WitCH 74: Aargh!

November 10, 2021

4:48 pm

complex numbers, exams, polynomials, Specialist Mathematics, VCAA, VCE, WitCH

This one comes from the 2021 Specialist Mathematics Exam 2 (not yet online). The general discussion on the exam is here, where some concerns on this question have been raised. It’s bad enough to be a WitCH, however, and so here we are.

UPDATE (14/11/21)

The main issue, of course, is that Part (a) is completely stuffed. Telling us that one root of a cubic is real and that the other two are complex tells us absolutely nothing about the complex roots. In particular, the three roots may still all be real, and anything, meaning there is simply no “relationship” to determine between any two roots. Specifically, Part (a)(ii) has two solutions, one of the form $\boldsymbol{\left(z^2 +9\right)(z-a)}$ and one of the form $\boldsymbol{\left(z^2 - 9\right)(z-a)}$ .

There is plenty more in the question to criticise, but let’s stop to ask: how can such a screw-up occur? How can these screw-ups continually, predictably occur? There is simply no way that a competent mathematician would read Part (a) and not immediately see that the question was stuffed. The logical conclusion is that VCAA does not have its exams vetted by competent mathematicians. Which is insane. The exams should be written by competent mathematicians. VCAA’s monumental indifference is inexcusable.

Now, the other criticisms, which are not minor, but only appear minor in comparison to the Everest of idiocy we’ve just described.

Let’s begin at the beginning, with the godawful subscripts. One sentence in and already the question is a muddy mess. Why the hell not call the roots of the polynomial α and β and γ, like normal human beings? Oh, yeah, because you clowns decided to use α and β and γ as the polynomial coefficients. Why? Notation matters and you should always have a reason for using one form or the other. But, anyway, call the roots a and b and c, or u and v and w. Whatever. But give the damn things some proper damn names. Why is this so hard?

Don’t write $\boldsymbol{\alpha \in R}$ . It is pompous, it is unnecessary and it is confusing. Just state that the polynomial has real coefficients or something.

If you are asking for the “relationship” between two numbers then you are almost certainly asking the wrong question. In this instance you have forced yourself to use cutely meaningless language because you are testing – falsely, as it happens – a triviality.

Writing $\boldsymbol{\left|z_2+z_3\right|=0}$ is pointless and way, way too cute. Test proper mathematical thought, not the detection of cheap tricks.

Seriously, you can’t think of a single coherent complex numbers question worth 9 marks?

OK, so you have two, boring, questions in one. Why on earth refer to the complex number in the second question as $\boldsymbol{z_4}$ ? Remember that thing about notation mattering?

The second question is a mess. There is no particular reason for students to hunt for an exact imaginary intercept in Part (b), nor to assume that the intercept is anything nice. If you want it, ask for it.

Part (c) then requires the imaginary intercept you didn’t ask for, and the absence of which you’ll probably blame on the students in your typical smugly incompetent manner. The imaginary intercept is apparent, but it is unclear whether this was required (even though it was required in 2(c)).

What a mess.

UPDATE (09/05/22)

The exam is here and, finally, the exam report is here. Of course the report simply lies about the answer to 2(a). It eventually notes there is “an alternative solution” , but this is way too little, way too late and way too dishonest. It is still unclear what detail was expected in the answer to 2(b).

24 Replies to “WitCH 74: Aargh!”

tom says:

November 10, 2021 at 5:50 pm

Hah, I didn’t realise this during the exam. Is the problem that $(z_2,z_3)=(3i,-3i)$ and $(z_2,z_3)=(3,-3)$ both fit the criteria?

I’m also slightly peeved at the use of $R$ rather than $\mathbb{R}$ (and similarly for other well-known sets), but that’s probably personal pedantry more than anything.

Reply
Sai says:

November 10, 2021 at 5:59 pm

An interesting title. My first thought when I looked question was “oh, they want us to recognise that $z_2 = \bar{z_3}$ as that leads to the quadratic expansion $(z^2 - 2\text{Re}(z_2)+|z_2|^2)$ which should lead to real $\alpha, \beta, \gamma$ .” Then, I realised if $\text{Im}(z_2) = \text{Im}(z_3) = 0$ , the real parts can take whatever value they want, for instance $(z-z_1)(z-1+0i)(z+1+0i)$ . However, it makes me wonder what they expect in the following question, which asks to determine the coefficients, as $z_2 = 3$ and $z_3 =-3$ also satisfy the equations. I’m not sure if any of what I’ve pointed out seems erroneous, but it really feels like they’ve let slip something that shouldn’t be there.

Reply
1. marty says:
  
  November 10, 2021 at 6:01 pm
  
  I give good title. Sai, can you suggest a specific polynomial that indicates what’s wrong?
  
  Reply
  1. Sai says:
    
    November 11, 2021 at 12:28 pm
    
    I’ve been beaten to it, but another polynomial satisfying the properties of a ii would be $p(z) = \frac{1}{5} (z-3) (z+3) (5 z-23)$ , with a separate set of values $\alpha, \beta, \gamma$ . Unless I misinterpreted the question… While I’m here (and in the other thread), I’ll plug the itute solutions, which came to the consensus of interpreting $z_2,z_3$ to be purely imaginary: https://www.itute.com/wp-content/uploads/2021-VCAA-Specialist-Mathematics-Exam-2-Solutions.pdf
    
    Reply
    1. John (No) Friend (of VCAA) says:
      
      November 11, 2021 at 1:20 pm
      
      Sai, you want (5z + 3) not (5z – 23) (you’ve used p(2) = +13).
      
      To answer the question “Do these solutions count?”
      
      I’m sure the Report will have some sanctimonious weasel-worded reason why they don’t. I’ll be interested to see what it says – especially since (a)(ii) does not have a “Hence determine …”.
      
      If VCAA has any decency it will say something – still weaselly, but less weaselly – like:
      
      “It was expected that students would assume that z2 and z3 were not real and hence the relationship between them is that they were complex conjugates of each other. However, since it was reasonable to think that z2 and z3 could also be real, a variety of answers were accepted for (a)(i). In (a)(ii) it was expected that students would use part (i) to find … However, the answers … when z2 and z3 are real were also accepted.”
      
      Are you listening VCAA?? Here’s a way to save face (one of your two faces, anyway).
      
      More interesting to me is whether itute will amend its solutions, and whether the MAV solutions (written by Asses) will explicitly discuss the two possible solutions. From what I’ve seen in the past, the MAV will bury its head in the sand.
      
      Reply
      1. Sai says:
        
        November 11, 2021 at 2:08 pm
        
        Thank you for the correction, I am eternally scatterbrained. It’s an infuriating conclusion, either way, damned if you do, damned if you don’t (include the solutions) because it feels dishonest to say that they “don’t count”. The fundamental annoyance I have with this question is that complex to them potentially implies *non-zero real part*, and if it doesn’t then they have a major issue to resolve.
        
        Reply
John (No) Friend (of VCAA) says:

November 10, 2021 at 6:41 pm

Q2(a)(i): If z2 and z3 are real then there’s no relationship between them. That’s why I think VCAA attach different meanings to C depending on the context. In this context VCAA clearly intend C to mean non-real numbers so that z2 and z3 can be complex conjugates of each other – the ‘expected’ answer. Otherwise the question makes no sense. But do they REALly intend this meaning …?

There’s actually a contradiction of meanings within the question itself! The first usage of $\displaystyle z \in C$ clearly allows $\displaystyle z_1 \in C$ . Therefore $\displaystyle R \subset C$ is intended. But the second usage clearly does not intend this.

Then there’s (a)(ii), which allows the solution $\displaystyle z_2 = 3$ and $\displaystyle z_3 = -3$ or vice versa:

$\displaystyle f(z) = \left( z + \frac{3}{5} \right) (z - 3) (z + 3)$

and the values of $\displaystyle \alpha$ , $\displaystyle \beta$ and $\displaystyle \gamma$ are readily computed.

Clearly z2 and z3 are not complex conjugates of each other in this case and the polynomial invalidates any prior question asking for $\displaystyle the$ relationship between them.
But …

The ‘VCAA solution’ we see in the Examination Report will use the ‘expected’ relationship from part (a)(i). In other words, part (ii) will require that z2 and z3 are NOT real and so $\displaystyle R \not\subset C$ is intended.

Further crap is left for others to discuss.

Reply
Alasdair the Annoyed says:

November 10, 2021 at 8:34 pm

A typical example of inexactness and sloppiness just where precision is not merely a virtue, but necessary. (Actually, when is precision not a virtue? I can’t think of anywhere.) To define $z_1\in\mathbb{R}$ and $z_2,z_3\in\mathbb{C}$ is bad enough at the start, since $\mathbb{R}\subset\mathbb{C}$ as has been pointed out. Since all reals are complex numbers by definition, the arbitrary distinction leads into a noisome mess.

What they should have done, maybe, was something like defining $z_k=x_k+y_ki$ with $y_1=0$ and $y_2,y_3\ne 0$ . This is in fact what they meant – so why not write it? Or they could have gone without their precious question a(i) and defined $z_1,z_2,z_3$ to be all complex numbers with $z_2$ and $z_3$ being complex conjugates.

I haven’t looked at (b) or (c).

And as usual there’s a confusion about what is actually being tested, and why. How vital is it to know that the roots of a polynomial with real coefficients occur in complex conjugate pairs? I get the feeling that the modus operandi of VCAA is a sort of “let’s make it more complicated and see if we can catch them out” which is bad pedagogy, and worse sense. Questions like these are muddled, confusing, and mostly pointless. And they reinforce the notion that mathematics is about a heap of disconnected tricks.

Reply
John (No) Friend (of VCAA) says:

November 11, 2021 at 6:57 am

Since things are a little slow here, I’ll mention the preamble to part (b):

“Consider the $\displaystyle point$ $\displaystyle z_4 = \sqrt{3} + i$ .” (My empahsis).

Since when is a complex $\displaystyle number$ a point? The $\displaystyle representation$ of a complex number (on an Argand diagram) is a point. $\displaystyle z_4$ is a number NOT a point.

“Consider the complex number $\displaystyle z_4 = \sqrt{3} + i$ .” is what $\displaystyle should$ have been said. And while I’m on my high horse … WHY the useless lines on the Argand diagram radiating from the origin?? And the concentric circles?? This is a generic polar grid that some dumbass pinhead chose from the VCAA stock images without thinking (and which the torpid vettors rubber stamped). It misleads and imposes white noise and static on what students are asked to do in parts (b) and (c). Maybe the stupid car in Question 4 can drive laps around the origin on it.

Further crap is left to others to discuss.

Reply
1. SRK says:
  
  November 11, 2021 at 7:35 am
  
  Using $z_4$ was weird – as far as I could tell, there was no connection between part a. and parts b. and c.
  
  I’m not understanding your complaint about the polar grid – how is it misleading? It seems to help with sketching the ray, because it’s easy to locate the endpoint using its polar coordinates, and then draw the ray parallel to the appropriate grid line.
  
  I was a bit puzzled by 2 marks for part c. ii. One of the few generous mark allocations in the paper.
  
  Reply
  1. John (No) Friend (of VCAA) says:
    
    November 11, 2021 at 8:39 am
    
    I find the polar grid a complete hindrance. Whatever extra assistance some might think it gives to drawing the ray (and I don’t see how it gives any extra assistance) is significantly outweighed by it getting in the way of clearly labelling the minor segment with the details needed to calculate the area. It is unnecessary white noise.
    
    I think there’s sufficient calculation required for part c(ii) to justify 2 marks. For me, 2 marks for part (b) is overly generous.
    
    Reply
    1. SRK says:
      
      November 11, 2021 at 10:48 am
      
      Yes, you’ve convinced me about c. ii. I think when I did the paper I just eyeballed the y-intercept and did a quick mental calculation that it seemed right. But obviously a student doing the exam would want to check more carefully.
      
      Reply
  2. John (No) Friend (of VCAA) says:
    
    November 11, 2021 at 12:57 pm
    
    Re: Defining z4.
    
    I’m guessing:
    1) they wanted to make the coordinates of the terminus obvious, and
    2) they didn’t want to write $\displaystyle \sqrt{3} + i$ in the ray equations.
    
    Weird is one word for it. Dumb, unnecessary, stupid, pointless … are other words.
    
    Reply
marty says:

November 11, 2021 at 9:09 am

Thanks everyone for commenting, and SRK for initially flagging the awfulness, on the discussion post. I’ll update soon, but I have a question.

Obviously the main issue here is with (a), which is just a disaster. But, along the lines people are discussing, I’m unsure what is reasonably expected for (b) and (c). Unless I’m missing something, the key to doing (c) is that the ray in (b) passes through (0,2). That it does so is not hard to see, but how are students expected to know to look for it in (b)? Is the polar grid intended to help, or be a clue? Or is this intended to be a Stupid CAS Trick? The lack of guidance in the question puzzles me.

Reply
1. John (No) Friend (of VCAA) says:
  
  November 11, 2021 at 10:25 am
  
  That’s actually a very good question, Marty!
  
  A couple of thoughts:
  
  Although the question doesn’t (but should) ask for all axes intercepts to be labelled (particularly since the y-intercept is required for part (c)(ii)), a reasonable student should realise that it’s implicitly required in order to make the gradient of the ray consistent with the scale given on the axes (that is, to get the ‘shape’ mark). And there’s several ways of getting it:
  
  1) The polar grid can help – sort of – if you want to get the correct slope of the ray. Simple geometry shows that the ray must be perpendicular to the radial line at 2pi/6, and this clearly requires the ray to pass through z = 2i. But if this is meant to be a clue to find the y-intercept, it’s a cryptic clue worthy of Batman.
  
  I think there are more obvious approaches, none of which are helped by the polar grid (in fact, I’d say the grid is a distinct hindrance), if finding the y-intercept is on your radar …
  
  2) Simple trigonometry of the right-angle triangle with vertices at i, bi (b > 1) and the terminus of the ray: There’s an interior angle of Pi/6 at the terminus, an adjacent side of length Sqrt[3] and so the opposite side clearly has a length of 1.
  
  3) It’s simple to get the cartesian equation of the line the ray lies on, and then the y-intercept is trivial.
  
  4) The ray obviously intersects the imaginary axis and a logical intercept to check is
  z = 2i. Not hard to see that it works.
  
  5) A CAS readily draws the ray and the y-intercept is obvious.
  
  But there’s no doubt about it, the question should have explicitly commanded students to find the y-intercept.
  
  Reply
  1. marty says:
    
    November 11, 2021 at 10:54 am
    
    Thanks, John. That helps. One follow-up question. You begin by saying one would expect students to give the correct intercept so as to be consistent with the scale, or the shape mark. I don’t get that. It seems to me that if you ask someone to draw a line at 150 degrees then drawing a line at roughly 150 degrees is plenty shaped enough. So, just to be clear:
    
    a) Are you suggesting it likely that students who fail to zap through (0,2) will be docked a point?
    
    b) Are you suggesting that, even given the vague wording of (b), that this docking would reasonable?
    
    Reply
    1. SRK says:
      
      November 11, 2021 at 11:30 am
      
      From what I’ve heard – but would appreciate confirmation / refutation of this – if VCAA supply a scaled axes on which to draw a graph, “key features” need to be positioned correctly, even if the question does not specifically ask for them to be labelled. So perhaps rays which don’t go through (0, 2) – even if they are pretty close to parallel with the 150° radial line – may have 1 of 2 marks withheld.
      
      I wonder what might happen if a student annotated their graph with an appropriate angle measure but their ray still didn’t quite go through (0, 2)…
      
      Reply
      1. John (No) Friend (of VCAA) says:
        
        November 11, 2021 at 11:40 am
        
        Huh! I didn’t even stop and think that the ray has to be parallel to the 5pi/6 radial line!! (Add this insight to my list above). Shows how helpful the polar grid was for me. I recant and admit that it $\displaystyle does$ help, I just didn’t see it. I hope it helped the students, I’d be interested to hear …
        
        (I still think it clutters up the diagrams, but I’ll agree that it’s swings and roundabouts).
        
        Reply
        
        Sai says:
        
        November 11, 2021 at 12:35 pm
        
        I thought the question was straightforward, identify the hole lies on $|z|=2$ , specifically the radial line one assumes corresponds to pi/6, then use a ruler to ensure its parallel to the radial line corresponding to 5pi/6.
        
        Reply
        
        John (No) Friend (of VCAA) says:
        
        November 11, 2021 at 12:46 pm
        
        And I love how the area to be found is always some pissant segment that’s almost impossible to draw. A ruler and “aids for curve sketching” are pretty much essential. Even using these aids, the area is pissant and you end up having to draw a larger and exaggerated picture to label everything. When drawn to scale on the VCAA diagrams, the required area makes a stamp look a wide open prairie.
        
        And speaking of VCAA love affairs, I’ve had a gutful of the love affair with the apting “Find the area of the segment” questions.
        
        Reply
        
        Prince Lancelot says:
        
        November 11, 2021 at 4:43 pm
        
        John,
        
        Despite MM2 and SM2 allowing “rulers” and “aids for curve sketching”, there is one prohibited item – “Mathomat”.
        
        From what I have heard from various students, at a number of exam venues, the invigilators do not allow students to bring in a pair of compasses, which I think is not right. The reason was “it is a sharp and dangerous item”…
        Without a pair of compasses, I doubt if many students would be able to draw a nice circle…
        
        In fact, I strongly believe Mathomat is much more helpful. It contains 30-60-90 degrees triangle and 45-45-90 degrees triangle, as well as all other sorts of great, accurate shapes.
        
        Reply
    2. John (No) Friend (of VCAA) says:
      
      November 11, 2021 at 11:34 am
      
      a) I don’t know. One could argue that the students will get penalised via part (c) if they don’t have their ray passing through (0, 2). In fact, I think many students will have gone back and changed their graph in (b) after starting (c)(ii) …
      
      But VCAA is petty and vindictive enough to penalise it in part (b) regardless of what happens in part (c). From the (hypothetical) Examination Report:
      
      “Many students indicated that their ray was inclined at 5pi/6 to the horizontal, but did not draw it passing through (0, 2). Students are reminded that even when the labelling of features such as axes intercepts is not required, the graph that is drawn must still be consistent with those features.”
      
      b) What a question to ask!!
      Judging from past experience, the marking scheme will require a terminus labelled with an open circle, and a labelled angle that indicated the correct direction of the ray. The direction would need to be consistent with passing through (0, 2). However …
      
      If I was forced to write a marking scheme for part (b), I’d say that the ray did not need to pass through (0, 2) (but should come ‘close’) IF part (c) (ii) was a mess. I think part (ii) will be a mess if they don’t have it passing through (0, 2), so I wouldn’t dock them in part (b).
      
      I think a marking scheme has to take a holistic approach at times, and this is one of those times.
      
      I really think the polar grid is a hindrance and a distraction in all this. It’s extraneous crap.
      
      Reply
      1. Alasdair the Annoyed says:
        
        November 11, 2021 at 11:50 am
        
        Yes, I agree – the polar grid is a hindrance and grossly misleading. And the question remains – what is actually being tested here? I’m sure that the learning outcomes, or whatever, could be assessed in a more straightforward and intellectually honest manner without all of this specious rubbish.
        
        Reply
        
        John (No) Friend (of VCAA) says:
        
        November 11, 2021 at 1:00 pm
        
        We’ve had a couple of suggestions at another Marty blog for name changes to this subject. You’re suggesting $\displaystyle Specious$ $\displaystyle Mathematics$ …? As good a name as any.
        
        Reply

UPDATE (14/11/21)

UPDATE (09/05/22)

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