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.
