207 replies to this topic

[sprintdominator]

Hey there, has any one gotten a mathematics hl type 1 portfolio on roots of unity, complex numbers?

[sheldonxp]

Yes I'm doing that, but I can't figure out how to prove the conjecture. Any ideas?

[Desy Glau]

is this the newest task?? (for 2013)

can I have a PDF of it? or just scan the task paper.. I can try to help you guys

[DDuino]

Hello Guys...i have got to start my type one maths HL...is anybody having it for complex numbers?if yes let me know how are you doing-finding it.

[DDuino]

yeah i am struggling as well with it...i dont know as well...any ideas out from it???thanks guys...

[Wide Eyed Wanderer]

I did the one involving patterns in linear equations - but I can help you if you attach a pdf or something (:

[Desy Glau]

[Kaiwen]

Oh my, mine is dued in 2 days and I still haven't decide on which to do yet ;P mainly because I don't even know what's the Moivre theorem

[Desy Glau]

the Moivre's Theorem:

(rcisθ)ⁿ = rⁿcis(nθ)

Edited by dessskris, May 13, 2011 - 06:42.

[Homira Rezai]

can someone please give me some hints on the following question... its HL maths

A). use de moivre's theorem to obtain solution to the equation Z^3-1= 0.


(I suck at maths HL)

thank you, help will be appreciated!

[bomaha]

what does (generalize and prove your results for zⁿ = a + bi, where /a+bi/ = 1) mean?
It's bullet point no. 11.

How can I generalize for it, because /a+bi/ = 1, means that zⁿ = 1 or -1 or i or -i.

Thanks in advance. I need help urgently.

[timtamboy63]

It's essentially just generalise the roots of unity. So what are the roots of unity for zⁿ? If you've done the other parts, I can't see this being too hard. Proving it would be harder, but definitely possible.

Remember that 1 = cis(2k pi)

Edited by timtamboy63, Jun 01, 2011 - 14:41.

[bomaha]

It's essentially just generalise the roots of unity. So what are the roots of unity for zⁿ? If you've done the other parts, I can't see this being too hard. Proving it would be harder, but definitely possible.

Remember that 1 = cis(2k pi)

The generalizing was easy, but I couldn't prove the conjecture for the distance between roots and the generalization for the roots of z^n = a+bi, where /a+bi/ = 1

How can you prove them?

I heard that the conjecture can be proved analytically but I don't know what does analytically mean.

Edited by bomaha, Jun 01, 2011 - 16:22.

[timtamboy63]

I'll give it a go when I get some free time, but I suspect it might have something to do with

cis(x) = e^ix

Edited by timtamboy63, Jun 02, 2011 - 13:10.

[Desy Glau]

sorry for not replying to your PM. analytically means algebraically. you found the distance for every value of n using cos, right? now you do EXACTLY the same thing, just with the variable n instead of numbers.

[DDuino]

Bomaha,

generalizing.. basically what you do you dont need to prove really...you just put down an example and you should be fine...think of the magnitude od a+bi when it equals 1...so that if you have done the other parts it should be very easy to come to the conclusion..the unit circle thing helps a lot for the answer...if you are still not ask again...

[bomaha]

I have a question about pullet point no. 6 (Factorize zⁿ-1 for n = 3, 4, and 5.)

To what extent do I need to factorize?

[DDuino]

factorize always leaving a bracket (x-1)...so that one of the answers will always be 1...and see whether you can get it...this might help you to structure your conjecture

[Desy Glau]

you need to factorise it to get a multiplication of n linear functions.
so when n=3 you need to get (z+a)(z+b)(z+c)=0

[Gnick Wright]

Hi guys,
I was wondering what sort of conjectures you could form from this? And how to prove such conjectures. My conjecture was that as N's value approaches infinity, the area of the polygon formed by the vectors between the points approaches pi. Thanks.