<s>[INST] 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.
 [/INST] 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.
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