JCCDQBHWHCN029 Complex Numbers: Exponential Polar Form

[original source unknown]

Introduction

     Just as in my previous article, we are unable to trace the original source of this question.  This question contains an interesting link between logarithms and complex arguments.  For those interested, this link will be explored deeper at university level in the topic of Complex Analysis.

Prelude Stage 1:  Understanding the Problem

What is given?

     We are given  z  which is some quotient expression with complex numbers in rectangular form.  The denominator is the square of some complex number.  The numerator is actually  -1 + i  in a “reversed” form (be careful!).

What are you supposed to do / find?

     Find the modulus and prove that the argument is ` 5pi /12 `.

     Note that exact answers (using fractions, surds, logarithms, … etc but not decimals) are required, as with most maths questions at JC level.  You cannot use your calculator to show the steps of this problem, although you could numerically check your answers if you want.

What topic / sub-topic is this question on?

     Modulus and argument of complex numbers, cyclic roots (n^th roots of a complex number), exponential (polar) form.

Prelude Stage 2:  Planning the Method of Attack

How shall we do this?  What options are available?  Is there a short-cut?

     We could expand and simplify the bottom part, and then rationalise the denominator using the technique of complex conjugates.  After further simplification, we get something in rectangular form (something like ??? + ??? i).  Then find the modulus and the argument.  This method is do-able, but it is pretty tedious.

     We could also convert everything first to polar form, which is more suited for multiplication, division and powers.  There are two flavours of polar forms:
     (i)   trigonometric  r (cos q + i sin q) and

     (ii)  exponential  re^iq

I prefer to use the exponential polar form because it is more compact and yet it contains the same information  (r and q).  Then we can use the formulas for manipulating exponential polar forms.  This seems to be the ideal method, because we can “kill two birds with one stone”: we can handle the modulus as well as the argument calculations at one go.

Prelude Stage 3:  Execution

figure 2 - converting to polar form

     One should try to sketch or mentally visualize the complex numbers  i – 1  and  Ö3 + i  on Argand diagrams.  These complex numbers form special triangles with special angles (30º, 45º, 60º) with the horizontal axis, giving exact values for the moduli and arguments.

i – 1  forms a right-angled triangle with sides 1 and 1 and hypotenuse Ö2, making 45º with the horizontal in the left upper quadrant.  The angle turned from the real positive axis is thus 135º or 3p /4  radians

Ö3 + i  forms a right-angled triangle that makes  30º  or  p /6  radians with the positive real axis.

For checking, you can use your calculator to get the arguments, but you will get decimal answers.  After dividing by  p, you may convert the result to a fraction.  Thus you can verify the exact fraction of  p  for both arguments.

Now we proceed with the exponential polar form calculations.

figure 3 - solving the prelude

     Line #1 is the original expression which we convert to exponential form (line #2) as worked out earlier.  In line #3, we do modulus and argument (i.e. angle) calculations in a single step, with the modulus on the left of the ‘e’, and the argument on the right in brackets after the ‘i’.  The subtraction ‘–’ in the argument is because of the fraction/division in previous line.  Anything that appears below the fraction line will have their arguments negated.  The  p/6  is multiplied by 2 because of the squaring in the denominator.  All these are in accordance to these powerful exponential form rules:

     ·     ` r_1 e^(i theta_1) \cdot r_2 e^(i theta_2) = (r_1 r_2) e^(i (theta_1+ theta_2))  `

     ·     ` r_1 e^(i theta_1) divide r_2 e^(i theta_2) = (r_1 r_2) e^(i (theta_1- theta_2))  `

     ·     ` (r e^(i theta) )^n = (r^n) e^(i (n theta))  ` 

     After simplifying, we obtain line #4.  We should check that ` (5pi)/(12) ` is in the range

(-p, p] the principal range for complex arguments.  If not, we would need to adjust by adding or subtracting some multiple of  2p.  Finally, we just pick out the modulus (line #5) and the argument (line #6) to answer the question properly.

Prelude Stage 4:  Evaluation

Are we done for this part?  Is there another way to do this?

     Yes, we have answered the question as required, and we did it by combining the calculation with the proof.  We would also have done these separately, but it would have been longer to write out the solution.  This solution is short and sweet.

Is the answer correct?

     We can use the graphing calculator to verify that the modulus is correct.  Calculate and store the expression as ‘Z’.  Then we subtract Ö2/4  from the modulus (or absolute value) of ‘Z’ to see if we get zero.  This would show that those two expressions are equal.  The picture below shows how you can do it using Texas Instruments TI-84.  You can do something similar using the Casio fx-9860G.

figure 4 – numerically verifying the answer using a calculator

Yes!  Got it!!  As for the proof of the argument  5p/12, we just need to check that our steps were logical.

Part (i)  -- Piece of Cake

     This cyclic roots part of the question is solved by the standard method.  It is fairly straightforward, although the numbers are a bit ugly.  I shall just present my solution here below.

figure 5 – solution to part (i)

Part (ii) Stage 1:  Understanding the Problem

What are we supposed to find?

  

     We are to find the exact values of  a  and  b,  hence almost certainly, no decimal answers are accepted.  [This is because exact answers almost always entail an infinite number of decimal places, which our calculators are not able to supply.]  Although the  z³  term looks like we are need to solve for  z (as would be the usual case), let us remember that the  z  is actually given and known.  That is why we need to read the question carefully to avoid getting ensnared.

Part (ii) Stage 2:  Planning the Method of Attack

     The question looks a bit unusual, but do not be frightened.  We observed that the RHS involves an exponential expression, which suggests the use of the exponential polar form.  As for the LHS, since  z  is known,  we can actually calculate  z³.  In what form?  Rectangular or Polar?  In exponential polar form, as suggested by the RHS and also because this form is compact and especially convenient for the calculation of powers.  Let’s try this approach and see.

Part (ii) Stage 3:  Execution

figure 6 – solution to part (ii)

     For the LHS, we use the information taken from the earlier prelude to re-express  z  in polar form (line #2).  The key observation is that on the RHS, the + in the exponential index changes to a multiplication as we split  e^a+ib  into two parts.  Treating  e^a  as one entity (learn to observe and recognise ‘chunks’ as one), we see that the RHS is actually in exponential polar form  re^iq,  with  r = e^a  and  q = b.  In line #3, we apply the exponential law  (re^iq)ⁿ = r ⁿ e^{i (nq)}.  However, we note that  ` (5pi)/(4) `  is outside the required range.  It is more than  p.  So we need to adjust it by subtracting 2p.  We need to add or subtract as many times  2p  as necessary to keep the argument between  -p  and  p.  Here, we subtract just  2p.  [ Remark: although  ` (5pi)/(4) ` is  not the same as ` -(3pi)/(4) `,  ` e^(i(5pi)/(4)) ` is  exactly the same as ` e^(-i(3pi)/(4)) `, because they refer to the same complex number i.e. they are represented by the same point on the Argand diagram.  The difference is how much you turn clockwise or anti-clockwise from the positive real axis to get to that point.  ]

     After doing this adjustment (line #4), we are all set up to compare the moduli (highlighted in green) and the arguments (highlighted in pink) on both sides (lines #5 to #7).  We are using the fact that if  ` r_1 e^(i theta_1) = r_2 e^(i theta_2) `  then  ` r_1 = r_2 `  and    ` theta_1 = theta_2 `,  provided both  q ₁  and  q ₂  lie in the range  (-p, p].  Remember that in maths, it is useful to be able to recognise chunks.  In this context, the relevant chunks are those that give the moduli and the arguments, as highlighted.  To uncover  a,  we take logarithms (line #6).  Line #8 just writes the answer nicely.

Part (ii) Stage 4:  Evaluation

Are we done?  Is the answer correct?

     Yes, we have found the values of  a  and  b  and  b  is in the correct range,  as required.  We can verify the answers numerically using the calculator.  For example, like this:-

figure 7 – checking part (ii) answers

     Take full advantage of the graphing calculator’s ability to store complex numbers into variables like ‘Z’, ‘A’ and ‘B’.

Part (ii) Stage 5:  Reflection

What did we learn by solving this problem?

     We learnt that the exponential (polar) form of a complex number is a very powerful and compact way to solve complex number problems.  It is able to handle both the modulus (distance from 0) and the argument (angle with the positive real axis) simultaneously.  It is convenient for manipulating products, quotients and powers.  We learned to apply the following rules for manipulating the exponential polar form of complex numbers.

     ·     ` r_1 e^(i theta_1) \cdot r_2 e^(i theta_2) = (r_1 r_2) e^(i (theta_1+ theta_2))  `

     ·     ` r_1 e^(i theta_1) divide r_2 e^(i theta_2) = (r_1 r_2) e^(i (theta_1- theta_2))  `

     ·     ` (r e^(i theta) )^n = (r^n) e^(i (n theta))  `

     ·     if  ` r_1 e^(i theta_1) = r_2 e^(i theta_2) `  then  ` r_1 = r_2 `  and    ` theta_1 = theta_2 `,
            provided both  q ₁  and  q ₂  lie in the range  (-p, p]

     Using the last formula, we learned to compare moduli and arguments to solve for unknowns.

     [ Remark: Because of Euler’s Formula e^iq = cos q + i sin q, the idea behind De Moivre’s Theorem  (cos q + i sin q)ⁿ = cos (nq) + i sin (nq)  is actually subsumed by the third formula which implies that  (e^iq)ⁿ = e^{i( nq)}  when  r = 1.]

     We applied metacognition (self-monitoring and self-checking) to all the 5 stages of the maths problem-solving process.  We used heuristics like making comparisons, making observations, recognising relevant chunks, comparing.  We also practised being careful by checking all our answers and verifying that we answered the question in the form required, and verifying that the complex-argument answers are within the correct range.

If you encountered a similar problem in future, would you be able to solve it?

     [Please say “Yes!”, but make sure you can do it.]

Can you explain to a friend how to solve this sort of maths problem?

Can you set a similar question for yourself or your friend to solve?

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  e^2kpi/5  term, with a plus or minus 1.  We apply the ‘Half-Power Trick’ (line #2).  Half of the power (square root) of  e^2kpi/5  is  e^kpi/5.  Although this term    e^kpi/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  e^kpi/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.