# Complex vector question

• May 6th 2011, 07:18 AM
minifhncc
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 $z_2-z_1$ or $z_1-z_2$. 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 $z_1=r_1 cis \theta_1$ and $z_2=r_2 cis \theta_2$. If $z_1$ and $z_2$ are parallel then prove that $z_1 = k z_2$ for k real.

(ii) Let the points A, B, C and D be represented by the complex numbers $z_1$, $z_2$, $z_3$ and $z_4$ 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 $z_1$, $z_2$, $z_3$ and $z_4$.

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

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

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...
• May 7th 2011, 05:17 AM
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 $z_1$ and $z_2$ 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.
• May 7th 2011, 05:26 AM
minifhncc
Hello,

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

Thanks again
• May 7th 2011, 05:28 AM
minifhncc
Quote:

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 $z_1$ and $z_2$ 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...
• May 8th 2011, 03:28 AM
HappyJoe
Quote:

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 $z_1$ and $z_2$ 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.
• May 8th 2011, 06:32 AM
minifhncc
I think it means to take it generally.

ie. $z_1$ and $z_2$ in part (i) aren't the same vectors in the following parts.
• May 8th 2011, 08:09 AM
Plato
Quote:

Originally Posted by minifhncc
I think it means to take it generally.
ie. $z_1$ and $z_2$ 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 $A:z_1,~ B:z_2,~ C:z_3,~\&~ D:z_4,$ are the vertices of a parallelogram in counter-clockwise order the side $AB$ can be represented as the vector $<\mathif{Re}(z_2)- \mathif{Re}(z_1), \mathif{Im}(z_2)- \mathif{Im}(z_1)>$.

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

Does that help?
• May 8th 2011, 08:50 AM
minifhncc
Quote:

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

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
• May 8th 2011, 09:01 AM
Plato
Quote:

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.
• May 8th 2011, 09:49 AM
HappyJoe
We want to find k in $z_1=kz_2$, is that right?

In part (iv), what is your interpretation of $z_1$ and $z_2$, that is, which part of the parallelogram do they represent?
• May 8th 2011, 03:48 PM
minifhncc
$z_1$ and $z_2$, 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 $z_3-z_4=k(z_2-z_1)$. Upon solving it I get $k=\pm 1$. But for both these values of k I don't get the required expression. I only get it for one value ie. k=-1...
• May 8th 2011, 09:02 PM
HappyJoe
Quote:

Originally Posted by minifhncc
$z_1$ and $z_2$, 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 $z_3-z_4=k(z_2-z_1)$. Upon solving it I get $k=\pm 1$. 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 $a$ and $b$, there exists only one $k$, such that $a=kb$, so them asking for two such values of $k$ is strange.

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