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Complex numbers question (1 Viewer)

_ShiFTy_

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z1, z2 and z3 lie on a circle. The origin also lies on the circle
Prove 1/z1, 1/z2 and 1/z3 are collinear

All i could gather from this question is that:
|z1 - w| = |z2 - w| = |z3 - w| where w is the centre of the circle
 
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no_arg

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Assuming usual naming around the circle
angle 0 z1 z3=angle 0 z2 z3 (Why)
Convert above to a statement re args
show that arg(1/z1-1/z2)=arg(1/z2-1/z3)(mutatis mutandi)
 
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I

icycloud

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Construct the diameter of the circle passing through the origin and the center of the circle. Have one variable point Z. Construct 1/D and 1/Z (where D is the fixed diameter). Prove OZD is similar to O(1/D)(1/Z). Can you finish the proof from there?
 

no_arg

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It's very interesting to see such varying approaches on this one.
Anyone have any more proofs that are distinctly different?
 

YBK

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Buchanan, that's confusing :S

what do you mean with all those weird symbols..?

and your first post's initial step... where did all that come from?
 
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It's determinants.

It actually makes the problem so trivial you barely need to think about it at all.
 
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YBK

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buchanan said:
It's determinants.

It actually makes the problem so trivial you barely need to think about it at all.
are we supposed to learn that?
or can we learn that :D
 

who_loves_maths

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lol, Iruka's simple proof of the collinearity of the inverse of all points on the circle really renders your restricted and unnecessarily complex method quite amusing Buchanan.

But on the issue of other elegant proofs of this result no_arg, you can think about [hint] rotating the circle about the origin a fixed radians.
 
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Once it's been proved for 3 points it immediately follows for all points on the circle.

It's not my fault you don't know anything about determinants.

But anyone who does would know that my solution is actually quite simple. I just used a few simple arithmetical properties of complex numbers and determinants and the result easily follows.
 
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acullen

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You wont need determinants until the first semester of university maths.

But for a 2x2 matrix A
det(A)=
|a b|
|c d|
=ad-bc

And for a 3x3 matrix called say... B
det(B)=
|a b c|
|d e f|
|g h i|
=a(e*i-f*h)-b(d*i-f*g)+c(d*h-e*g)

an interesting note is that if det(X) != 0 then X-1 exists where X*X-1=I (Identity matrix).
 
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It used to be in the 4 unit course before the sentinels of mediocrity watered it down in 1980.

Anyway, no-arg wanted alternative solutions. No mention was made of having to stick to an inferior HSC syllabus. So I didn't!:)
 
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acullen

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I think we are the only state/territory in the country that doesn't teach matrix algebra at the highest high school level. Ah well, it wasn't too hard to pick up at uni.

I would dread being an Advanced Maths student though, they skip an entire year of uni maths and hit the Multivariate stuff head on (atleast at UoW). It is assumed they know all the first year content (which they obviously wouldn't if they were never taught anything about matrices, differential equations, etc). It's kind of funny, I'm finding second year maths pretty easy, and here are these adv maths students stressing over trying to understand what's going on.
 

acullen

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They are those who applied through UAC for the B.Sc Mathematics (Advanced) course and scored sufficiently well in the HSC. They do a subject called MATH110 which teaches them in 1 semester what they would normally learn in a year of first year maths minus the fundementals from high school. This is all whilst doing 2nd year mathematics subjects concurrently.

4 Unit HSC maths students can do either providing they score a sufficient UAI, but the numbers in B.Sc Mathematics are higher than the Adv course.

We also have several streams for first year maths depending on the background from the HSC. However, a B.Sc Maths student is assumed to have atleast 3 Unit knowledge so they don't typically apply to them.
 
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