Prove the representation of complex numbers is parallelogram

Question

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.

Answer 1

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.

Answer 2

$ABCD$ 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.