complex number (1 Viewer)

[OLDMAN]

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quote freaking_out:
actually, the fact that the question is posted by "OLDMAN", turns me off straight away.
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Criticism noted. An examiner's dilemma : pose a challenge, but be fair. Only a few weeks left, time to boost confidence. Hope this one fits the bill.

w^3 =1. Prove that z_1, z_2,-wz_1-w^2z_2 form the vertices of an equilateral triangle, z_1,z_2 arbitrary complex numbers and w not=1.

 

[McLake]

nice Q (I leave it for HSC students)

 

[ND]

w^3-1=0
(w-1)(w^2+w+1)=0
w^2+w+1=0 when w=/=1
w = -w^2-1

Let A, B, C rep complx numbers z_1, z_2,-wz_1-w^2z_2 resp.

-wz_1 - w^2z_2 = -z_1(-w^2-1) - w^2z_2
= z_1 + z_1w^2 - w^2z_2
= z_1 + w^2(z_1-z_2)
= z_1 + cis(-2pi/3)(z_1-z_2)

Let D represent z_1 + (z_1-z_2)

Now on a diagram, D is the vector BA + z_1. /_BAD = 180*.

Now vector AC is vector AD rotated through 120*. So /_BAC = /_BAD - 120* = 60*.

edit: Can you show me a better way of explaining it instead of using D? I've never been good at putting these things into words.

 

[McLake]

What's wrong with the "D"?

 

[OLDMAN]

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Quote McLake:
nice Q (I leave it for HSC students)
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Thanks McLake. Though it has a distinct recycled feel to it: Q8 b 1997.

ND shows mastery of complex vectors - a topic many students assume they know, but assumption falls apart with a Q8 type problem.
Another proof can be obtained by proving sides are all equal.

 

[McLake]

yes, many people underestimate how hard Complex Numers can be. Flicking back throught the paast decade of exams, about half the Q7/8 are Complex Numebers ...

 

[ND]

Originally posted by McLake
What's wrong with the "D"?

I just wonder if there's a better (simpler) way of putting it all into words.

 

[McLake]

Originally posted by ND
I just wonder if there's a better (simpler) way of putting it all into words.

Looks pretty good to me (creating points is fine)

 

[freaking_out]

Originally posted by OLDMAN

...Another proof can be obtained by proving sides are all equal.

can u show us how u prove it that way?

 

[ND]

Originally posted by freaking_out
can u show us how u prove it that way?

Well C can be represented by both 'z_1 + cis(-2pi/3)(z_1-z_2)' and 'z_2 + cis(2pi/3)(z_2-z_1)'. You have a go at showing AB=BC=AC (remember to use the modulii).

 

[freaking_out]

Originally posted by ND
Well C can be represented by both 'z_1 + cis(-2pi/3)(z_1-z_2)' and 'z_2 + cis(2pi/3)(z_2-z_1)'. You have a go at showing AB=BC=AC (remember to use the modulii).

finally, i've actually went somewhere with OLDMAN's question (even though ND pretty much did everything). anyway, for dumb students like me here is the finishing of the above proof:

correct me if i'm wrong but u find the "other" version of C by observation of the diagram. so:

OC=z_1 + cis(-2pi/3)(z_1-z_2).........1
or
OC=z_2 + cis(2pi/3)(z_2-z_1)...........2

AC=OC-z_1
sub in equation 1
AC=|cis(-2pi/3)| |(z_1-z_2)|
= |(z_1-z_2)|

similarily: BC=|(z_1-z_2)|

AB= |z_1-z_2| (obviously )

therefore the triangle is equilateral since AB=BC=AC.

btw, would they lead us into this question, if we were to have this on the HSC?

 

[ND]

Originally posted by freaking_out
correct me if i'm wrong but u find the "other" version of C by observation of the diagram.

Actually it's from substituting w^2=-w-1 into -wz_1-w^2z_2, then subbing w=cis(2pi/3). But yeh, that's all correct.

edit: Just one thing: in an exam, i wouldn't write "similarly" in that case, cos you get BC=|(z_2-z_1)| then have to make it |-(z_1-z_2)|=(z_1-z_2).