Proving that if $|z_1|=|z_2|=|z_3|=|z_4|$ and $z_1+z_2+z_3+z_4=0.$ then $z_1,z_2,z_3,z_4$ are points of some rectangle ...or two pairs of points are identical.
The assignment is done as follows:
$$z_i=|z|e^{i \varphi_i}=re^{i \varphi_i}, i=1,\ldots,4$$
the assignment goes on to prove that $$\varphi_1-\varphi_2=\pm(\varphi_3-\varphi_4) \\ \varphi_2-\varphi_3=\pm(\varphi_1-\varphi_4) \\ \varphi_1-\varphi_3=\pm(\varphi_2-\varphi_4)$$
then this is unclear: $$|z_1-z_2|=r|e^{\varphi_1}-e^{i\varphi_2}|=2r\left|\sin \frac{\varphi_1-\varphi_2}{2}\right|$$ 
How??
 A: Hint: $e^{i\varphi_1} - e^{i\varphi_2} = e^{i\varphi}(e^{i\delta} - e^{-i\delta})$, where $\varphi = \frac{\varphi_1+\varphi_2}{2}$ and $\delta = \frac{\varphi_1-\varphi_2}{2}$.
A: First, note that $|e^{i\theta_1}|=1$ So:
$$|e^{i\theta_2}-e^{i\theta_1}| = |e^{i(\theta_2-\theta_1)}-1|$$
Now, let $x=\theta_2-\theta_1$. Then:
$$
e^{ix}-1 = \cos x-1+ i\sin x$$ and thus $$\begin{align}|e^{ix}-1|^2 &= (\cos x-1)^2+ \sin^2 x\\
&=(cos^2x+\sin^2x+1)-2\cos x\\
&=2(1-\cos x)
\end{align}$$
Now you need to remember that $\cos 2y = 1-2\sin^2 y$, with $y=x/2$, you get:
$$|e^{ix}-1|^2 = 4\sin^2(x/2)$$
or:
$$|e^{ix}-1| = 2|\sin(x/2)|$$
A: Here is a geometric approach:
First a lemma whose proof is straightforward:
Suppose $c \in \mathbb{C}$ with $0<|c|<1$, then there are exactly two pairs $(w_1,w_2)$ satisfying
$c = {1 \over 2}(w_1+w_2)$ with $|w_1|=|w_2| = 1$. The solutions are
$((1+i \lambda)c,(1-i \lambda)c) $ and $((1-i \lambda)c,(1+i \lambda)c) $, where $\lambda = \sqrt{{1 \over |c|^2}-1}$. (The $z$s should be $w$s in the little picture.)

By dividing through, we can presume that all of the $z_k$ lie on the unit circle.
Let $c={1 \over 2}(z_1+z_2)$. Then we must have ${1 \over 2}(z_3+z_4) = -c$.
If $|c|=1$, then strict convexity of the unit disk implies $z_1 = z_2$, and similarly, $z_3 = z_4$.
If $c=0$, then $z_1 = -z_2, z_3 = -z_4$ and we are finished.
Otherwise $0 < |c| < 1$, and the above lemma shows that the set of points is 
given by $\{(\pm 1\pm i \lambda)c \} = \{(\pm |c|\pm i \lambda |c|){c \over |c|} \}$, which is easily seen to be a rotated rectangle.
A: WLOG, assume $0=\varphi_1\leq\varphi_2\leq\varphi_3\leq \varphi_4< 2\pi$ - this can be achieved by permuting $z_i$'s and dividing them by $z_1$. We'll show that $z_1+z_3=0=-(z_2+z_4)$ which corresponds to a rectangle - possibly degenerate - as required.
Suppose $z_1+z_3 \neq 0$. Then equating absolute values and arguments of $$z_1+z_3=-(z_2+z_4)$$ yields $$\varphi_3=\varphi_2+\varphi_4 +\pi \mod 2\pi$$ $$cos(\varphi_3)=cos(\varphi_4-\varphi_2)$$
which has a single solution $\varphi_3=\pi,\varphi_2=\pi/2,\varphi_3=3\pi/2$ in our assumed ranges and has already been accounted for. Q.E.D.
