on a common line.
Step 3 · Parametrize the line
$$ \boxed{z=z_1+t(z_2-z_1), \qquad t\in\mathbb R} $$
Step 4 · Check the endpoints
At $z=z_2$ the quotient is undefined — excluded. At $z=z_1$ the quotient equals $0$, which is real, so $\text{Im}(0)=0$ — $z_1$ is included (corresponding to $t=0$).
Verification
| Point | $\frac{z-z_1}{z-z_2}$ | Real? |
|---|---|---|
| $z=z_1$ ($t=0$) | $0$ | yes ✓ |
| $z=\frac{z_1+z_2}2$ ($t=\frac12$) | $-1$ | yes ✓ |
| $z=2z_2-z_1$ ($t=2$) | $2$ | yes ✓ |
Comparison with the perpendicular case
| Condition | Meaning | Locus |
|---|---|---|
| $\text{Re}\left(\frac{z-z_1}{z-z_2}\right)=0$ | quotient purely imaginary — vectors perpendicular | circle with diameter $z_1z_2$ |
| $\text{Im}\left(\frac{z-z_1}{z-z_2}\right)=0$ | quotient real — vectors parallel | line through $z_1,z_2$ |
Result
The line through $z_1$ and $z_2$, minus the excluded point $z_2$ — the natural companion to Problem 90's circle, since parallel and perpendicular are complementary conditions on a quotient's argument.