If $A,B,C,D$ are complex numbers on the unit circle with $A+B+C+D=0$, then they form a rectangle

Let $A, B, C, D$ be points on a unit circle. Prove that if $A+B+C+D=0$, then $A,B,C,D$ make a rectangle. (Use complex numbers.)

How do I prove this? I tried to use the dot product of 2 adjacent sides, but I got an ugly trig expression.

I suppose the assumption is that $A,B,C,D$ are all distinct, otherwise it is not necessarily true.

Here is a pure complex number only proof.

Assume that $A+B \ne 0$ and $A + D \ne 0$. We will show that this implies that $A + C = 0$.

Since $$A+B+C+D = 0 \quad \quad (1)$$ we must have that $$\overline{A} + \overline{B} + \overline{C} + \overline{D} = 0$$ where $\overline{z}$ is the conjugate of $z$ and thus

$$\frac{1}{A} + \frac{1}{B} + \frac{1}{C} +\frac{1}{D} = 0 \quad \quad \quad (2)$$

$(1)$ and $(2)$ imply that $$A + B = -(C+D) $$ and $$\frac{A+B}{AB} = -\frac{C+D}{CD}$$

and thus $$AB = CD\quad \quad \quad (3)$$ (because $A+B \neq 0$).

Similary because $A + D \ne 0$, we get $$AD = BC\quad \quad \quad (4)$$

Now $(3)$ and $(4)$ imply (just divide) that $B^2 = D^2$ and hence $B+D = -(A+C) = 0$.

Now rotate the plane around the origin so that $\overline{A} = D$. (This is always possible).

Since rotation is just multiplying by some non-zero $w$, we still have that $A+C = 0$

Thus we have that $D = \overline{A} $, $C = -A$ and $B = -\overline{A}$ and thus $A,B,C,D$ form a rectangle.

Since $A,B,C$ and $D$ are nonzero, the sum of some two of them must be nonzero. Without loss of generality, let $A+B=2x\ne0$. Then $C+D=-2x\ne0$. By rotating the four points (i.e. by multiplying on both sides of the two equations by $e^{-i\arg x}$), we may assume WLOG that $x$ is real. Hence $A,B,C,D$ must take the following forms: \begin{align*} A&=x+iu,\\ B&=x-iu,\\ C&=-x+iv,\\ D&=-x-iv, \end{align*} where $u$ and $v$ are real numbers. As $|A|=|C|=1$, it follows that $|u|=|v|=\sqrt{1-x^2}$. Hence $BACD$ is a rectangle if $v=u$, or $BADC$ is a rectangle if $v=-u$.

The sum of two unit vectors lies on the line that bisects the angle between them, and the length of the sum determines the angle.

Having two equal and opposite such sums forces the existence of a symmetry relating one pair of summands to the other. Four points on a circle that can be divided into two pairs related by a symmetry, form a rectangle.

Maybe I am missing an extremely simple solution with complex numbers, but this seems to be a pure geometry problem where complex numbers do not help much. Of course you can prove the geometry statements using complex numbers, as an exercise.

Well, if you really want a proof which uses complex numbers...

If $A+B \not =0, A+C \not =0$, $$(\frac{A+B}{2}) \cdot (A-B)=0$$ $$(\frac{A+B}{2}) \cdot (C-D)=(-\frac{C+D}{2}) \cdot (C-D)=0$$ (Here we are using dot product)

Since $A+B \not =0$, then the vector represented by $(\frac{A+B}{2})$ is perpendicular to $AB,$ and $CD$, so $AB//CD$. Similarly $AC//BD$, since $A+C \not =0$. Thus $ABDC$ is a paralellogram, so $A-B=C-D$ (since $A-B \not =D-C$), giving $A+D=B+C=0$.

Thus either $A+B=0, A+C=0,$ or $A+D=0$.

By symmetry it suffices to consider when $A+B=0$, then $C+D=0$. $(A-C) \cdot (B-C)=(A-C) \cdot (-A-C)=0$ so $AC \perp BC$. Similarly the other 3 angles are also right angles, so we get a rectangle.