1. ## Complex vector question

Hi, here's a question from an exam that I don't know how to do. I can do (i) and (ii). For (ii) I get $\displaystyle z_2-z_1$ or $\displaystyle z_1-z_2$. But I don't know what they want for (iii) and (iv).

I was thinking of using the fact that the diagonals of a paralleogram bisect each other, but that doesn't seem to work out...

(i) Let $\displaystyle z_1=r_1 cis \theta_1$ and $\displaystyle z_2=r_2 cis \theta_2$. If $\displaystyle z_1$ and $\displaystyle z_2$ are parallel then prove that $\displaystyle z_1 = k z_2$ for k real.

(ii) Let the points A, B, C and D be represented by the complex numbers $\displaystyle z_1$, $\displaystyle z_2$, $\displaystyle z_3$ and $\displaystyle z_4$ respectively (in clockwise order). Find two possible vectors representing side AB.

(iii) If A, B, C and D are the vertices of a parallelogram then using parts (i) and (ii) find an expression for side DC in terms of AB using the vectors $\displaystyle z_1$, $\displaystyle z_2$, $\displaystyle z_3$ and $\displaystyle z_4$.

(iv) Using a property of a parallelogram find the two possible values of k.

(v) Show that for both these values of k that $\displaystyle z_1-z_2-z_3+z_4=0$.

Any ideas?

Thanks

EDIT: Does this belong in the pre calculus section? If so, many apologies. My university considers complex numbers as an algebra topic...

2. Some of these problems seem strange.

For (iii), the vector representing the side DC is equal to the vector sum DA + AB + BC, which I guess is an expression using AB and the four given vectors.

(iv): Hm? We know that the two vectors $\displaystyle z_1$ and $\displaystyle z_2$ are parallel. Doesn't this mean that the points A and B lie on the same straight line through the origin? I don't see how this restricts the value of k in any way.

3. Hello,

Sorry I made a typo the line in the last part should be "$\displaystyle z_1-z_2-z_3+z_4=0$."

Thanks again

4. Originally Posted by HappyJoe
Some of these problems seem strange.

For (iii), the vector representing the side DC is equal to the vector sum DA + AB + BC, which I guess is an expression using AB and the four given vectors.

(iv): Hm? We know that the two vectors $\displaystyle z_1$ and $\displaystyle z_2$ are parallel. Doesn't this mean that the points A and B lie on the same straight line through the origin? I don't see how this restricts the value of k in any way.
I think it's hinting at the fact that sides DC and AB are parallel, and I need to use that result somehow...

5. Originally Posted by minifhncc
I think it's hinting at the fact that sides DC and AB are parallel, and I need to use that result somehow...
As I see it, the vectors $\displaystyle z_1$ and $\displaystyle z_2$ don't have anything to do with the side DC, do they? They represent the point A and B, and for the two vectors to be parallel, the two points A and B have to lie on the same straight line through the origin.

6. I think it means to take it generally.

ie. $\displaystyle z_1$ and $\displaystyle z_2$ in part (i) aren't the same vectors in the following parts.

7. Originally Posted by minifhncc
I think it means to take it generally.
ie. $\displaystyle z_1$ and $\displaystyle z_2$ in part (i) aren't the same vectors in the following parts.
From reply #2, “Some of these problems seem strange”. I absolutely agree with that. Moreover, whoever wrote certainly knows what he/she means. But that does not mean that we do. That said, here are some observations.

If $\displaystyle A:z_1,~ B:z_2,~ C:z_3,~\&~ D:z_4,$ are the vertices of a parallelogram in counter-clockwise order the side $\displaystyle AB$ can be represented as the vector $\displaystyle <\mathif{Re}(z_2)- \mathif{Re}(z_1), \mathif{Im}(z_2)- \mathif{Im}(z_1)>$.

Because we have a parallelogram, it must be true that $\displaystyle \mathif{Re}(z_3)- \mathif{Re}(z_4)= \mathif{Re}(z_2)- \mathif{Re}(z_1),~$
$\displaystyle \&~ \mathif{Im}(z_3)- \mathif{Im}(z_4)= \mathif{Im}(z_2)- \mathif{Im}(z_1)$.

Does that help?

8. Originally Posted by Plato
Because we have a parallelogram, it must be true that $\displaystyle \mathif{Re}(z_3)- \mathif{Re}(z_4)= \mathif{Re}(z_2)- \mathif{Re}(z_1),~$
$\displaystyle \&~ \mathif{Im}(z_3)- \mathif{Im}(z_4)= \mathif{Im}(z_2)- \mathif{Im}(z_1)$.

Does that help?
Yes I know that the diagonals of a parallelogram bisect each other... but how do I get the two values of k ...? Wouldn't that expression only give one value of k?

Thanks

9. Originally Posted by minifhncc
Yes I know that the diagonals of a parallelogram bisect each other... but how do I get the two values of k ...? Wouldn't that expression only give one value of k?
Well, my reply has absolutely nothing to do with the diagonals of a parallelogram.

10. We want to find k in $\displaystyle z_1=kz_2$, is that right?

In part (iv), what is your interpretation of $\displaystyle z_1$ and $\displaystyle z_2$, that is, which part of the parallelogram do they represent?

11. $\displaystyle z_1$ and $\displaystyle z_2$, I think, are just two arbitary vectors.

ie. I'm thinking that since sides AB and DC are parallel, we need to find k such that $\displaystyle z_3-z_4=k(z_2-z_1)$. Upon solving it I get $\displaystyle k=\pm 1$. But for both these values of k I don't get the required expression. I only get it for one value ie. k=-1...

12. Originally Posted by minifhncc
$\displaystyle z_1$ and $\displaystyle z_2$, I think, are just two arbitary vectors.

ie. I'm thinking that since sides AB and DC are parallel, we need to find k such that $\displaystyle z_3-z_4=k(z_2-z_1)$. Upon solving it I get $\displaystyle k=\pm 1$. But for both these values of k I don't get the required expression. I only get it for one value ie. k=-1...
This is what Plato said.

You are right that for a pair of parallel vectors $\displaystyle a$ and $\displaystyle b$, there exists only one $\displaystyle k$, such that $\displaystyle a=kb$, so them asking for two such values of $\displaystyle k$ is strange.

Unless of course they also want a $\displaystyle k$ such that $\displaystyle z_4-z_1=k(z_3-z_2)$, where we are looking at sides $\displaystyle AD$ and $\displaystyle BC$, which they just might do, even though the problem text takes care not to give anything away.