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I have a question which asks:

If vectors $v_1$ and $v_2$ have associated complex numbers $z_1$ and $z_2$ respectively then express, in terms of $z_1$ and $z_2$, the fact that the two vectors are a) parallel and b) perpendicular. Then, using this information, find the conditions necessary for 4 points, $z_1,z_2,z_3,z_4$ to constitute a parallelogram.

This should be super easy but I'm getting hung up.

My attempt at a solution:

If the vectors are perpendicular, ${z_1}\cdot{z_2} = 0$ so $x_1x_2+y_1y_2=0$ or $\frac{\operatorname*{Re}(z_1)}{\operatorname*{Im}(z_1)}=-\frac{\operatorname*{Im}(z_2)}{\operatorname*{Re}(z_2)}$. I think this is fine, but I could be wrong.

If two vectors are parallel, their slopes should be equal, namely $\frac{\operatorname*{Im}(z_2)}{\operatorname*{Im}(z_1)}=\frac{\operatorname*{Re}(z_2)}{\operatorname*{Re}(z_1)}$. I also believe this is correct, but I could still be mistaken.

The parallelogram part is where I'm getting confused. Suppose for convention that the line joining $z_1$ and $z_3$ is parallel to the line joining $z_2$ and $z_4$. Similarly for the line joining $z_1$ and $z_2$, and $z_3$ and $z_4$. For there to be a parallelogram, I know that the lengths of the sides must be equal, so $|z_4-z_2| = |z_3-z_1|$ and $|z_4-z_3| = |z_2-z_1|$. This is fine. However, how do I make sure the vectors constituting the parallel sides are, indeed, parallel?

Is it okay to use the parallel condition I used above if the vectors dont start at the origin? So, should I say that $\frac{\operatorname*{Im}(z_4-z_2)}{\operatorname*{Im}(z_3-z_1)}=\frac{\operatorname*{Re}(z_4-z_2)}{\operatorname*{Re}(z_3-z_1)}$ for one pair of sides,and a similar expression for the other pair?

If I can make anything clearer please let me know.

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    Without entering in your calculations (sorry, no time enough), I can affirm that there is no simpler condition for $z_1,z_2,z_3,z_4$ to form a parallelogram than : $z_1+z_3=z_2+z_4$.2017-01-29
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    Why is this condition sufficient to form a parallelogram? That doesn't seem obvious to me at all. I'm not saying you're wrong I just don't get why that is all that is necessary2017-01-29
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    It would make more since to me if it said $z_3-z_1=z_4-z_2$2017-01-29
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    If you divide LHS and RHS by 2, this condition reads: "midpoint of $z_1z_3$ = midpoint of $z_2z_4$". Another proof: it is equivalent to $z_1-z_4=z_2-z_3$ (equality of vectors defind by opposite sides)2017-01-29

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If the vectors are perpendicular, ${z_1}\cdot{z_2} = 0$ so $x_1x_2+y_1y_2=0$ or $\frac{\operatorname*{Re}(z_1)}{\operatorname*{Im}(z_1)}=-\frac{\operatorname*{Im}(z_2)}{\operatorname*{Re}(z_2)}$.

This can also be written as $\operatorname{Re}(z_1 \bar z_2) = 0\,$, which is the same as $\arg(z_1)-\arg(z_2) = \pm \pi /2\,$.

If two vectors are parallel, their slopes should be equal, namely $\frac{\operatorname*{Im}(z_2)}{\operatorname*{Im}(z_1)}=\frac{\operatorname*{Re}(z_2)}{\operatorname*{Re}(z_1)}$.

This can also be written as $\operatorname{Im}(z_1 \bar z_2) = 0\,$, same as $\arg(z_1)-\arg(z_2) = 0$ or $\pi$.

The parallelogram part is where I'm getting confused.

A quadrilateral is a parallelogram iff two opposite sides are parallel and equal. For $z_1\,z_2\,z_3\,z_4$ to be the vertices of a parallelogram (in this order) the necessary and sufficient condition is $z_2-z_1=z_4-z_3\,$. Another way to write it is $z_1+z_3=z_2+z_4\,$ which corresponds to the condition that the diagonals intersect in their respective midpoints.

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    Thank you for the detailed answer. I guess I have a hard time picturing how the subtraction of the complex numbers $z_2-z_1$ yields the line formed by the two points. I get it for vectors, but for complex numbers not so much. If you have the time to explain this, it would be appreciated. Either way, great answer, thanks.2017-01-29
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    @johnmorrison It works just as for vectors, once you reckon that complex number $z$ is associated to the vector $\overrightarrow{OZ}$ in the complex plane. Then $z_1-z_2$ is associated to $\overrightarrow{OZ_1}-\overrightarrow{OZ_2}=\overrightarrow{Z_1 Z_2}\,$.2017-01-29
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    Thank you very much! You're my hero2017-01-29