WitCH 67: Rooted

August 26, 2021

12:28 pm

complex numbers, complex roots, NHT, principal value, Specialist Mathematics, VCAA, VCE, WitCH

Our final 2021 NHT WitCH is the final question from the Specialist Exam 1.

UPDATE (29/08/21)

Sigh. We really hate updating WitCHes.* We have about a hundred of these things to do, and we’re generally thinking of the updating as a task for retirement or, better, for a reincarnation. However these NHT WitCHes, and this one in particular, have highlighted a degree of confusion. So, alas, it seems worthwhile sorting out the issues while the issues are in mind.

To give the conclusion first, the exam question is screwed, for the various reasons suggested in the comments below. VCAA may attempt to weasel out of it in their standard weaselly way, but the question is undeniably screwed. Pinpointing the way(s) in which the question is screwed, however, requires care. The essential ingredients are in the comments, and thank you to all who contributed to the discussion. We’ll now attempt to write it all in a more comprehensive and more linear fashion.

We’re trying to make sense of, and hopefully solve, the equation

$\color{blue}(*) \qquad \boldsymbol{\sqrt{1+i\sqrt{a}} + \sqrt{1-i\sqrt{a}} = \sqrt{2a}\,.}$

The key issue, of course, is how (and whether) to make sense of the notation $\boldsymbol{\sqrt{c}}$ when $\boldsymbol{{c}}$ is possibly complex.** In general we have two “roots” from which to choose and, unlike the case when $\boldsymbol{{c}}$ is a nonnegative real number, there is no natural way to choose. That leaves us with three possible approaches for other $\boldsymbol{{c}}$ :

a) Don’t define $\boldsymbol{\sqrt{c}}$ at all;

b) Concoct some rule to define $\boldsymbol{\sqrt{c}}$ to be one of the two choices of “root”;

c) Define $\boldsymbol{\sqrt{c}}$ to be both possible roots.

We’ll go through each approach in turn, considering both $\boldsymbol{(*)}$ , and whether and how the definition or non-definition fits in with the Specialist curriculum.

With approach (a), if $\boldsymbol{\sqrt{a}}$ is real and non-zero then $\boldsymbol{\sqrt{1 + i\sqrt{a}}}$ and $\boldsymbol{\sqrt{1 - i\sqrt{a}}}$ are simply undefined. So, $\boldsymbol{(*)}$ makes no sense except for $\boldsymbol{{a=0}}$ , which also doesn’t satisfy the equation. In sum, the equation is either non-sensical or has no solution, take your pick. Either way, the exam question is screwed (unless they really expect students to answer “No solutions”).

Approach (a) is also seemingly closest to the intentions of the Specialist curriculum. No use of or general definition for $\boldsymbol{\sqrt{c}}$ appears in either the Victorian curriculum or the Australian curriculum or Cambridge. Complex roots do arise implicitly, in solving quadratic equations and the quadratic formula, and in $\boldsymbol{{n}}$ th roots (p 82), but the focus is upon the “roots” of equations, roots of $\boldsymbol{z^n ={c}}$ and so forth. Cambridge, for example, only ever employs the notation $\boldsymbol{\sqrt{c}}$ when $\boldsymbol{{c}}$ is real, and where a natural and fortuitous $\boldsymbol{\pm\sqrt{c}}$ is built into the scenario.***

Approach (b), amounting to defining a principal root of a complex number, is standard, is similar to defining a principal argument of a non-zero complex number, and is not remotely part of VCE mathematics. As such, the exam question is screwed.

Still, the notion of principal root exists, so can we use approach (b) to interpret and to solve $\boldsymbol{(*)}$ ? The answer is “No”, and “Yes”, and “Hell no”.

The first problem is that the principal root is not meaningful. Yes, one chooses a specific root for each complex number, but the actual choice depends upon the context. It is common, for example, to choose the principal root to have nonnegative real part. If, however, the negative real axis happens to be where the interesting stuff is going on then that choice of principal root will likely work very poorly; instead, in such a scenario one might choose the principal root to have nonpositive real part.

Still, if one makes a particular choice of principal roots then, yes, $\boldsymbol{(*)}$ makes sense and one can hunt for solutions. Taking the principal roots to have nonnegative real part, for example, we can manipulate $\boldsymbol{(*)}$ in the manner that was presumably intended, beginning by squaring both sides:

$\color{red}\boldsymbol{(1+i\sqrt{a}) + 2\sqrt{1+i\sqrt{a}}\sqrt{1- i\sqrt{a}} + (1-i\sqrt{a}) = 2a}\,.$

Noting the conjugates and the difference of two squares on the left, we can simplify to

$\color{red}\boldsymbol{2 + 2\sqrt{1+a} = 2a}\,.$

Shifting the 2 to the right, dividing by 2 and squaring once more, we get

$\color{red}\boldsymbol{{1+a} = (a -1)^2}\,.$

Expanding the square, this easily gives $\boldsymbol{{a=0}}$ and $\boldsymbol{{a=3}}$ . Checking these possibilities, we have already noted that a = 0 is not a solution to $\boldsymbol{(*)}$ . However, and keeping in mind the choice of principal root, a little polar arithmetic shows that $\boldsymbol{{a=3}}$ is a solution:

$\color{red}\boldsymbol{\aligned{ \mbox{\bf LHS} &= \sqrt{1+i\sqrt{3}}+\sqrt{1-i\sqrt{3}} \\[1\jot] &=\sqrt{2{\bf cis}\!\left(\frac{\pi}{3}\right)} +\sqrt{2{\bf cis}\!\left(-\frac{\pi}{3}\right)}\\[1\jot] &= \sqrt{2}{\bf cis}\!\left(\frac{\pi}{6}\right) +\sqrt{2}{\bf cis}\!\left(-\frac{\pi}{6}\right)\\[1\jot] &=\sqrt{2}\left( \frac{\sqrt3}2+\frac{i}2\right)+\sqrt{2}\left(\frac{\sqrt3}2-\frac{i}2\right)\\ &=\sqrt6\\ &= \mbox{\bf RHS} \endaligned\,.}}$

So, with a particular and very non-VCE choice of principal root, $\boldsymbol{(*)}$ makes sense and has a solution. Are we therefore justified in going that route? Hell no.

A mathematician would never write something like $\boldsymbol{(*)}$ , simply expecting the choice of principal root to be understood. Indeed, a mathematician would seldom write such an equation even if the principal root were understood. Indeed indeed, a mathematical would seldom write $\boldsymbol{{\sqrt{c}}}$ to denote the principal root.

The central, sad fact is that “principal root” is not a useful algebraic notion. The key property we would want from such a root is

$\color{blue}\boldsymbol{(**) \qquad\sqrt{b} \times \sqrt{c} = \sqrt{bc}\,.}}$

Alas, standard examples such as $\boldsymbol{b=c=-1}$ show that $\boldsymbol{(**)}$ will not always hold true, no matter how the principal root is defined. Now, as it happens, $\boldsymbol{(**)}$ was used in our equation fiddling above, and its application was valid. But the validity had to be checked (which we did). The very fact that the principal root cannot generally and automatically be manipulated in the manner of $\boldsymbol{(**)}$ is why $\boldsymbol{{\sqrt{c}}}$ is not and should not be used to denote a principal root.

In summary, the exam question is screwed under approach (b), even beyond VCE.

Finally, approach (c) takes $\boldsymbol{{\sqrt{c}}}$ to stand for both square roots. This creates ambiguity since, for example, $\boldsymbol{{\sqrt{4}}}$ might equate to 2 as usual, or to ±2. In practice, the meaning of $\boldsymbol{{\sqrt{c}}}$ for $\boldsymbol{{{c}}}$ a positive real is usually clear in context.

Of course, this double-valued interpretation of $\boldsymbol{{\sqrt{c}}}$ is not part of VCE, and thus the exam question is screwed. We’ll now consider whether $\boldsymbol{(*)}$ can nonetheless be solved with this double-valued interpretation.

Since $\boldsymbol{{\sqrt{c}}}$ is now double-valued, any equation involving such roots must be interpreted in a multiple-valued manner: the set of values on the left equals the set of values on the right. So, for example, $\boldsymbol{(**)}$ above is equating all possible roots on the left with all possible roots on the right. And, so interpreted, $\boldsymbol{(**)}$ is indeed always true.

Similarly, $\boldsymbol{(*)}$ now makes sense, but it must now be interpreted in a multiple-valued manner. As we shall see, there is still a slight ambiguity, but it turns out not to matter how that ambiguity is resolved.

Because $\boldsymbol{(**)}$ holds true, we can manipulate $\boldsymbol{(*)}$ just as above. It follows, just as above, that the only possible solutions are $\boldsymbol{{a=0}}$ and $\boldsymbol{{a=3}}$ . Of course $\boldsymbol{{a=0}}$ still fails, but what about $\boldsymbol{{a=3}}$ ?

Plugging $\boldsymbol{{a=3}}$ into the left side of $\boldsymbol{(*)}$ gives us four quantities in total, corresponding to the two choices of roots (and noting we are told that $\boldsymbol{{\sqrt{a}}}$ is a (single) real number). Precisely, the left side of $\boldsymbol{(*)}$ evaluates to $\boldsymbol{ \frac1{\sqrt{2}}\left( \pm{\sqrt3} \pm{i}\right)}$ .

What about the right side of $\boldsymbol{(*)}$ ? Well, one might want to stretch the resulting $\boldsymbol{{\sqrt{6}}}$ to be double-valued, but it most definitely is not four-valued. That is, plugging in $\boldsymbol{{a=3}}$ does not give left = right.

In summary, under approach (c), $\boldsymbol{(*)}$ makes sense, is the kind of equation that a mathematician may write, and has no solutions.

And, we’re done. We had contemplated closing with a sermon, of just how maddening this all is. Not just VCAA, which of course has absolutely no excuse for this nth instance of incompetence, but the whole damn thing. The whole reduction of an absolutely beautiful topic to meaningless ritual. But, we figured there’s no point. Regular readers will know what we think. And, pretty much no one else gives a shit.

*) Yeah, yeah, it’s our own damn fault. But one way or another we’re gonna pin the blame on Gladys.

**) This was discussed in some depth in WitCH 2, where we hammered Cambridge‘s idiotic introduction to complex numbers, and see also the companion WitCH 3.

***) Even then, Cambridge stuffs it up.

UPDATE (10/09/21)

The exam report has appeared. It simply assumes that (**) holds for the square roots, and goes on its merry computing way, winding up with a = 0 or a = 3. Then,

SInce a = 0 does not satisfy the original equation, [therefore] a = 3.

Idiots.

93 Replies to “WitCH 67: Rooted”

Glen says:

August 26, 2021 at 1:04 pm

Sneaky! Looking forward to seeing what others write about this one.

Reply
1. marty says:
  
  August 26, 2021 at 1:09 pm
  
  Who are you suggesting is being sneaky? Them or me?
  
  Reply
  1. Glen Wheeler says:
    
    August 26, 2021 at 5:26 pm
    
    Of course I mean that the question is sneaky!
    
    Reply
    1. marty says:
      
      August 26, 2021 at 5:28 pm
      
      I think you’re giving the writers way too much credit.
      
      Reply
SRK says:

August 26, 2021 at 2:32 pm

As far as I can tell $\sqrt{z},\; z \in C$ is undefined in VCE maths.

Also, I think the intended solution method involves $\overline{z}^n = \overline{z^n}$ which I think is true only for integer values of n.

Reply
1. John (No) Friend (of VCAA) says:
  
  August 26, 2021 at 7:31 pm
  
  Correct.
  
  Reply
Alasdair says:

August 26, 2021 at 3:20 pm

Since $a$ only occurs in the question in its square root, as $\sqrt{a}$ , why not simply write the equation as

$\sqrt{1+ia}+\sqrt{1-ia}=\sqrt{2} a$

Of course every complex number $z$ has two distinct square roots $\pm w$ where $w^2=z$ . It’s hard to know what they’re getting at here. I imagine what they expect people to do is to square both sides, then put the remaining square root on one side of the equation, square again and solve for $a$ . However, if you do that (using the given form of the equation), you get a = 0, 3. Both can be true depending on whether you take positive or negative square roots on the left.

I doubt whether VCE maths discusses the concept of the principal branch of a multivalued function, such as the complex $n$ -th root.

It could be observed that since the two expressions in the square roots on the left are complex conjugates, (since $a$ is understood to be real), then the left hand side is either real or purely imaginary. It can’t be imaginary as the right hand side is real. This means that $\sqrt{1\pm i\sqrt{a}} = p\pm qi$ from some $p$ and $q$ ….

Reply
1. Red Five says:
  
  August 26, 2021 at 3:35 pm
  
  a = 0 is clearly not a solution.
  
  Checking that any other value of a (especially a negative one…) satisfies the original equation is a lot more work than 3 marks for the whole question seems to suggest.
  
  My instinct tells me there is meant to be a simple solution here that doesn’t require checking whether values satisfy the original equation or not…?
  
  I can’t find it.
  
  I squared both sides, twice (as I would when finding an ellipse equation for example) then solved the resulting quadratic, ignoring the a = 0 solution.
  
  It felt like a lot of algebra (not a new thing for NHT papers) but then I’m not at all confident that my one remaining solution is valid either!
  
  Reply
  1. SRK says:
    
    August 26, 2021 at 3:44 pm
    
    Doesn’t $\sqrt{a} \in R$ guarantee that $a \geq 0$ ?
    
    Reply
    1. Red Five says:
      
      August 26, 2021 at 3:52 pm
      
      Hence my desire to find a way to check my working…!
      
      I’ve clearly missed something. Not sure what yet.
      
      Reply
      1. Red Five says:
        
        August 26, 2021 at 3:59 pm
        
        OK, missed a negative sign…
        
        a = 3 seems to be the only valid solution. Still really puzzled about how to check for validity.
        
        Reply
        
        Banacek Spaces says:
        
        August 26, 2021 at 4:24 pm
        
        As Alasdair mentioned, if you allow -1 for the square root of 1, then a=0 is an option.
        Though 1+i sqroot(0) simplifies to real we still have 2 roots of it?
        I rejected a=0 at my first go, applying arithmetic root to LHS.
        
        Reply
        
        Red Five says:
        
        August 26, 2021 at 4:30 pm
        
        As it is written though, and others have commented, is there not an implication in the question that it is the *positive* root?
        
        Reply
        
        Sai says:
        
        August 26, 2021 at 4:34 pm
        
        Somewhere along the line you’ll run into the expression $\sqrt{a+1} = a-1$ which yields an implicit condition that $a-1>0$ and thus $a> 1$ as $\sqrt{ \cdot } \geq 0$ .
        
        Reply
        
        John (No) Friend (of VCAA) says:
        
        August 26, 2021 at 5:20 pm
        
        Aha ha ha! That’s funny! In VCAA exams you don’t have to check for validity when there’s only one possible solution! VCAA want a solution. $a = 3$ is the only possible solution. Therefore $a = 3$ is the solution. That’s$ the VCAA test for validity!
        
        Yep, it stinks. But that’s the VCAA way.
        
        Reply
2. John (No) Friend (of VCAA) says:
  
  August 26, 2021 at 4:46 pm
  
  I agree that the $\sqrt{a}$ is cumbersome.
  
  To avoid confusion, let’s write the equation as $\displaystyle \sqrt{1 + ib} + \sqrt{1-ib} = \sqrt{2} b$ .
  
  Using the method described gives $b = 0$ and $b^2 = 3$ .
  
  Re: “Both can be true depending on whether you take positive or negative square roots on the left.”
  
  If a consistent convention is used for each square root, then $b = 0$ can be rejected. This leaves $b^2 = 3$ .
  
  Re: “I doubt whether VCE maths discusses the concept of the principal branch of a multivalued function, such as the complex n-th root.”
  
  Correct. And it’s likely the exam writers and vettors have limited understanding of this. There are two square roots of $\displaystyle \alpha + i \beta$ and there’s no good way of distinguishing which square root is meant when writing $\displaystyle \sqrt{\alpha + i \beta}$ (unlike real numbers where there is a ‘convention’).
  
  Reply
Sai says:

August 26, 2021 at 3:30 pm

Uh…. nth roots and complex number will lead to something that has multiple values, unless one implicitly assumes that the principal root is the one being used. While I’m here, is $\sqrt[n]{x}$ typically the real root of a number? From my studies of complex analysis, it seems that $x^{\frac{1}{n}}$ is favored for finding the nth roots of a number. That being said, the way I did it feels unusual, as I presumed $\sqrt{ab} = \sqrt{a} \sqrt{b}$ (which doesn’t hold over $\mathbb{C}$ ), which still led me to the answer. I’m really not sure how they wanted students to do it…

Reply
Red Five says:

August 26, 2021 at 4:41 pm

A more subtle point – what skill are they actually trying to test here?

Reply
1. John (No) Friend (of VCAA) says:
  
  August 26, 2021 at 5:06 pm
  
  RF, you know the answer! There is no Specialist skill that is being tested here except $(1 + ib)(1 - ib) = 1 + b^2$ . The contextual use of complex numbers is window dressing for a typical Yr 11 maths question in algebra. It’s a dumb question. As I’ve commented elsewhere, the exam smells of being written by people whose experience is mainly Maths Methods.
  
  Reply
marty says:

August 26, 2021 at 5:05 pm

Hands up who thinks a = 3 is a solution to the equation.

Reply
1. SRK says:
  
  August 26, 2021 at 5:46 pm
  
  *nervously raises hand* Letting a = 3 on both sides of the equation, taking the principal values of the square roots, gives LHS = RHS. Or so it seems to me…
  
  Reply
2. Red Five says:
  
  August 26, 2021 at 6:12 pm
  
  I’m going to be tricky and say I think $a=3$ could be a solution to the equation.
  
  If it is not, there are no solutions with how the question is written and how I believe VCAA thinks it should be interpreted.
  
  Of course, I don’t know what (or if) VCAA thinks.
  
  Reply
  1. marty says:
    
    August 26, 2021 at 6:30 pm
    
    Not an answer. Is a = 3 a solution, and why or why not?
    
    Reply
    1. John (No) Friend (of VCAA) says:
      
      August 26, 2021 at 6:37 pm
      
      No.
      Because the given condition is ill-defined.
      
      Reply
      1. marty says:
        
        August 26, 2021 at 7:35 pm
        
        JF, I’m not sure what you mean by “the condition”.
        
        Reply
        
        John (No) Friend (of VCAA) says:
        
        August 26, 2021 at 7:43 pm
        
        The square root stuff = $\displaystyle \sqrt{2a}$ equation/condition/restriction/constraint.
        
        Reply
        
        marty says:
        
        August 26, 2021 at 7:50 pm
        
        OK thanks.
        
        Reply
    2. Red Five says:
      
      August 26, 2021 at 6:57 pm
      
      OK, I’m going to be brave and say yes, $a=3$ is a solution.
      
      If $a=3$ then the left hand side becomes $\sqrt{2cis(\frac{\pi}{3})}+\sqrt{2cis(\frac{-\pi}{3})}$ which becomes $\sqrt{2}(\frac{\sqrt{3}}{2}}+\frac{1}{2} i)+\sqrt{2}(\frac{\sqrt{3}}{2}}-\frac{1}{2} i)=\sqrt{6}$
      
      Reply
      1. Red Five says:
        
        August 26, 2021 at 6:59 pm
        
        Now, tell me why I’m wrong…
        
        Reply
        
        John (No) Friend (of VCAA) says:
        
        August 26, 2021 at 7:35 pm
        
        $\displaystyle \sqrt{z}$ does not mean $z^{1/2}$ , $z \in C$
        
        Reply
        
        Red Five says:
        
        August 26, 2021 at 7:42 pm
        
        Fair point.
        
        So at worst, the solution $a=3$ is meaningless and at worst it is wrong?
        
        Reply
        
        John (No) Friend (of VCAA) says:
        
        August 26, 2021 at 7:53 pm
        
        It’s the question that’s wrong. The given equation/constraint/condition is ill-defined. No point trying to find an answer to an ill-defined question. Any potential ‘answer’ is pointless. In another comment I have stated the question that was probably intended.
        
        Reply
        
        marty says:
        
        August 26, 2021 at 7:55 pm
        
        It seems defined to me, or at least definable.
        
        Reply
        
        John (No) Friend (of VCAA) says:
        
        August 26, 2021 at 7:59 pm
        
        Would you consider my alternative question (tongue-not-in-cheek) as making the VCAA question definable?
        
        marty says:
        
        August 26, 2021 at 8:06 pm
        
        No. I think you’ve asked a clear question, which is also clearly not the exam question, suitably interpreted.
      2. Banacek Spaces says:
        
        August 26, 2021 at 7:12 pm
        
        The lhs evaluates to 4 different things, one of them being sqrt(6). Not the other 3. So it is 1/4 of a solution. A bit better for a=0. Then lhs = 2, -2 or 0 ( this one twice) – so a=0 is sort of a half of a solution.
        But i made this mathematics in the last 1/2 of an hour. May be quarter of an hour.
        
        Reply
3. John (No) Friend (of VCAA) says:
  
  August 26, 2021 at 6:33 pm
  
  begin{poking out tongue}
  I will put up my hand that $b^2 = 3$ is a solution to the following question:
  
  Let $z$ be a square root of $1 + ib$ , $b \in R$ . Find the values(s) of $b$ for which $\displaystyle z + \overline{z} = \sqrt{2} b$ .
  end{poking out tongue}
  
  Reply
4. Anonymous says:
  
  August 27, 2021 at 10:16 am
  
  I would say yes, as we would assume that the principle root was being used.
  Whether this is part of the study design and testable is a different question altogether. (I was under the impression that it was, but it appears that I was wrong.)
  
  Reply
5. Terry Mills says:
  
  August 28, 2021 at 12:27 pm
  
  I think that a=3 is one solution.
  
  Reply
  1. marty says:
    
    August 28, 2021 at 12:28 pm
    
    Prove it.
    
    Reply
    1. Terry Mills says:
      
      August 28, 2021 at 1:32 pm
      
      I interpreted your question to mean this. If I substitute $a=3$ into the given equation, do I get equality? The LHS involves the square roots of two complex numbers, and each of these has two values. So I have four equations to consider. If my arithmetic is correct, then one of the equations is true, but not the other three.
      
      Reply
      1. marty says:
        
        August 28, 2021 at 1:37 pm
        
        I’m not sure “solution” means what you think it means.
        
        Reply
        
        Terry Mills says:
        
        August 28, 2021 at 5:36 pm
        
        If someone gave me a complicated polynomial equation $p(x)=0$ and asked me: “Is $x=3$ a solution?” I’d evaluate $p(3)$ .
        
        Reply
        
        John Friend says:
        
        August 28, 2021 at 5:49 pm
        
        Quite right. And if someone gave me a cow, I’d milk it. But you can’t milk a bird.
        
        Reply
        
        marty says:
        
        August 28, 2021 at 6:13 pm
        
        So, evaluate p(3). You’re the one who claimed 3 is a solution. Prove it.
        
        Reply
        
        Terry Mills says:
        
        August 28, 2021 at 6:38 pm
        
        I can’t be bothered with the LaTeX, so let me explain.
        
        The RHS is $\sqrt{6}$ . I found the two values of each of the square roots of the two complex numbers of the LHS. Then I had four possible LHS. Only one of them was equal to $\sqrt{6}$ .
        
        Reply
        
        marty says:
        
        August 28, 2021 at 6:41 pm
        
        Sheesh. I’m in the middle of writing about a thousand words of update to this thing, complete with carefully TeXed equations, but you can’t be bothered …
        
        In any case, you still do not appear to understand what “solution” means.
        
        Reply
        
        Terry Mills says:
        
        August 29, 2021 at 6:13 pm
        
        Mea culpa.
        
        marty says:
        
        August 29, 2021 at 6:36 pm
        
        Culpa accepted (so to speak).
Alasdair says:

August 26, 2021 at 7:22 pm

Simple arithmetic shows that $(\sqrt{3}\pm i)^2 = 2\pm 2\sqrt{3}i$ , hence – and disregarding any cumbersome thinking about principal branches of $z^{1/2}$ – we can write

$\sqrt{1\pm\sqrt{3}i}=\frac{1}{\sqrt{2}}(\sqrt{3}\pm i})$

Add this for each sign gives the VCE result. So in answer Marty’s question – it is correct for certain branches of the square root.

Reply
1. marty says:
  
  August 26, 2021 at 7:34 pm
  
  That is also not an answer. We have an equation, and it is suggested that $a=3$ is a solution to that equation. Is that true?
  
  You may regard the “equation” as undefined, or you may clarify how the terms in the equation are defined. But once and if the terms are defined, then a = 3 is either a solution or it is not.
  
  Reply
2. John (No) Friend (of VCAA) says:
  
  August 26, 2021 at 7:41 pm
  
  But not for other branches of the square root … So, overall, it’s not correct.
  
  The question VCAA intended (but didn’t know they intended) is what I tongue-not-in-cheek posted earlier.
  
  Reply
  1. marty says:
    
    August 26, 2021 at 7:42 pm
    
    John, no “overall”. Is a =3 a solution of the equation or not?
    
    Reply
    1. John (No) Friend (of VCAA) says:
      
      August 26, 2021 at 7:55 pm
      
      I’ve already answered (no) and said why.
      
      Reply
      1. marty says:
        
        August 26, 2021 at 8:01 pm
        
        Yep, sorry, I missed that (although I disagree with your reason). But I’m hammering every “kinda”, “sorta” “overall” qualifier I see.
        
        Reply
Alasdair says:

August 26, 2021 at 7:25 pm

My last command needs editing, but I can’t seem to do it. It’d be nice if there was a preview function for LaTeX before posting…

Reply
1. marty says:
  
  August 26, 2021 at 7:28 pm
  
  Done. Also, I’m not sure why you can’t edit. You should have an hour after posting to do so.
  
  Reply
marty says:

August 26, 2021 at 7:38 pm

People pondering this mess may wish to read this WitCH.

Reply
1. Glen says:
  
  August 27, 2021 at 12:15 am
  
  Awww :(.
  
  This one was really fun while it lasted.
  
  Reply
  1. John (No) Friend (of VIT) says:
    
    August 27, 2021 at 9:53 am
    
    And it all counts as PD for ticking the Vampire’s box.
    
    Reply
    1. Glen says:
      
      August 27, 2021 at 10:41 am
      
      That one went way over my head! Sorry JF.
      
      Reply
      1. John (No) Friend (of VIT) says:
        
        August 27, 2021 at 11:11 am
        
        A Quick Question About the VIT
        
        Reply
      2. Red Five says:
        
        August 27, 2021 at 11:28 am
        
        Victorian teachers once per year (around this time) have to login to the VIT “portal” and tick two boxes: 1. Have you undertaken at least X days of teaching in the last 12 months? 2. Have you undertaken at least Y hours of professional learning in the last Z months?
        
        Then you pay them and they let you teach for another 12 months.
        
        If you’re really lucky, they ask you for more details on what your PD actually was…
        
        Reply
        
        John (No) Friend (of VIT) says:
        
        August 27, 2021 at 11:55 am
        
        And if you’re on a really lucky streak, they make you wait 7 months or more before approving your re-registration and sending out your Registration Card.
        
        Reply
        
        Glen says:
        
        August 27, 2021 at 12:58 pm
        
        What a scam!
        
        Reply
        
        Terry Mills says:
        
        August 27, 2021 at 11:51 pm
        
        You can list your PD on the VIT website.
        
        Reply
        
        John (No) Friend (of VIT) says:
        
        August 28, 2021 at 7:19 am
        
        TM, only a greenhorn or a fool would do that. Some reasons (there are many others):
        
        1) Why expend unnecessary effort? Even if it’s only small. Especially for these bloodsuckers.
        
        2) Why $volunteer$ information that might get used against you? (“What, you thought X was valid PD? OK, time for us to justify our gratuitous existence by demanding more details and other such nonsense).
        
        3) The next thing you know, you’re getting spam for all sorts of PD. How secure is $any$ information given to these bloodsuckers …?
        
        4) Why give them the dumb idea of making this compulsory? If one teacher can do it, let’s make all teachers do it.
        
        I live by a very simple maxim: Never volunteer information to anyone. And if it’s compulsory that you do, give the bare minimum, kicking and screaming.
        
        Reply
        
        Terry Mills says:
        
        August 28, 2021 at 2:50 pm
        
        I have heard teachers say that, when asked what PD they have done, they cannot find the information. So I put the information, with details, on VIT. Whether VIT regards my efforts as valid PD is up to them.
        
        Reply
        
        John (No) Friend (of VIT) says:
        
        August 28, 2021 at 3:19 pm
        
        There’s a simple solution: A signed Statutory Declaration stating the PD done and the dates when it was done.
        
        And if the Vampires don’t like that, they can instigate legal action and be made fools of. I think in the current circumstances, public sentiment is very much in favour of teachers and very much against bureaucratic, officious, megalomaniac statutory authorities.
        
        Reply
        
        Terry Mills says:
        
        August 28, 2021 at 6:45 pm
        
        I can’t resist stealing the ball. When I was applying to be a teacher, I had to fill in a stat dec for something. I went downtown to have my signature witnessed and wandered past a new legal office in town. Why not go to the top? I explained to the young lady behind the desk that I wanted someone to witness my signature as I signed the document. The receptionist explained that the lawyer was not in the office. In any case, the lawyer was new to town, and did not know anything about my background, and so would not be able to witness it. Anyway, she explained, you will need a police check to become a teacher, and so you won’t need a stat dec for anything. So I went to the pharmacist across the road; done!
Red Five says:

August 27, 2021 at 7:44 am

OK, follow-up question to the audience:

How do you think VCAA will mark this question?

I’m guessing VCAA has at least one solution in mind and so is looking to see that to pay the final answer mark…?

(My guess is two marks for working, one for answer)

Follow-up question: if Marty (or any interested party with a better Mathematics knowledge than mine – not hard to find) was to write to VCAA and explain the wrongness of this question, would they do anything about it?

Reply
1. marty says:
  
  August 27, 2021 at 8:02 am
  
  Good questions, RF. I’m flat out today but will try to update the post tonight, to clarify what is really going on.
  
  Reply
2. John (No) Friend (of VCAA) says:
  
  August 27, 2021 at 9:50 am
  
  Hi RF.
  
  Re: “How do you think VCAA will mark this question?”
  
  VCAA’s refusal to disclose their marking scheme (probably out of fear that it would expose VCAA to litigation) means that only the Assessor knows for sure. However, one can (pointlessly) speculate (this does not represent my own professional opinion of the question, it represents my best guess for how VCAA will mark the question):
  
  M1: $\displaystyle 1 + \sqrt{1 + i\sqrt{a}} \sqrt{1 - i\sqrt{a}} = a$ .
  
  M1: $\displaystyle 1 + a = (a - 1)^2$ .
  
  A1: ‘Correct’ answer a = 3.
  
  Re: “if … any interested party … was to write to VCAA and explain the wrongness of this question, would they do anything about it?”
  
  No. VCAA has a proven track record for denying the existence of any error on an exam once the exam has been sat. The best that happens with November exams is that the Examination Report $might$ get amended to include a very dilute acknowledgement of an error (this can be a fight that takes over a year, or it might happen much sooner if someone high enough in VCAA takes an interest). Then VCAA $might$ provide a mealy-mouthed excuse for the question. Since the Examination Report for the NHT exams is particularly pauce, nothing would be said. VCAA will never admit when it is wrong. At $best$ , VCAA would provide a tepid mealy-mouthed excuse for the question.
  
  Reply
3. Glen says:
  
  August 27, 2021 at 11:07 am
  
  Hi RF! I think Marty kind of gave it away with the link to the earlier WitCH, but anyway the essential problem (from where I sit anyway) goes back to the way the square root is defined and what the meaning is behind this equation. At least from my perspective, any appeal to a “principal branch” of the square root is like saying that we can integrate pretty much any function we find just because later the notion of Lebesgue measure (may) be introduced and we will see that the Lebesgue integral equals the Riemannian integral whenever the latter is defined. So we can just pretend like our Riemannian integration works for things that aren’t e.g. continuous.
  
  The square root has a meaning to these poor kids and it hasn’t got anything to do with a principal branch. We should not be allowed to use any “further” mathematics, or definitions that haven’t been made, etc. I think that’s the way the question should be treated.
  
  …. Yes, I’m still resisting writing up anything specific! Haha :).
  
  Reply
  1. Red Five says:
    
    August 27, 2021 at 11:33 am
    
    Yeah… I figured something along those lines (Measure Theory didn’t cross my mind though!)
    
    My question remains though, if there IS NO SOLUTION, how the can VCAA mark as right or wrong anything a student writes?
    
    3 marks out of 40 is not inconsequential either in a world where 1% can be the difference between making the cut for University course X and not making it.
    
    Ultimately, it is the school/teacher that will cop the blame for a student underperforming though…
    
    …hence part of me cares less about what the CORRECT answer is to a question like this and more about what VCAA CONSIDERS CORRECT at the time of marking the damn thing.
    
    Reply
    1. John (No) Friend (of VCAA) says:
      
      August 27, 2021 at 12:15 pm
      
      RF, I’ll give you a money-back guarantee that:
      
      1) VCAA sees no problem with the question, and
      2) the $VCAA$ marking scheme and answer are very similar to what I’ve posted.
      
      We’ll know (at least the VCAA answer) in a few years time when the 2021 Examination Report gets published.
      
      You’re correct that “Ultimately, it is the school/teacher that will cop the blame for a student underperforming though…” But I doubt a teacher or student will see a problem with the question, so ‘all is well’. No harm, no foul …
      
      However, pity the NHT teacher(s) that did not tell their Specialist Maths student(s) that discrete random variables (including binomial random variables) and continuous random variables (not restricted to normal random variables) are implicitly examinable (as occurred in the 2021 NHT Exam 1 Question 3 and Exam 2 Question 5 (a) and (b)). This is an ambush that’s been lying in wait since probability and statistics was first imposed onto the Specialist Maths curriculum in 2016. This is what will really screw those students over (particularly the ones that did Maths Methods the previous year).
      
      Reply
      1. Red Five says:
        
        August 27, 2021 at 1:57 pm
        
        I suspect the probability questions will be a future blog post.
        
        VCAA will defend the question by saying that knowledge of Methods is a requirement of Specialist.
        
        Or say nothing at all.
        
        Reply
        
        marty says:
        
        August 27, 2021 at 3:29 pm
        
        Hi RF. I had pondered a separate post on the probability questions but have now decided otherwise. I’ll write a general post on the NHT exams, in the next day or so, and I’ll quickly hammer the probability nonsense there and then.
        
        Reply
        
        John (No) Friend (of VCAA) says:
        
        August 27, 2021 at 4:01 pm
        
        Hi RF. We can talk in the past tense because VCAA was contacted. The response was $almost$ as you predicted:
        
        1) Knowledge of Methods (Units 1 – 4) is a requirement of Specialist Units 3-4. I don’t need to point out that this is a big disadvantage for students who did Maths methods the previous year, and that none of the relevant formulae are on the Formula Sheet.
        
        2) Knowledge of Specialist Units 1-2 is a requirement of Specialist Units 3-4. The implication is that short topics/content such as
        i) Sequences and Series (and we all know that none of the relevant formulae are on the Formula Sheet),
        ii) Proof by Induction
        are potentially examinable.
        
        The implications of 1) and 2) are very troubling. None of this is explicit in the Stupid Design. The VCAA response is clearly bullshit. But the NHT Exams have let the genie out of the bottle and so the response must be taken seriously. Whether VCAA would get away with this sort of ambush (and that’s exactly what it is, an ambush) on the November exams is debatable. But if the same idiots are writing the November exams …
        
        Anyway, I’ll defer to Marty posting further on this lunacy.
        
        In fairness to VCAA, its response occurred in a timely way. No doubt because someone higher than a subject manager facilitated the enquiry.
        
        Reply
        
        Red Five says:
        
        August 27, 2021 at 4:14 pm
        
        All I will say is this:
        
        Nowhere does VCAA say that Methods 3&4 (as a prerequisite or co-requisite for Specialist 3&4) has to be done in the same calendar year as Specialist 3&4.
        
        Theoretically, someone who took the Methods 3&4 exam in 2015 could sit the Specialist 3&4 exam in 2023. This would have (theoretically – we live in hope) been two study designs ago.
        
        So… will you only assess skills that have been on every iteration of the study design since the creation of the rules surrounding prerequisites???
        
        If not, the question is hideously unfair, even if it is defendable based on VCAA logic.
        
        Reply
        
        marty says:
        
        August 27, 2021 at 4:20 pm
        
        The question cannot be defended, even using VCAA logic. But please leave it for now. I will post on it soon.
        
        Reply
      2. Terry Mills says:
        
        August 28, 2021 at 12:01 am
        
        Re: Q3: I don’t like the phrase “binomially distributed random variable”. In all my years, I have never seen that expression.
        
        Reply
        
        marty says:
        
        August 28, 2021 at 8:17 am
        
        Terry, as I suggested to the others, please save it. I will post generally on the NHT exams soon.
        
        Reply
  2. SRK says:
    
    August 27, 2021 at 12:35 pm
    
    Glen, completely agree. While I’m happy to defend my view that a = 3 is a solution (on one suitable disambiguation of the question), it’s completely unfair to expect students to answer the question with that in mind. (In fact, I have explicitly told my students to NOT use the notation $\sqrt[n]{z}$ when z is complex, because of this issue. Fortunately my students don’t sit the NHT exams, but seeing a question like this come up is concerning).
    
    Reply
Alasdair says:

August 28, 2021 at 6:54 pm

The trouble is that the expression is not actually an equation because the LHS isn’t unambiguously defined. If we were dealing with real numbers, then we’d be OK, as you can unambiguously define $\sqrt{x}$ to be the value $y\ge 0$ for which $y^2=x$ , and so for example an equation such as

$\sqrt{x-2} + \sqrt{x+3} = x-1$

has meaning and can be solved. But when we are working with complex numbers, as here, the symbol $\sqrt[n]{z}=z^{1/n}$ is shorthand for the set of all values $w$ for which $w^n=z$ , and there are always $n$ distinct such $w$ .

In the example given, if we put $a=3$ we find that

$(1\pm i\sqrt{3})^{1/2} = \pm (\sqrt{3/2}\pm i\sqrt{1/2})$

and so the LHS actually evaluates to four different values:

$\pm \sqrt{6}$ , $\pm i\sqrt{2}$ .

It is true that one of those values is indeed equal to $\sqrt{2a}$ , but in treating the original expression as an equation there is a huge abuse of notation, in trying to squeeze four values into one.

The question tries to restrict the domain by claiming that $\sqrt{a}$ must be real, but that doesn’t solve the problem of multivalued complex square roots, and choosing the one that makes the “equation” work. Because you could just as well ask for the value $a$ for which

$\sqrt{1+i\sqrt{a}} + \sqrt{1-i\sqrt{a}} = -\sqrt{2a}$

for which again $a=3$ “works”. But that would mean, comparing both equations, that $\sqrt{2a}=-\sqrt{2a}$ which it, err, doesn’t.

Thus the problem is not so much whether $a=3$ is or is not a solution, because the original expression is not an unambiguously defined equation.

Reply
1. John Friend says:
  
  August 28, 2021 at 7:19 pm
  
  lol! Thanks Alasdair. You’ve just shown mathematically that a cow is not a bird. (See my relevant post somewhere above!)
  
  Reply
Red Five says:

August 28, 2021 at 8:33 pm

I would like to take a slight detour and ask the following:

If VCAA had instead said “verify that $a=3$ is a solution.”

1. Would this make the question better or worse?

2. Would this make the question less wrong?

Both genuine questions from someone who has spent the last 24 hours re-reading lecture notes from Complex Analysis and trying to understand principal values a bit better…

…more than one bottle of wine may have been involved.

Reply
1. marty says:
  
  August 28, 2021 at 8:39 pm
  
  1. No change. 2. No.
  
  RF, in an hour or two I’ll post a detailed update, which will hopefully make sense of it all.
  
  Reply
  1. Red Five says:
    
    August 28, 2021 at 9:01 pm
    
    OK. Looking forward to it.
    
    Seriously.
    
    Reply
SRK says:

August 29, 2021 at 11:46 am

Thanks for the update Marty

Reply
Red Five says:

August 29, 2021 at 12:41 pm

Marty – there are a lot of people who read this blog who may not comment that often for one reason or another, but these WiTCHes (and most importantly those that relate to VCAA exams) are one of the main places I visit now to further my own Mathematics education (yeah I know…)

Sure, there are free websites which have correct definitions of these things and I still have all my University textbooks, but these WiTCHes really highlight *how* things can be made wrong by some level of misunderstanding in a way that books or websites that only contain correct statements/definitions/examples really cannot achieve (sure, if VCAA/Cambridge/NAPLAN/Gladys etc didn’t make as many mistakes there wouldn’t be a need for this blog but we all know that is not going to stop any time soon)

So thanks. I know I get a lot of of all this and I’m not the only one.

Reply
1. John Friend says:
  
  August 29, 2021 at 1:05 pm
  
  Hear, hear, RF. I totally agree.
  
  I’ll add that a real positive of Marty’s blogs is the culture and environment: Anyone can ask a question, everyone is treated equally, and there is a genuine freedom for open discussion. I really value the free-wheeling environment. It’s an environment where everyone can learn something. And it’s always fun!
  
  If VCAA visited (and it probably does lurk) * $and$ * was open-minded enough to *learn* from the experience, everybody would win.
  
  Thanks for the update, Marty. We all know how much time and effort it takes for such thoughts to be formulated and then clearly articulated.
  
  Reply
Glen says:

August 29, 2021 at 5:17 pm

Great update Marty.

Reply

UPDATE (29/08/21)

UPDATE (10/09/21)

Related posts:

93 Replies to “WitCH 67: Rooted”

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