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# Another complex number question watch

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1. Bit of help pls.

Question

In an argand diagram, the complex numbers zz1=x1+iy1 and z2 = x2+iy2 are represented b the points A and B respectivel. Prove that the omplex number z3 (where z3 = 5z2 - 4z2) and represented by point C in the Argand diagram lies on the line passing through A and B.

So I've tried to come up with a line using points A and points B (and y-y1=m(x-x1).

I've then tried to put C into the equation to show that everything works - although I can't get it to.

2. Try treating them as vectors. The condition you need is
3. I can understand that the vector approach would work - but is this the best method.

Surely my method would work-out, unless there's a quick step somewhere that I'm missing.
4. Well, I think it's a lot better than your method. Though of course yours would work. Essentially you're finding the equation of the line AB, then the equation of the line AC, and showing they're the same line, right? What equations did you get for them? Alternatively, you're finding the equation of the line AB and showing the point C lies on it - in which case, what equation did you get for AB?
5. Okay - that one's done.

I'm now trapped here, I can't for the life of me get the rearrangement - I think I'm missing something at the start, as I get z = just 2 brackets.
Attached Files
6. Complex Question.doc (23.0 KB, 280 views)
7. I presume you multiplied through by the denominator, collected the z terms on the LHS and divided stuff?

Do that, then multiply the fraction on the RHS by the conjugate of the denominator.
8. that makes sense - many thanx, it works
9. Sorry, but I'm still unsure of my answer to the first one.

See the attachement for what I've done - but to be honest, I'm now convinced that I've got it correct. Am I not just proving 1 = 1
10. Forgot this
Attached Files
11. Complex Question.doc (34.5 KB, 83 views)
12. Of course. That's what you wanted to happen, you wanted both sides to be equivalent. You didn't set it out very nicely, but it's what you wanted to happen.
13. What do you mean I didn't set it out nicely?

I thought I did - although I'm tryig to use equation editor on a mac, and I want my pc keyboard back

If you would lay it out differently - what would you do? and is what I've done enough to get the marks? It just seems a bit easy.
14. (Original post by martinkings)
What do you mean I didn't set it out nicely?

I thought I did - although I'm tryig to use equation editor on a mac, and I want my pc keyboard back

If you would lay it out differently - what would you do? and is what I've done enough to get the marks? It just seems a bit easy.
It is easy.

Think about it. If you have the equation of a curve f(x, y) = 0, how do you prove that the point (a, b) lies on the curve? You show that f(a, b) = 0 too. If it doesn't equal 0, that point doesn't lie on the curve.

Putting this into context, you had the equation of a line in some form, involving an x and a y and a load of constants. You put x = 5x_2 - 4x_1, y = 5y_2 - 4y_1, and showed that that equation still held. Put it this way - you were either going to get 0=0 out of it (in which case, the point (5x_2 - 4x_1, 5y_2 - 4y_1) was on that line), or 0=1 (in which case that point clearly wasn't on that line, because if it was, you would've ended up concluding something we know to be false).

As a simpler example, let's take the line y = x + 1. We know the point (3, 4) is on the line, because LHS = y = 4 = 3+1 = x+1 = RHS. But we know the point (4, 4) isn't on the line because LHS = y = 4 = 3+1 x+1 = RHS. There's a break in the logic. This is setting it out nicely. All you've done is the logical equivalent:

(3, 4):
y = x + 1
4 = 3 + 1
0 = 0 - true => point lies on the line.

(4, 4):
4 = 4 + 1
0 = 1 - false => point doesn't lie on the line.

But you can see why I'm slightly uneasy with the way you set it out, right?

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