# WitCH 118: The Chaos Factor

I was gonna go with The Sot-Weed Factor, but that was too cute a title, even for me.

We’re now getting to the VCAA-related material, which prompted this whole series. The last two sections of Nelson‘s complex numbers chapter are on factors and roots of polynomials. Below are excerpts on factorisation. (For the sake of interpretation, note that: the factor and remainder theorems are stated reasonably clearly, but of course with no hint of a proof; these two theorems are followed by two standard “worked examples”; the working of all the worked examples is painfully earnest and slow, but is close enough to correct.)

## 71 Replies to “WitCH 118: The Chaos Factor”

1. Red Five says:

So… do the authors not consider Reals to be a subset of Complex?

That is what it looks like.

I also think the fundamental theorem of algebra could have been more clearly stated. “At least one” is vague at best and if reals are not considered a subset of complex numbers then… problem. So there is a hint of contradiction here.

But as with all of these WiTCHEs, there is probably a lot more…

1. marty says:

Don’t go global. Take each statement as it comes, and figure out what it says literally, and also, if you can, what was intended.

2. Joe says:

Is it correct to say that {z^3 – z^2 – z + 1 = 0} has 3 roots? I would have said that it has 2 roots, one of multiplicity 1, and one of multiplicity 2, and that the sum of the multiplicities is equal to the degree of the equation. I’d also much rather speak in terms of zeroes of a polynomial than roots of an equation.

1. marty says:

It’s a bit of abuse of the word “root”, but it’s pretty standard to say something like “n roots, counting multiplicity”. So, the Nelson wording is in the ballpark of ok, but somewhere in left field. And yes, equations have solutions, not zeroes.

3. Joe says:

Some more thorough complaints:

The first sentence has the implication that you can only use the quadratic formula when the discriminant is negative, which is of course false.

They should say at the start what domain the coefficients are in, and what domain the solutions are in. It seems to be that they want us to treat real polynomials over the complex numbers. Fine, but they should state that.

Well, they do state it, but at the end: “For an equation with real coefficients, k, u + v, and uv must all be real.” This implies that the preceding stuff did not assume the coefficients were necessarily real: but then, we can have an equation with real roots: . This has a real root, but a negative discriminant.

They don’t state that a, b, c, and d, are necessarily all real.

Doesn’t the term “real quadratic expression” imply that the codomain of the quadratic expression is the reals? Then it seems stupid to allow for a complex domain. They should have just said “a quadratic expression with real coefficients”, or even better “a complex quadratic expression with real coefficients.”

I was under the impression that “a X polynomial” just meant a polynomial with coefficients from X, the domain could be something else.

1. marty says:

Thanks, Joe. That was a great job on Part 1. On to Part 2 …

(EDIT: Actually, checking now, I think you missed at least three errors …)

I also find it hilarious that they write endlessly about a quadratic (in x? in z?) without bothering to explicitly write the quadratic.

2. hjm says:

If you consider quadratic polynomials with complex (here with the correct meaning, i.e. in C including R as a subset) coefficients, does it even make sense to say that the discriminant (which is a generic complex number) “is negative”?

So, the confusion already starts when the text considers that case that Δ is negative without stating that we are talking about quadratics with real coeefficients.

(I also tend to disagree with Joe’s example. The discriminant of is (the complex number) . Is it “negative”?)

1. marty says:

In that first box, the text is obviously considering complex roots of real polynomials. So, I don’t think there is genuine confusion in talking about negative Δ, even if it is madness for them to not state explicitly what they are doing.

I think Joe’s negative discriminant example is fine, and very good.

4. Javert says:

– Real numbers are complex.
– “Suppose the degree of p(z) is a polynomial equation”.
– I’d be much less confused if all instances of “polynomial equation” was replaced with “polynomial” (and with the associated ‘= 0’s removed). This would considerably improve the FTA exposition (though I’m still unsure of the relevance of the last six lines there).
– Constants are arguably simpler polynomials than quadratics.

1. marty says:

Thanks, Javert. Not only would you be much less confused, but the text would be much more correct. Although I think it’d take a lot more than that to make the FTA exposition remotely stomachable.

5. Red Five says:

Complaint 1:

The quadratic formula gives solutions to a quadratic equation, it does not give the factors of a quadratic expression.

OK, you can work out the factors from the solutions easily enough, but that is not what the text says.

Complaint 2:

Using the quadratic formula in worked example 12a seems pointless and also not helpful.

I would prefer a complete-the-square approach:

Such an approach reinforces a few of the more fundamental ideas of complex numbers and avoids the issues from earlier sections concerning what might actually mean.

1. marty says:

Thanks, RF. For clarification, 12a is done in the text by completing the square. Which makes the introductory whatever-it-is even more nuts.

Yes, the QF approach is a little clumsy here, because they are factorising not finding roots. And, yes, the QF is a little dodgy here, since it spits out . But the complex root comes with an automatic ± and so, more by luck than skill, it is ok. But in any case, the introductory whatever-it-is doesn’t even use the QF. Ironically, they wade into a tedious approach to the complex conjugate thing, presumably precisely because they are scared of .

Much saner would be to revisit the derivation of the quadratic formula, along the lines of your computation above, and then spell out the negative discriminant scenario.

6. marty says:

I think it might be worth providing a quick hint/example to guide this one.

To be blunt, you WitCH-hunting guys (like almost everyone) don’t always read with sufficient care, and I want you to take *real* care with this one. This WitCH is getting to the heart of VCAA’s complex madness, and I want to not have to point this stuff out every time it occurs over the next decade. I’m trying to teach you all to read mathematics (and anything) with a proper critical eye, and to be able to write your critiques with accuracy and clarity. I would like to retire one day.

The first sentence of the first excerpt, referring to an invisible quadratic, has been hammered by Javert and RF, above. Now consider the second sentence:

Since Δ < 0, the roots are not real; they may be purely imaginary or complex.

Once once is commanded to read the statement with care, it should be immediately obvious that the statement is screwed. Not quite wrong, but very close and definitely screwed. And, it is screwed in exactly the kind of manner in which VCAA screwed up the complex exam questions.

So, now your task: find all such statements in the above excerpts that are similarly screwed.

7. Red Five says:

OK. I think I understand the mission.

TO check: is writing

Line 1 – the quadratic formula can be used to factorize quadratics when is negative.
This should read: When is negative, the quadratic equation has no real solutions, but can still be used (carefully) to find the non-real solutions.

A way to fix the problem?

1. marty says:

That’s mission creep. Rewriting every bad line is way, way, way too much.

Simply noting the bad and/or wrong lines in a clear and accurate manner is huge enough. And I’m asking in particular for you/everyone to take/make careful note of the text’s perverted and inconsistent use of “complex”. That is the fundamental awfulness behind VCAA’s exam questions.

In any case your rewording is a hell of a lot better, but: you don’t say what can be used (carefully). The sentence reads as if the equation can be used, which doesn’t make sense. So, you mean TQF and/or CTS, and this/these should be stated.

1. Red Five says:

OK, so where the book says at least one of is complex… this is crap because will always both be complex. What matters is the situation when is non-real.

Is that more along the lines you are wanting?

1. marty says:

Exactly.

8. Anothermouse says:

Nelson displays a fundamental and embarrassing misunderstanding of what the set of complex numbers C is and what the various subsets of C are. These subsets include real numbers R and non-real numbers C\R (the union of these two subsets ‘obviously’ forms C). Further, subsets of C\R can include the set of purely imaginary numbers ‘iR’ (that is, non-real numbers whose real part is zero). It is interesting or not that . It is interesting to wonder how much influence Nelson has/had over the writing of VCAA exams or vice versa (which came first, the Nelson or the VCAA? The VCAA exams seem to be the victim of a full nelson).

1. Red Five says:

Did you mean zero or the empty set in your final intersection statement?

I assume the former, but should that then be {0} (or did take away your accolades?)

1. Anothermouse says:

Thanks Red. I meant {0}, the set with 0 as the sole element. (I don’t know if my accolades were taken away or if I was careless with the code).

2. marty says:

Thanks a-mouse, although I’m not sure that it is exactly “misunderstanding”. As I remarked, there’s absolutely no consistency in how the term “complex” is used. So, it seems less an issue of misunderstanding and more simply not giving a stuff to use mathematical terms in a properly correct manner.

The pondering of your last sentence is exactly my pondering, and was the initial inspiration for this WitCHfest.

1. Anothermouse says:

They understood it here: https://mathematicalcrap.com/2024/01/11/witch-115-not-so-complex/#

But then they either forgot or, as you say, no consistency and “not giving a stuff to use mathematical terms in a properly correct manner.” Either way its lazy writing and proofreading that has diabolical consequences (for students and teachers using Nelson and possibly for the quality of VCAA mathematics exams). It will be interesting to see what changes (or lack of) are made in the next edition. (I think the on-line eBooks can be edited in real-time. That’s where I’d expect to see any changes in the short term).

1. Red Five says:

Maybe the two sections had different authors…

1. marty says:

Very unlikely.

1. Red Five says:

I’ll listen to any alternative theories that meet the evidence.

1. Anothermouse says:

It is far more likely that the chapter had a single writer than multiple writers. That’s how textbooks are written. However, it is possible that one of the current writers ‘updated’ the chapter from an older edition which may (or may not) have been written by someone else. If so, they (both) did a poor job. Regardless, the proof reader is responsible for ensuring consistency and correctness throughout the entire chapter (and the entire textbook). They failed in this task. The result is a chapter that propagates shoddy and sloppy mathematics to teachers and students (and possibly exam writers) and this has consequences.

1. marty says:

I agree with most of that, but I wouldn’t necessarily blame the proofreader. I think the text is unproofreadable.

1. Anothermouse says:

I cannot disagree. It is a very frustrating experience proofreading the unproofreadable and should never be attempted. It makes me wonder if the proofreader threw their hands up in despair and sent it back to the editor with the comment that it required significant rehabilitation (with maybe a few suggestions). And maybe was ignored (sometimes the ego of a writer can get in the way of constructive criticism).

2. marty says:

Of course they understand that the complex numbers contain the reals. But they are shit-lazy writers, and shit-lazy thinkers, and they cannot give up the colloquial use of “complex” to mean “not real”. They simply do not care that if their later use contradicts their earlier use.

Will the text be fixed? I don’t even see how the text could be fixed, even if there were some genuine desire to do so, and I am skeptical that there is any genuine desire to do.

9. Red Five says:

Complaint 2:

The third-to-last box is all crap.

The fundamental theorem of algebra is not correctly stated in the previous box, and this box then tries to prove the fundamental theorem of algebra using something resembling the fundamental theorem of arithmetic (but fails to do so by failing to mention the factorisation is unique).

I am of the opinion that stating an incorrect fact then using this fact to prove the theorem you say you were stating earlier is quite a depth of crap.

It should really be deleted to improve the section somewhat.

1. marty says:

Why do you think FTA is incorrectly stated? It’s very badly stated, but I’m not sure it’s incorrect.

But, yes, the whole FTA proof box is appalling. And pointless. Why megafuss the existence of n roots if you can’t provide even a clue why the first root exists?

1. Red Five says:

I believe saying “at least one complex root” is a tautology when the polynomial in question is non-constant.

The FTA links the number of solutions (over the complex numbers) to the degree of the polynomial.

The way the FTA is stated here, I do not believe achieves this.

1. marty says:

I’m a bit confused by what you’re saying but I think I disagree.

FTA says that any non-constant polynomial over C has a root. (That is, the coefficients are in C and we’re hunting roots in C.) That is anything but tautologous. It’s too difficult to give any good sense in high school why it is true. Then, the business of a degree n polynomial having n roots (counting multiplicity) is an easy, albeit very important, extension.

I guess you can consider FTA to mean the whole degree n thing, but it is preferable not to. So, I think the text’s statement of FTA is, although appalling, legally ok.

2. Red Five says:

To clarify: I think the statement of the FTA gets there eventually but the first sentence is wrong and the second sentence that miraculously concludes the FTA fails to link properly to the first statement.

Is this crap or just messy writing? I don’t know.

1. marty says:

1. Red Five says:

Yep. Not a problem. Still think the text could do a lot better in a lot less space.

10. Red Five says:

Worked example 16 – the example is not wrong as such and maybe giving a polynomial for which or are so obviously factors is a good place to start but it would have been nice to see some use of the rational root theorem and a discussion about whether (or why) it applies to polynomials with non-real coefficients.

This is probably more of a nit-pick than a proper complaint though as I’m sure most textbooks follow similar lines in their chosen examples.

VCAA has been known to start with “show is a solution to ” so it is at least consistent there, to some degree.

1. marty says:

You’re missing the point. You forgot your mission.

2. Anothermouse says:

My complaint about example worked 16 (a) is that there are two z^2 terms. Why?
And unfortunately I couldn’t avoid seeing worked example 15 (a). Is it my imagination or is that example just plain stupid? What working was given?

1. marty says:

Geez. I hadn’t even noticed the screwy way of writing 16a. God knows why they did it. The first line of the solution is to group the two terms.

15a they do by De Moivre, first getting the roots of -i.

I think everyone is missing, or at least not stating, an important point about the post-FTA boxes. Remember your mission.

1. Anothermouse says:

Do you mean the second sentence “Suppose the degree of p(z) is a polynomial equation ….”, which makes no sense.

1. marty says:

Me? No. Of course the “equation” shouldn’t be there. I’m talking about the message of 15 and 16. You guys have to read more carefully. (Remember what you tell your kids when they’re confronted with a word problem?)

1. Joe says:

Is the “message” that factorisation is about guessing games?

1. marty says:

Nope. By “message” I don’t mean Nelson’s intended message. I mean an unintended message.

You know when you ask a kid to read a word problem, and they ignore the preamble, which happens to have the critical information?

1. Anothermouse says:

“Factorise ….” over C.

1. marty says:

You ignored the preamble.

1. Anothermouse says:

We might have differing views on what the preamble is. My view is that the preamble (NOT the ‘Heading’) to worked example 15 and 16 is “Factorise”.

2. marty says:

But the title is important.

3. Anothermouse says:

Do you take umbrage with the title “Factorising a complex …”.
Should the title be less misleading and more precise.

4. marty says:

Will reply as a new comment.

11. Craig says:

In Witch 2 you said real numbers and imaginary numbers are not subsets of complex numbers. But isn’t that what Nelson is getting wrong? Commented about this before and still don’t quite get what’s going on.

1. marty says:

Yeah, I did and I regret it. What I wrote in WitCH 2 was technically correct, but also confusing and unhelpful. In school, it is best to think of R as a subset of C.

The issue was discussed in the comments above and below here. The point is that, formally, complex numbers are pairs of real numbers: so, done properly, 3 + 4i is really (3,4). But then the complex number 3 + 0i is not the same as the real number 3: the first is (3,0) and the second is 3. It can all be sorted, but any discussion shouldn’t alter that, functionally, the reals are a subset of the complexes.

These things matter, and it’s worth touching upon it, even in high school. Fitzpatrick and Galbraith, for example, does the usual a + bi introduction (but a million times better than the current school texts), and then has an asterisked section doing the ordered pairs stuff. They explicitly discuss the issue that (a,0) is not the same thing as a.

Cambridge, by contrast, has bugger all, and it was that bugger all that got up my nose. But I strayed too far into pureism.

12. marty says:

Absolutely I take umbrage with the title “Factorising a complex quadratic/cubic”. What precisely does the title mean?

1. Red Five says:

Factorising means to write as a product of factors (to me it does anyway).

The text seems to suggest that once you have a quadratic, use the quadratic formula to find the factors.

So if we want to factorise we (I assume) write and then extract and write and then we have a problem. Or two, depending on your interpretation of “problem”.

The alternative approach is to write and then convert to polar form and use De Moivre’s theorem but the text specifically says you can use the quadratic formula.

EDIT: WAIT… I see a possible other issue… maybe the one you were alluding to:

If a quadratic is the complex polynomial then it cannot be factorised as the factors themselves would need to be simpler polynomials.

1. marty says:

No, that’s not the point. You guys must read, and you must remember your mission.

Suppose you’re in NelsonLand. What, precisely, is a “complex quadratic/cubic”? Give examples.

1. Red Five says:

Does this have anything to do with the line “the coefficients and variable are complex numbers.”?

1. marty says:

You guys have to do work. You’re like the students who want the teacher to give all the answers.

1. Yu says:

A quadratic polynomial is a second degree polynomial. Its roots are in C. Each root may have non-zero real part and zero imaginary part, zero real part and non zero imaginary part, or both parts are non zero.

You don’t need to complex it.

1. marty says:

13. Anonymous says:

I dislike the proof that if a quadratic equation with real coefficients has one non-real root u, then its conjugate u* must also be a root. Leaving aside the fact that this follows from the quadratic formula, isn’t it more instructive to show how the equality au^2 + bu + c = 0 remains unchanged when you take conjugates? (To be fair, I’m not sure if the textbook discusses the rules for operating with conjugates, but that would seem like a big omission.)

1. marty says:

Yes, the conjugate proof here is absurd. In the next section, the text states and proves the conjugate root theorem, in the manner you describe: see WitCH 119. I think it’s fine to give a different approach for quadratics here, but, as has been discussed above, there are much more natural approaches.

14. Yu says:

I noticed two “wrongs” in Box 1 last line:
…… of the form k(z-u)(z-v)=k(x-a-ib)(x-a+ib). The factors are complex conjugates.

First, z becomes x.
Second, it is wrong to label the factors as complex conjugates. Complex numbers can be conjugates but not the factors of a quadratic polynomial.

Also in Box 1 under the heading Factorising polynomials, lines 5 and 6 read “For an equation with real coefficients, k, u+v and uv must be real.
This conclusion is for an equation instead of a polynomial.

1. marty says:

Thanks, Yu. Yes, I agree with you on all of that. Shockingly sloppy.

15. Red Five says:

(This is the start of my reply to the question above about what Nelsonites may or may not “think”):

The text talks about Real polynomials. This seems to imply two things:

1. The coefficients are all real.
2. Any solutions to are also real.

Then the text talks about Complex polynomials and makes mention of two similar (but not the same) things:

1. One or more of the coefficients need to be non-real. (Actually, the text simply says the coefficients are complex numbers which does not immediately mean that one or more of the coefficients need to be non-real)
2. One or more of the solutions to needs to be non-real. (With a condition similar to the previous that saying the variable is “complex” does not mean it cannot be real).

It seems, upon my reading anyway, that these two “types” of polynomials are mutually exclusive. However it is not difficult to find a polynomial with real coefficients that cannot be resolved into linear factors.

It could therefore be the case that a “complex polynomial” is a generalised case of “real polynomial” but I suspect this is now getting off-topic too much.

1. marty says:

Ugh! You gotta learn to focus. You have the ingredients there, but it’s all mushed together.

Just give the definition, as best you can and as clearly as you can as succinctly as you can, for a Nelsonite “complex polynomial”.

No essays. A 1-sentence definition.

1. Red Five says:

A polynomial with one or more complex coefficient.

Or: A polynomial with one or more non-real coefficient.

The choice depends on how “complex” is being interpreted at the time.

1. marty says:

Exactly. But that “one or more” is adding an extra level of insanity, and inconsistency.

1. Red Five says:

(Trying to use only one sentence): If all “real” numbers are also “complex” numbers then all “real” polynomials are also “complex” polynomials.

1. marty says:

Yes, they should be (modulo issues of whether the variable x is permitted to take on complex values). But not in NelsonLand.

1. Red Five says:

So, does “complex polynomial” have a meaning in NelsonLand?

A consistent one, that is.

I’d prefer a useful definition, but I’ll settle for consistent.

1. marty says:

Almost, but not quite. For their examples, they consistently seem to use it to mean “not all real coefficients”. But the discussion of the factor theorem and remainder theorems seems to use it to mean what it oughta mean: a polynomial with complex (including maybe real) coefficients.