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.
 A: 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 $v=-u$.
A: 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.
A: 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.
A: Kinda late to the party, but here's another relatively simple solution. First I'll enumerate the numbers as $z_k$ for $k = 1\dots 4$. Since they are all on the unit circle, they can all be written as $e^{i\theta_k}$. Now, consider the equation
$$
x + y = z + w
$$
The solution set to this equation is spanned by $x = \alpha z + \beta w$ and $y = (1 - \alpha) z + (1 - \beta) w$, where $\alpha$ and $\beta$ are parameters. Thus, we get
$$
\sum_kz_k = 0 \Longleftrightarrow\ e^{i\theta_1} + e^{i\theta_2} = -e^{i\theta_3} - e^{i\theta_4} \\
e^{i\theta_1} = -\alpha e^{i\theta_3} - \beta e^{i\theta_4} \\
e^{i\theta_2} = (\alpha - 1)e^{i\theta_3} + (\beta - 1)e^{i\theta_4}
$$
From here it is easy to see that the only way for $e^{i\theta_1}$ and $e^{i\theta_2}$ to be on the unit circle is if $\alpha = 1,\ \beta = 0$ or $\beta = 1,\ \alpha = 0$, (which can be proven by the Pythagorean Theorem). Without loss of generality we can choose $\alpha = 1,\ \beta = 0$. Then
$$
e^{i\theta_1} = -e^{i\theta_3} \\
e^{i\theta_2} = -e^{i\theta_4}
$$
and
$$
\theta_1 = \theta_3 + \pi \\
\theta_2 = \theta_4 + \pi
$$
which form the corners of a rectangle on the unit circle.
A: Geometric Proof:
Let $M$ and $N$ be the midpoints of $AC$ and $BD$, respectively.  The condition $A+B+C+D=0$ implies that the midpoint of the segment $MN$ is the circumcenter $O=0$.  We want to show that $M=N$.  Assume contrary that $M\neq N$.  Hence, $O\neq M$ and $O\neq N$.
As $OM\perp AC$ and $ON\perp BD$ (the line passing through the center of a circle and the midpoint of a chord of this circle is perpendicular to the given chord), $MN$ is perpendicular to both $AC$ and $BD$.  Consequently, $AC\parallel BD$, which can happen only when $ABCD$ is a degenerate quadrilateral with $\{A,C\}=\{B,D\}$.  However, in this case, we still have $M=N$, a contradiction.
Now, we have $M=N$.  Then, $M=N=O=0$.  That is, $AC$ and $BD$ are diameters of the circumcircle of the quadrilateral $ABCD$.  Ergo, the angles at $A$, $B$, $C$, and $D$ of this quadrilateral are all right angles.  Thus, $ABCD$ is a rectangle.
A: 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.
A: another way is to say $A=cosQ1+isinQ1,B=cosQ2+isinQ2,C=cosQ3+isinQ3,D=cosQ4+isinQ4$, we assume $0 \leq Q1\leq Q2 \leq Q3\leq Q4 \leq 2\pi$, which sould not effect the final result. the our target is to proof $Q3-Q1=Q4-Q2=\pi$
so we can get:
$cosQ1+cosQ2+cosQ3+cosQ4=0,sinQ1+sinQ2+sinQ3+sinQ4=0$, that is:
$sinQ1+sinQ3=-(sinQ2+sinQ4)$......[1] 
$cosQ1+cosQ3=-(cosQ2+cosQ4)$......[2]
if [1]=0 , we can get sinQ1=-sinQ3 and sinQ2=-sinQ4, according to our assumption, we can get $Q3=Q1+\pi$ and $Q4=Q2+\pi$,so ABCD is rectangle.
if [2]=0, we have $Q3=Q1+\pi $ or $Q3=\pi-Q1$ and $Q4=Q2+\pi $ or $Q4=\pi-Q2$.
if $Q3=Q1+\pi$ and $Q4=Q2+\pi $, then QED
if $Q3=\pi-Q1$ and $Q4=\pi-Q2 $, we put in [1] and get $sinQ1=-sinQ2$,$Q2=\pi+Q1$,that is $Q2 \geq Q3$,only when $Q1=0$ ,then $Q2=Q3=\pi,Q4=2\pi$,which is a very special case for the rectangle.
if $Q3=\pi-Q1$ and $Q4=Q2+\pi$, put in {1], we have $sinQ1=0$, then $Q1=0$ and $ Q3=\pi$, which also means $Q3-Q1=\pi$ QED
if  $Q3=\pi+Q1$ and $Q4=Q2-\pi$, we have $Q1=Q2=0, Q3=Q4=\pi$ which is also a special case for the rectangle.
if [1]and [2] are both none zero, 
[1] canbe $2sin\dfrac{Q1+Q3}{2}cos\dfrac{Q3-Q1}{2}=-2sin\dfrac{Q2+Q4}{2}cos\dfrac{Q4-Q2}{2}$ .....[3] 
[2] can be $2cos\dfrac{Q1+Q3}{2}cos\dfrac{Q3-Q1}{2}=-2cos\dfrac{Q2+Q4}{2}cos\dfrac{Q4-Q2}{2}$ ......[4]
and $\dfrac{[3]}{[4]}$,we get $tan\dfrac{Q1+Q3}{2}=-tan\dfrac{Q2+Q4}{2}$
since $0 \leq \dfrac{Q1+Q3}{2} \leq \dfrac{Q2+Q4}{2} \leq 2\pi$,
then we must have 
$\dfrac{Q2+Q4}{2}-\dfrac{Q1+Q3}{2}=\pi$......[5] 
or 
$\dfrac{Q2+Q4}{2}+\dfrac{Q1+Q3}{2}=\pi$......[6]
in both case : put it in [3] and [4], then [3]+[4],we have $cos\dfrac{Q3-Q1}{2}=0$ which casue [1] and [2] be zero. so it is imposible. that is all.
A: Say A+B!=0 and A+D!=0 and consider the quadrilateral with vertices 0,A,A+B,A+B+C=-D. It's a rhombus bc all sides have length 1. So A and C are parallel, as are B and D, forcing A=-C and B=-D. The angles in the original quadrilateral A,B,C,D are all 90 as the diagonals are diameters of the circle.
A: Consider four complex numbers $A,B,C,D$ on the unit circle as shown in Figure 1, such that $$\tag{1} |A-D|\lt|B-C|.$$

$\tag{Fig. 1}$
By the parallelogram law for addition $$\tag{2}|A+D|\gt |B+C|\implies A+B+C+D\neq0.$$
Looking again at Figure 1, and imagining that $$|A-B|=|C-D|,$$ then if we extend the segments $AB$ and $CD$ they will meet on the line through $A+D$ and the lengths of these extensions will be equal (to see this consider the rotation of the plane $z\mapsto uz$ which maps $1/2(A+D)$ onto the real line and argue from symmetry). From Thales theorem (that a line drawn parallel to one side of a triangle cuts the other two proportionally) $$\tag{3} |A-B|=|C-D|\implies AD\parallel BC.$$ Since $(3)$ also holds when $AB$ is parallel to $CD$, it follows that, if $A,B,C,D$ are vertices of a cyclic quadrilateral, with opposite sides equal, then they form a cyclic parallelogram, which is a rectangle. This completes the proof, since assuming $A,B,C,D$ do not form a rectangle $\implies$ opposite sides are not all equal $\implies A+B+C+D\neq0$ which is the contrapositve of what we wanted to prove.
$\square$
