IA (HL) Portfolio Type I -- Patterns from Complex Numbers

[dessskris]

- generalise your results (the solutions, NOT the distance between solutions) for z^n=a+bi where the magnitude of that complex number is 1, or z^n=cis(θ). generalise means you observe from your previous investigations and then conclude.

- prove it using de Moivre's theorem

- work out the solutions to z^n=r*cis(θ)

[K.Hosny.M]

- generalise your results (the solutions, NOT the distance between solutions) for z^n=a+bi where the magnitude of that complex number is 1, or z^n=cis(θ). generalise means you observe from your previous investigations and then conclude.

- prove it using de Moivre's theorem

- work out the solutions to z^n=r*cis(θ)

Thanks for your support but actually i didn't got the point

please re-phrase your statements so that i can understand what do you mean

one more thing, please give me an example to clarify the point

thanks again

[dessskris]

omg... should i learn your language and speak in it? i thought it was clear enough!

first point: random example of generalisation:

when n=1, z=2

when n=2, z=8

when n=3, z=18

when n=4, z=32

when n=5, z=50

so, z=2n^2

second point: it's proving, I CANNOT show you how because it means I'm giving you the answer. anyway it's very simple!

third point: working out solutions means solving. like when you need to work out the solutions to y=(x-1)(x+2), the answer, or the solution, is x=1 and x=-2.

I can't understand why you didn't understand the words I used. it's either my english is so bad I can never explain anything or you did NOT think about those words carefully and only skimmed through my post.

[K.Hosny.M]

omg... should i learn your language and speak in it? i thought it was clear enough!

first point: random example of generalisation:

when n=1, z=2

when n=2, z=8

when n=3, z=18

when n=4, z=32

when n=5, z=50

so, z=2n^2

second point: it's proving, I CANNOT show you how because it means I'm giving you the answer. anyway it's very simple!

third point: working out solutions means solving. like when you need to work out the solutions to y=(x-1)(x+2), the answer, or the solution, is x=1 and x=-2.

I can't understand why you didn't understand the words I used. it's either my english is so bad I can never explain anything or you did NOT think about those words carefully and only skimmed through my post.

Thank you

Now i got your point

One last thing

the generalizing and proving i have to find a relation between and (z)

right ?

[dessskris]

yeah find the formula/equation of Z in terms of n.

Edited November 7, 2011 by King Glau

[syked]

Hi I'm having a problem with point 5 in part A.

It says to repeat the steps for the equations z^4-1=0 and z^5-1=0.

How do I draw the lines?

It was straightforward for z^2-1=0 which only produces three points. But for z^4-1=0 and z^5-1=0?

Do I take one point and join it to all other points?

or

Do I take one point and join it to the ones adjacent only?

or

Do I join all points adjacent to each other to form a polygon?

The conjecture will obviously vary by which step I take. Hopefully I am allowed to ask this...

Thank you

[dessskris]

it's basically the same thing. I don't get why people don't know how to do it, it's just like using de Moivre's theorem!! isn't that very straightforward?? if you have z^3=something you have 3 roots, what about z^4 and z^5? obviously you have 4 and 5 roots respectively!! join only adjacent roots, to form polygons.

[syked]

it's basically the same thing. I don't get why people don't know how to do it, it's just like using de Moivre's theorem!! isn't that very straightforward?? if you have z^3=something you have 3 roots, what about z^4 and z^5? obviously you have 4 and 5 roots respectively!! join only adjacent roots, to form polygons.

Oh ok thank you!

I just thought that the "instructions" were very vague

[chocomuncher]

hello!

When proving the conjecture, can we just use de moivre's theorem? and how do you use graphing software to test your conjecture...?

thanks

[chocomuncher]

Hi I'm having a problem with point 5 in part A.

It says to repeat the steps for the equations z^4-1=0 and z^5-1=0.

How do I draw the lines?

It was straightforward for z^2-1=0 which only produces three points. But for z^4-1=0 and z^5-1=0?

Do I take one point and join it to all other points?

or

Do I take one point and join it to the ones adjacent only?

or

Do I join all points adjacent to each other to form a polygon?

The conjecture will obviously vary by which step I take. Hopefully I am allowed to ask this...

Thank you

Hi I am very confused about this as well. I joined the adjacent points to form a polygon, however I am not sure what kind of conjecture are we supposed to form...? :/ the value of z with changes in n? or the length of the polygon? or the area? In part B, it does not ask you to join the points, so I am not sure whether it has anything to do with the polygons at all.

[chocomuncher]

- generalise your results (the solutions, NOT the distance between solutions) for z^n=a+bi where the magnitude of that complex number is 1, or z^n=cis(θ). generalise means you observe from your previous investigations and then conclude.

- prove it using de Moivre's theorem

- work out the solutions to z^n=r*cis(θ)

thanks for the explanation, but do you have to explain why the magnitude is 1, and what happens when it is not 1....? like the last bullet says.

[dessskris]

if you want a good mark, yes. you don't need to explain in detail though. just mention it briefly.

[Matricaria]

- Use de moivre's theorem to obtain solutions for z^n=i for n=3, 4 and 5.

- Generalize and prove your results for z^n=a+bi, where |a+bi|=1.

- What happens when |a+bi|is not equal to 1

Can somebody help me with that?

[iHubble]

There is already a complete topic with many pages about "Patterns from complex numbers". Before posting, you should use the search bar at the top right of your page buddy! You'll find info about your task here.

[dessskris]

of course not, if you were asking for direct answers.

1) literally use de Moivre's theorem. I bet you've learnt it before, unless your teacher sucks! even if you haven't, I bet you have a textbook in which you can find the theorem. even if you don't, I bet you know how to use google to find this theorem. convert the number into cis form first.

2) generalise your results for z^n=cisθ and prove it using de Moivre's theorem again

3) solve z^n=r*cisθ

I swear to God I've explained the exact same thing in the thread linked by iHubble, so if you think the thread is not helpful it's either you haven't read the whole thread or you only skim through it.

I don't like re-explaining things, so next time please read the whole thread CAREFULLY first before asking.

[Matricaria]

of course not, if you were asking for direct answers.

1) literally use de Moivre's theorem. I bet you've learnt it before, unless your teacher sucks! even if you haven't, I bet you have a textbook in which you can find the theorem. even if you don't, I bet you know how to use google to find this theorem. convert the number into cis form first.

2) generalise your results for z^n=cisθ and prove it using de Moivre's theorem again

3) solve z^n=r*cisθ

I swear to God I've explained the exact same thing in the thread linked by iHubble, so if you think the thread is not helpful it's either you haven't read the whole thread or you only skim through it.

I don't like re-explaining things, so next time please read the whole thread CAREFULLY first before asking.

I didn't skim through the topic, I did read it carefully!

I don't have a problem using de Moivre's, and I'm done with part one as a whole...

It's just that I don't know how to apply generalization in this case, our teacher didn't explain that in class, and I couldn'r get any help outta the textbook!

Here's my solution for n=4 as an example:

z^4 - i = 0

==> z^4 = i

==> z = i^(1/4).

So this is equivalent to trying to find the 4 fourth roots of i.

In polar form, we see that:

i = cos(π/2) + i sin(π/2).

By De Movire's Theorem:

i^(1/4) = cos[(π/2 + 2πk)/4] + i sin[(π/2 + 2πk)/4]

==> i^(1/4) = cos(π/8 + πk/2) + i sin(π/8 + πk/2) for k = 0, 1, 2, and 3.

Thus:

k = 0 ==> z = cos(π/8) + i sin(π/8)

k = 1 ==> z = cos(5π/8) + i sin(5π/8)

k = 2 ==> z = cos(9π/8) + i sin(9π/8)

k = 3 ==> z = cos(13π/8) + i sin(13π/8)

[dessskris]

z^4 - i = 0

==> z^4 = i

==> z = i^(1/4)

rofl...

just write something like:

z^4 = i

z^4 = cis(θ/2)

z = ...

and continue from there

don't write i^(1/4) = ...

[dessskris]

I don't care what others think about me, I've got swagger.

you should put z=something because you're asked to solve for z. you don't put i=something because they don't ask you to express i in cis form or something. they ask for z not i.

just ask if you have any other question.

[Ziad Murad]

hey 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?

[dessskris]

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!