## Monday, January 9, 2012

### JCCDQBHWHCN027 Complex Numbers: Cyclic Roots

 [original source unknown]
Introduction
This question is taken from a certain book sold in Singapore which did not credit the original sources.  They seem to be interested in merely making a quick buck by replicating questions set by school teachers paid on public tax dollars.  I am not going to persecute them for copyright infringement, but my peeve is: there is no way for students and educators to make proper references.  “JCCDQBHWHCN027” is just my own made-up reference.  If anyone knows school (junior college) and examination year from which it is taken, please inform me.
Anyway, I am just quoting this question for educational purposes.  Besides showing the solution, I shall illustrate the all-important metacognitive (shown in dark red) and cognitive processes (shown in blue) used.  These processes are the secrets of “maths geniuses” and are supported by extenstive international research on mathematics education.  By mimicking these processes, you can become a  “maths genius” yourself!

Part (i)
Part (i) is just a straight-forward standard “roots of unity” problem.  The method of solution is probably given in most schools’ lecture notes.  The instructions “write down” means no working is needed and credit is given for the answers only.  But I shall show the working anyway.

 figure 1 - solution to part (i)
After putting to standard form (line #2), we convert unity (i.e. the number 1  which obviously has modulus r = 1 and argument q = 0) to polar exponential form (line #3).  We know that for this quintic (degree 5) equation there are many (five, actually) solutions.  The term “+2kp” with  k  being an integer is added to the argument to facilitate the search for the five solutions.  Line #4 takes the fifth root of the modulus (which remains as 1) and the arguments (angles) get divided by 5.  If you substituted every integer value of  k,  you would find that the values of  z  repeat themselves in cycles and that there are only  5  distinct values.  To get the 5 distinct values, you can substitute any 5 consecutive integer values for k.  The most convenient ones are the (positive or negative) integers closest to 0.  So the numbers to substitute are  k = 0, ±1 and ±2 (line #5) which yield the answers in the required form (line #6).  We check that the arguments are in the correct range -p < q < p.

Part (ii) Stage 1:  Understanding the Problem
Part (ii) is a bit more challenging.  This style of questioning used to appear two decades ago, so the teacher who set this must be recycling the ideas.  In mathematics, “roots” means “solutions”.  In this context, they are the values of  z  that make the given equation true.  The instruction is “show” i.e. prove or derive.  The final expression is actually given.  Your task is to demonstrate via a sequence of logical steps how one can arrive at that expression.

Part (ii) Stage 2:  Planning the Method of Attack
Usually parts of the exam questions are linked.  Be observant.  Try to look for clues.  It is helpful to ask yourself: “How is the equation in part (ii) linked to part (i)?  What is similar?  What is different? ”

So, what is/are similar?
Part (i) and part (ii) involve quintic equations in which  z  is the unknown complex number.

And what is/are different?
Part (ii) seems to contain two “chunks” with power of 5.  [they are called binomial powers]

Idea:
Try to rearrange part (ii) equation to look like something obtained earlier in part (i).

Part (ii) Stage 3: Execution
Let’s be brave, get on with it and see what we get.

 figure 2 - trying to solve part (ii)

Shifting things around (lines #1 to #3), we managed to make the RHS equal to 1, which is a kind of standard form for this sort of equation.  Observe that there is one big chunk raised to the power of 5 that is equal to 1.  To simplify matters, we call this big chunk ‘w’ (line #4, #5).  Since the equation of line #4 is the same as part (i), with ‘z’ replaced by ‘w’, we know something about w: we know that its possible values are exactly those we found in part (i).

Part (ii) Stage 4:  Evaluation

Are we done yet?  Have we solved the problem?
No!  Remember, we are trying to find  z,  not w.  Note: the ‘z’ of part (ii) is not the same as the ‘z’ of part (i).

It seems like we need to loop back to stage 2 (planning) of problem solving.  [Do not be alarmed.  This looping back and forth is quite normal, even for experts.]

Part (ii) going back to Stage 2 again: Planning
How to link up all the clues?  We know ‘w’  and we have an equation linking ‘z’ and ‘w’.  Since we want ‘z’, we can try to express ‘z’  in terms of  w.

Part (ii) Stage 3 (again): Execution
Let’s do it!
 figure 3 - part (ii) making z the subject
This process involves secondary school mathematics, which you should be familiar by now.  After cross-multiplying (line #1), we move the ‘z’ terms to one side (line #2).  After factorizing (line #3), we throw the ‘1 + w’ term (same as  ‘w + 1’) across and down to the other side.  We arrange to make ‘z’ appear alone on the LHS and the other stuff appear on the RHS (line #4).

Part (ii) Stage 4:  Evaluation
Are we done yet?
No!  We seem to be almost there.  But the expression needs to be converted to trigonometric form.  Go back to stage 2 (planning) once again.

Part (ii) Stage 2 once again:  Planning the Method of Attack
The most natural thing to do is to substitute the known expression for  w  into the latest equation.  We would then obtain some complicated quotient exponential expression.  This can be done via complex conjugation, but it would be unwieldy.  Is there a better method?  Yes!  We can use a trick based on Euler’s Formula, which I call the ‘Half-Power Trick’.  First, let us review some useful formulas based on Euler’s Formula.
 figure 4 - useful formulas based on Euler's Formula
Here, line #1 is Euler’s Formula.  Taking conjugates, we get line #2.  If we add line #1 and line #2, we obtain the formula in line #3.  If we take line #1 and subtract line #2, we get line #4.

Part (ii) Stage 3 (again): Execution
Let’s get our hands dirty.

 figure 3 - part (ii) The 'Half-power trick'

Line #1 is obtained by substituting ‘w’.  Both numerator and denominator contain an  e2kpi/5  term, with a plus or minus 1.  We apply the ‘Half-Power Trick’ (line #2).  Half of the power (square root) of  e2kpi/5  is  ekpi/5.  Although this term    ekpi/5  is not a ready common factor, we can force it out to be the common factor by writing it outside the brackets (shown in the above figure in red).  The terms inside the brackets are then obtained by dividing the original terms by  ekpi/5, shown in brackets in line #2.  Applying the Euler-related formulas, we get line #3.  From secondary school trigonometry, you should know that sine divided by cosine gives a tangent (line #4).

Part (ii) Stage 4
Are we done yet?
Yes!  We have shown a series of logical steps that leads to the desired expression.

Part (ii) Stage 5:  Reflection

After solving a mathematics problem, it pays to do some reflection about how we solved the problem.

What did we learn in solving this problem?

This challenging problem was tackled by linking with part (i), making observations and breaking down the problem into smaller problems.  Our first sub-problem as to find the link between (i) and (ii).  We made observations by asking “What is similar?”  and  “What is different?”.  We put the equation in part (ii) to be similar to the equation in part (i) by making the RHS equal to 1.  We then observed the link: one whole chunk of the LHS in part (ii) actually corresponds to the ‘z’ in part (i).  We called this chunk ‘w’.  We then made ‘z’ in terms of this ‘w’.  This was our second sub-problem.  The last sub-problem was to substitute the ‘w’ and change the expression into a trigonometric form.  This was done using the ‘Half-Power Trick’ and Euler-related Formulas.  We thus managed to solve the whole problem successively solving sub-problems.

If you encounter a similar problem in future, what will you do?

If we encounter a similar problem in future, we can apply the same general metacognitive processes (monitoring yourself as you go through stage 1, 2, 3, 4, looping back to stage 2 a few times if necessary and finally stage 5 reflection) and the same heuristic strategies (breaking down the problem into smaller problems, asking “What is similar?”  and  “What is different?”, observing for linkages, observing chunks.  Moreover, we can use these special techniques for complex numbers: ‘Half-Power Trick’ and the Euler-related Formulas.

Thank you for your patience reading up to this point.  I hope you have learned some things that you can apply in future, so that you become a better mathematics problem solver.