Prove the representation of complex numbers is parallelogram Question :
Let $z_1,z_2,z_3,z_4$ be the position vectors of the vertices for quadrilateral $ABCD$. Prove that $ABCD$ is a parallelogram if only if $z_1-z_2-z_3+z_4=0$
What i've done so far :
Proof of $p\Rightarrow q$
Let $z_1=x_1+iy_1,\quad z_2=x_2+iy_2, \quad z_3=x_3+iy_3, \quad z_4=x_4+iy_4$
$ABCD$ is a parallelogram $\Leftrightarrow\,\overrightarrow{AB}=\overrightarrow{CD}\, \land\, \overrightarrow{AD}=\overrightarrow{BC} $ 
$\begin{aligned}
\overrightarrow{AB}=\overrightarrow{CD}
&\Rightarrow |z_1-z_2|=|z_3-z_4|\\
&\Leftrightarrow x_1^2-2x_1x_2+x_2^2+y_1^2-2y_1y_2+y_2^2=x_3^2-2x_3x_4+x_4^2+y_3^2-2y_3y_4+y_4^2\\
\overrightarrow{AD}=\overrightarrow{BC}
&\Rightarrow |z_1-z_4|=|z_2-z_3|\\
&\Leftrightarrow x_1^2-2x_1x_4+x_4^2+y_1^2-2y_1y_4+y_4^2=x_2^2-2x_2x_3+x_3^2+y_2^2-2y_2y_3+y_3^2
\end{aligned}$
I'm stuck and see no conclusion here.
Proof $p\Leftarrow q$
I don't have an idea for this.
Please help me, thanks.
 A: 
$ABCD$ is a parallelogram $\Leftrightarrow\,\overrightarrow{AB}=\overrightarrow{CD}\, \land\, \overrightarrow{AD}=\overrightarrow{BC} $ 

That is almost correct. If $A, B, C, D$  are the vertices in clockwise orientation then it should be
$\overrightarrow{AB}=\overrightarrow{DC}\, \land\, \overrightarrow{AD}=\overrightarrow{BC} \ .$ 

$\overrightarrow{AB}=\overrightarrow{CD}
\implies |z_1-z_2|=|z_3-z_4|$

That is correct, but you lose information by comparing the vector lengths only. The whole task becomes easier if you compare the vectors themselves:
$$
\overrightarrow{AB}=\overrightarrow{DC} \iff z_2-z_1=z_3-z_4 \, , \\
\overrightarrow{AD}=\overrightarrow{BC} \iff z_4-z_1=z_3-z_2 \, .
$$
Now both conditions on the right-hand side are equivalent and equivalent to
$$
z_1-z_2+z_3-z_4=0 \, ,
$$
so this is the correct condition for $ABCD$ being a parallelogram.
A: With a different approach:
Let $z_1,z_2,z_3,z_4\in \mathbb{C}\setminus\{0\}$, $z_k=x_k+iy_k$.
$(\Rightarrow)$
$z_1-z_2-z_3+z_4=0$, which implies
$ Re(z_1-z_2-z_3+z_4)=Im(z_1-z_2-z_3+z_4)=0$.  
So,
$
  \begin{cases}
    x_1+x_4-(x_2+x_3)=0\\
    y_1+y_4-(y_2+y_3)=0
  \end{cases}
$ 
As the sum of the horizontal (real) and vertical (imaginary) components equal zero, what can you deduce?
Proof of $(\Leftarrow)$ is quite similar.
