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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.

Martin R
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user516076
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2 Answers2

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$ABCD$ 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.

Martin R
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  • Let me conclude it and correct me if i'm wrong. So, is a vector the difference between the numbers without modulus? – user516076 Oct 06 '19 at 10:16
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    @user516076: If you identify complex numbers with vectors in $\Bbb R^2$ (as you do here) then the addition of complex numbers corresponds to the addition of vectors. Consequently, a difference $z_2 - z_1$ of complex numbers is the vector from $z_1$ to $z_2$. – Martin R Oct 06 '19 at 10:21
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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.

M.P
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  • Actually, i still can't figure it out how the vectors stuff become Real part and Imaginary part? Is that mean you go from the right hand side and claim it into parallelogram? Please give me more hints. – user516076 Oct 06 '19 at 10:12
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    The imaginary number $z=x+iy$ can be considered as a vector $(x,y)$ with the length of $\sqrt{x^2+y^2}$. E.g. if you have $z_2-z_1=0$, it implies $x_2-x_1-0$ and $y_2-y_1=0$. I.e. $z_1$ and $z_2$ are the opposite vectors. In this case we 2 sets of 2 opposite vectors, forming a closed loop, namely a parallelogram. – M.P Oct 06 '19 at 10:36
  • it's $x_2-x_1-0$ or $x_2-x_1=0$ ?? – user516076 Oct 06 '19 at 10:40
  • Ok thanks. I think i'm understand now... – user516076 Oct 06 '19 at 10:42
  • Yes, $x_2-x_1=0$. It's a typo :) – M.P Oct 06 '19 at 10:43