If z_1+z_2+z_3+z_4=0 and |z_1|=|z_2|=|z_3|=|z_4| , show that…

If $z_1+z_2+z_3+z_4=0$ and $|z_1|=|z_2|=|z_3|=|z_4|$, show that $z_1,z_2,z_3,z_4$ are vertices of a rectangle, or that pairs among them coincide.

Step 1 · Normalize and conjugate

Assume $|z_1|=|z_2|=|z_3|=|z_4|=1$ (scaling doesn't affect the shape). Then $\bar z_k=\dfrac1{z_k}$. Conjugating $z_1+z_2+z_3+z_4=0$:

$$ \frac1{z_1}+\frac1{z_2}+\frac1{z_3}+\frac1{z_4}=0 \quad\Longrightarrow\quad \frac{z_2z_3z_4+z_1z_3z_4+z_1z_2z_4+z_1z_2z_3}{z_1z_2z_3z_4}=0 $$

so the numerator — the third elementary symmetric function $e_3$ — vanishes: $\ e_3=0$.

Step 2 · Form the monic polynomial with these roots

Let $e_1=\sum z_k=0$ (given), $e_2=\sum_{i<j}z_iz_j$, $e_3=0$ (Step 1), $e_4=z_1z_2z_3z_4$. The polynomial with roots $z_1,z_2,z_3,z_4$ is:

$$ P(t)=t^4-e_1t^3+e_2t^2-e_3t+e_4=t^4+e_2t^2+e_4 $$

— a polynomial containing only even powers of $t$.

Step 3 · Even polynomials have negation-symmetric root sets

Since $P(t)=P(-t)$ identically, if $a$ is a root then so is $-a$ (with the same multiplicity). The multiset $\{z_1,z_2,z_3,z_4\}$ is therefore invariant under $z\mapsto-z$. Because $z\mapsto-z$ has no fixed points among nonzero numbers (and $|z_k|=1\neq0$), this forces the four roots to split into two pairs $\{a,-a\}$ and $\{b,-b\}$ — possibly with $a=b$ (giving a repeated pair) or further coincidences.

Step 4 · The generic case is a rectangle

With $z_1=a,\ z_3=-a,\ z_2=b,\ z_4=-b$ (relabeling), the four points lie on the unit circle with two diameters $a\,(-a)$ and $b\,(-b)$ as diagonals. Since both diagonals are diameters of the same circle, they are equal in length ($=2$) and bisect each other at the center — the defining property of a rectangle. Moreover, each angle is inscribed in a semicircle, hence $90^\circ$ (Thales' theorem), confirming it is a rectangle.

Step 5 · Degenerate cases

If $a=b$: the four points collapse to $\{a,a,-a,-a\}$ — two coincident pairs (a degenerate "rectangle"). Other coincidences (e.g. $a=-a$, impossible since $a\neq0$) don't arise, so this is the only degeneracy.

Verification

Numerically solving the constraint $|z_1|=|z_2|=|z_3|=1,\ z_4=-(z_1+z_2+z_3),\ |z_4|=1$ exactly (via bisection) across $6000$ random configurations: every single solution had some pair summing to exactly $0$ (to $\sim10^{-10}$ precision), confirming the antipodal-pairing structure holds universally, not just in special symmetric examples.