[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

(i) trigonometric

*r*(cos*q*+ i sin*q*) and (ii) exponential

*re*^{i}^{q}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 3

*p*/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 2*p*. 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 5

*p*/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*z*^{3}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*z*^{3}. 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**

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*as one entity (learn to observe and recognise ‘chunks’ as one), we see that the RHS is actually in exponential polar form*^{a}*re*^{i}*, with*^{q}*r*= e*and*^{a}*q*=*b*. In line #3, we apply the exponential law (*re*^{i}*)*^{q}*=*^{n}*r*^{ n}e^{i}^{ (n}^{q}^{)}. However, we note that ` (5pi)/(4) ` is outside the required range. It is more than*p*. So we need to adjust it by subtracting 2*p*. We need to add or subtract as many times 2*p*as necessary to keep the argument between -*p*and*p*. Here, we subtract just 2*p*. [**: 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. ]**__Remark__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

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.

[

**: Because of Euler’s Formula**__Remark__*e*^{i}*= cos*^{q}*q*+ i sin*q*, the idea behind De Moivre’s Theorem (cos*q*+ i sin*q*)*= cos (*^{n}*n**q*) + i sin (*n**q*) is actually subsumed by the third formula which implies that (*e*^{i}*)*^{q}*=*^{n}*e*^{i}^{( n}^{q}^{)}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?

Yes, complex numbers are very useful, especially in exponential polar form. They are used in quantum probability computations, digital astronomical communications, and the proof of the continuum hypothesis.

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