Talk:High School Mathematics Extensions/Supplementary/Complex Numbers

Please help with new content and spelling and grammar errors. I'm sick of correcting my own mistakes already!

Don't let it get you down. We all have the same problem, there are just too few of us at the moment. The page so far look really nice. Theresa knott 14:18, 3 Nov 2003 (UTC)

I hope my edits aren't unwelcome; I realize some of them are pretty major. I've been trying to order things such that they make sense to somebody who is completely unfamiliar with the concepts involved. I didn't realize I was invading someone's turf... I just got bored studying for finals, so I started clicking 'Random Page' until I found something I felt I could contribute to. :-) Glenn Willen 02:13, 10 May 2004 (UTC)


 * Don't be daft! Wikibooks is a collaboration.Theresa knott 09:04, 10 May 2004 (UTC)

Namespacing nitpick
I can see right here one of the practical problems with book-specific namespacing (and also with "breadcrumbs" like "< High School Extensions"). This page _should_ be useful for at least one of the "standard" math books in addition to HSE (which one depending on whose curriculum standards you like), but namespacing makes that awkward. Breadcrumbs are nice, but they inhibit reuse... the quickest, easiest solution I can see is to namespace more broadly (i.e. "Math:Complex number") or not at all (just "Complex number" as this was previously named), and to provide multiple sets of breadcrumbs at the top of a multiply-used page... what do people think?

Suggested To Do list
- Do you want to borrow mine? Dysprosia 09:33, 4 Nov 2003 (UTC)
 * Argand diagrams
 * Historical controversy and the "meaning" of complex numbers
 * practical uses of complex numbers eg electronics
 * Julia and Mandelbrot sets (r3m0t stares, zooms in, stares again...)

Suggested To Do list
- Do you want to borrow mine? Dysprosia 09:33, 4 Nov 2003 (UTC)
 * Argand diagrams
 * Historical controversy and the "meaning" of complex numbers
 * practical uses of complex numbers eg electronics
 * Julia and Mandelbrot sets (r3m0t stares, zooms in, stares again...)

Links you might like to take a look at

 * http://mathforum.org/johnandbetty/frame.htm A childrens book of complex numbers.

Thanks User:Xiaodai

I'm finally doing some more serious contributing!
I'm beginning to get the warm and fuzzy Wiki spirit inside me... r3m0t 00:13, 2 Jan 2004 (UTC)


 * Keep it up r3m0t! I just want to say when i drew up those plans, they are not meant to be the headings, they are there to remind me of what I want to talk about. So you might want to get creative and use your own headings, divide the chapter your way and write about things the way you desire. Don't let those plans dictate to you what you should write, or in what order you should write it. Zhuo Jia Dai

Am I right in thinking that every integer is real? In other words, is a number such as 4 + 5i an integer? r3m0t (whoops, forgot to sign, nevermind)

yes. although 4 + 5i are sometimes called Gaussian integers, we mean integers when we say integers nothing else. Zhuo Jia Dai

Some very neat work on number system there
Congratulation on some very neat work on number system there, r3m0t. Keep it up. Zhuo Jia Dai

Set theory not suitable for this level
I dont know what is taught in high school around the world. But set theory isn't taught in Australian high schoool. So one part of this chapter has to be rewritten, slightly. Xiaodai 08:59, 10 Jul 2004 (UTC)

I learned basic set theory (Venn Diagrams, sets, intersections, unions) at the age of 9. It certainly is not too complicated to be taught at this level.--206.127.177.59 21:09, 2 Apr 2005 (UTC)

Yeah, some of the set stuff is really not hard at all. But we don't learn it at all in South African primary/secondary schools. --168.209.97.34 20:13, 23 Jun 2005 (UTC) (Taejo in English Wikipedia)

DeMoivre's theorem not suitable
The student may not have had exposure to the natural logarithms at this level. It would be acceptable to include information about polar complex plane in the form z = r cos theta + ir sin theta. So a complex number in polar coordinates is like ( r, theta ). --206.127.177.59 21:09, 2 Apr 2005 (UTC)

yeah i agree, it is too hard for me (lol im crap at math anyway)

Note on curricula: In South Africa logarithms are taught to Math "Higher Grade" students in the final year. Natural logarithms are not taught in high school. --168.209.97.34 20:13, 23 Jun 2005 (UTC) (Taejo on English Wikipedia)

SOMEONE HELP ME!!!!!!!!!!!!!!!!
alright im only in year 9 so dont get too technical. theres this question that goes: find all values of a such that there are two real solutions to the equation x^3+(a-1)x^2+(9-a)x-9=0 can someone just tell me how to do cubics???

and i dunno whats calculus my teacher said i need it DONT GET TOO HARD!!!

edit: sorry its supposed to be 1 after a (thanks for telling me)

First of all, don't use all capital letters LIKE THIS, this is considered shouting in internet culture. To your question, I think either you wrote the wrong question here or your teacher made a mistakes(or you're in a pre-university class, then). At first, I thought you could take one of these approaches a12a22 &minus; 4a0a23 &minus; 4a13 + 18 a0a1a2 &minus; 27a02
 * Use the formula directly(involved matrix actually)
 * Use Vieta's Substitution to work out the solution directly.
 * Differentiate it, then the discriminant of the resulting quadratic equation should be greater than zero, and one of the solution of the quadratic should also be the solution to the cubic, use vieta again to solve it.
 * Let one of the root be k, then divide the cubic equation with x-k, the resulting quadratic should be a prefect square, that is, the discriminant of it should be zero.

Now, the trouble is that all of the method above are far too complex for a grade 9 student, and even if there is an answer, it is too clumsy..(I tried, but found out the algebra expression grows very long, I also tried using a grphic calculator, and the result seems to be an irrational number) Well, once we're at cubic things never go nicely, vieta's substitution itself are too complex already itself.--Lemontea 10:51, 14 Jun 2005 (UTC)

Edit:Ok, tried method one on a equation solver and the answer match with what I found on the graph. a= 9+2.788273389 or 9-35.48344244, now both seems to be irrational numbers...

hmm, it isnt supposed to be hard. its in this noether student problem set, for year 9 (im in australia), and all the other problems are piece of cake. can you please just give me a link to vieta's substitution? i really dont no what it is. thanks

Well, in my country Vieta's substitution isn't in the core part of the course and is in the "enrichment" part. Anyway, be perpared that it is going to be a clumsy method, as in history, it is not developed until ~1541, long after the quadratic was solved.

There are more than one approaches to Vieta, but the first step is always to eliminate the quadratic terms. Let x = y + a, where a is a constant, substitute into the equation, and find "a" such that the quadratic term's coeifficient become zero.

There're two way to go after this, the first is to let y = p + q and compare coeifficient to find out p3 + q3 and (pq)3.

The second method is to let y = z + b/z. Now find the value of b such that it reduces into quadratic, which can be solved with ease.

Hope these helps.

Opps, didn't notice your correction on the problem, now it is much easier. First, expand:

$$x^3 + ax^2 - x^2 + 9x - ax - 9 = 0$$

Then factorize:

$$x^2(x-1) + 9(x-1) + ax(x-1) = 0$$

$$(x^2 + ax + 9)(x-1) = 0$$

Now either x = 1 or x2 + ax + 9 = 0. I think you should be able to do it now. --Lemontea 08:06, 24 Jun 2005 (UTC)

yea i done it, but only found -6 and 6, never thought of when one of the roots of the binomial would be one (damn)