Our final 2021 NHT WitCH is the final question from the Specialist Exam 1.
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
The key issue, of course, is how (and whether) to make sense of the notation when is possibly complex.** In general we have two “roots” from which to choose and, unlike the case when is a nonnegative real number, there is no natural way to choose. That leaves us with three possible approaches for other :
a) Don’t define at all;
b) Concoct some rule to define to be one of the two choices of “root”;
c) Define to be both possible roots.
We’ll go through each approach in turn, considering both , and whether and how the definition or non-definition fits in with the Specialist curriculum.
With approach (a), if is real and non-zero then and are simply undefined. So, makes no sense except for 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 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 th roots (p 82), but the focus is upon the “roots” of equations, roots of and so forth. Cambridge, for example, only ever employs the notation when is real, and where a natural and fortuitous 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 ? 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, makes sense and one can hunt for solutions. Taking the principal roots to have nonnegative real part, for example, we can manipulate in the manner that was presumably intended, beginning by squaring both sides:
Noting the conjugates and the difference of two squares on the left, we can simplify to
Shifting the 2 to the right, dividing by 2 and squaring once more, we get
Expanding the square, this easily gives a = 0 and a = 3. Checking these possibilities, we have already noted that a = 0 is not a solution to . However, and keeping in mind the choice of principal root, a little polar arithmetic shows that a = 3 is a solution:
So, with a particular and very non-VCE choice of principal root, makes sense and has a solution. Are we therefore justified in going that route? Hell no.
A mathematician would never write something like , 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 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
Alas, standard examples such as show that will not always hold true, no matter how the principal root is defined. Now, as it happens, 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 is why 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 to stand for both square roots. This creates ambiguity since, for example, might equate to 2 as usual, or to ±2. In practice, the meaning of for a positive real is usually clear in context.
Of course, this double-valued interpretation of is not part of VCE, and thus the exam question is screwed. We’ll now consider whether can nonetheless be solved with this double-valued interpretation.
Since 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, above is equating all possible roots on the left with all possible roots on the right. And, so interpreted, is indeed always true.
Similarly, 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 holds true, we can manipulate just as above. It follows, just as above, that the only possible solutions are a = 0 and a = 3. Of course a = 0 still fails, but what about a = 3?
Plugging a = 3 into the left side of gives us four quantities in total, corresponding to the two choices of roots (and noting we are told that is a (single) real number). Precisely, the left side of evaluates to .
What about the right side of ? Well, one might want to stretch the resulting to be double-valued, but it most definitely is not four-valued. That is, plugging in a = 3 does not give left = right.
In summary, under approach (c), 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.
***) Even then, Cambridge stuffs it up.
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.