Jump to content

Portfolio Type I -- Patterns from Complex Numbers


Recommended Posts

hey y'all!

i'm doin this portfolio at the moment, and i came up with the conjecture, but i need to check if it's correct.. Can I post it right here?

we're not allowed to post answers on the forum.

for which part btw? I usually can be trusted on this (checking) (coincidentally I'm not doing this portfolio so I won't copy your conjecture or anything) but I haven't worked out the conjecture in part A (because I'm too lazy). if you allow me enough time I can work it out tomorrow (won't take long, just cba to do it atm) and let you know if it's correct. if you trust me, that is!

Ah, and it's for part A

Link to post
Share on other sites

  • 3 weeks later...

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.

I think that you could come up with ANY conjecture... and since N is a positive integer, it is easier to prove it using Mathematical Induction.

PS: thats a nice conjecture.. :D

Link to post
Share on other sites

I think that you could come up with ANY conjecture... and since N is a positive integer, it is easier to prove it using Mathematical Induction.

PS: thats a nice conjecture.. :D

I would disagree. they're asking us to conjecture the lengths of the line segments in part A and roots of complex numbers in part B.

Link to post
Share on other sites

I've just started the IA

I've came up with some values of the distance between the roots when n=3,4,5,6,7,8

They are decimal numbers, I'm not sure if I'm on the right track and I have difficulty in finding the pattern and forming an equation

do you guys have any advice for me??

Thanks!

Link to post
Share on other sites

I've just started the IA

I've came up with some values of the distance between the roots when n=3,4,5,6,7,8

They are decimal numbers, I'm not sure if I'm on the right track and I have difficulty in finding the pattern and forming an equation

do you guys have any advice for me??

Thanks!

Yes, you are on the right track. The roots are equally spaced around the unit circle. So the distance between any two roots is the same. Using the same method of finding the distances between roots when n = 3,4,5,6,7,8, you just need to do the same but using the variable "n". Hope I helped.

Link to post
Share on other sites

I've just started the IA

I've came up with some values of the distance between the roots when n=3,4,5,6,7,8

They are decimal numbers, I'm not sure if I'm on the right track and I have difficulty in finding the pattern and forming an equation

do you guys have any advice for me??

Thanks!

Yes, you are on the right track. The roots are equally spaced around the unit circle. So the distance between any two roots is the same. Using the same method of finding the distances between roots when n = 3,4,5,6,7,8, you just need to do the same but using the variable "n". Hope I helped.

Thanks for helping!

I've got an idea on how to do it

I have a question on proving, is it necessary to start at k=1 for mathematical induction?

Or is there any other way to prove the conjecture?

Link to post
Share on other sites

I've just started the IA

I've came up with some values of the distance between the roots when n=3,4,5,6,7,8

They are decimal numbers, I'm not sure if I'm on the right track and I have difficulty in finding the pattern and forming an equation

do you guys have any advice for me??

Thanks!

Yes, you are on the right track. The roots are equally spaced around the unit circle. So the distance between any two roots is the same. Using the same method of finding the distances between roots when n = 3,4,5,6,7,8, you just need to do the same but using the variable "n". Hope I helped.

Thanks for helping!

I've got an idea on how to do it

I have a question on proving, is it necessary to start at k=1 for mathematical induction?

Or is there any other way to prove the conjecture?

I did not prove the conjecture using mathematical induction so I don't know. But you can prove it algebraically (analytically): doing the same as you did with numbers but with variables.

Edited by bomaha
Link to post
Share on other sites

  • 2 weeks later...

In the generalisation in part B i used induction to justify the generalisation! My portfolio's marked already..just had to work on my grammar and language i.e. make it more maths-y i guess! But other than that its fine! Btw is 35 pages too long for a portfolio? Mine's 35! probably coz I used 2 approaches for each length..geometric and algebraic!

Link to post
Share on other sites

hi guys

i just my portfolio task and i am kind of stuck on the factorisation for z^n - 1 for n =3 ,4 and 5

i have read through your post and i do feel kind of dumb that i still don't get it :(

my teacher suggested first using de Moivre and then putting your solutions into your factorisation.

it sort of failed. eighter i used a wrong approach or my solutions with De Moivre were wrong

Can you guys please help me?

Thanks

Link to post
Share on other sites

if

p><p>z=...?

express 1 in cis form. and then use de Moivre's Theorem.

but then for the factorisation, I think you literally need to factorise it. like, for n=2 you can do:

p><p>

(z+a+bi)(z+c+di)=0

then expand and just compare the coefficients. of course in this case you can spot that it must be (z+1)(z-1) but in other cases it's more difficult and hence you'll need to do that ^

Edited by Desy Glau
Link to post
Share on other sites

  • 2 weeks later...

ok... a random example.

when n=1, L=0

when n=2, L=1

when n=3, L=4

when n=4, L=9

when n=5, L=16

...

so after doing some calculation and working etc you can get L=n²-2n+1. you can do matrices or graphing, anything (perhaps algebraically since you're in HL) to get a conjecture.

Hi there, im currently doing this IA, and need to get it finished asap

I came up with a conjecture for finding the length of the distances between 2 adjacent vertices, of an n sided polygon for zn. when plotting the roots zn on an argand diagram in autograph, it looks like a n sided regular polygon, with angles 2 π/n or (360/n) ⁰ between the vectors from the origin to each n vertices. See pic url for the plots.

http://imageshack.us...otsofunity.png/

My conjecture, after checking with measurements from autograph, works when tested with more n values. However, i need to prove that for n being a positive integer, that z n gives a regular polygon with n vertices, so angles between are equivalent to 2 π/n or (360/n) ⁰. Yes we can visually see that on the graph, but how do we algebraically prove this?

Any help is incredibly appreciated. For anyone out there wanting an answer, Please dont ask me to give you my conjecture, as i spent ages on this IA.

Hello, what software did you use to graph this

Link to post
Share on other sites

Do we have to follow the bullet points on the portfolio? Can we mix up to a certain way that is neater to me and it makes more sense?

Also for example, in my case I believe that there was no need for factorizing z^n-1=0 for n =3,4,5

Can we skip it if there was no use for it in my portfolio.

Link to post
Share on other sites

Do we have to follow the bullet points on the portfolio? Can we mix up to a certain way that is neater to me and it makes more sense?

Also for example, in my case I believe that there was no need for factorizing z^n-1=0 for n =3,4,5

Can we skip it if there was no use for it in my portfolio.

Just do it, even if it has no use in your portfolio. You don't want to lose some points on something that is so easy.

Link to post
Share on other sites

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.

Guest
Reply to this topic...

×   Pasted as rich text.   Paste as plain text instead

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

×
×
  • Create New...