### Thoughts about 2016 A’levels H2 Mathematics Paper

I’ve covered some things in classes, with sufficient revisions and final lap papers set. So I thought we have a little breakdown. And of course, we should review what was weeded away in the 9740 H2 Mathematics Syllabus. After all, this is the very LAST time the can test them.

1. Recurrence Relations. I’ve harped on this last year too. Conjectures! Conjectures! You can read more about it here. An a question involving conjecture should start with a recurrence relation, then a conjecture, ending with a recurrence MI. Students should know how to do both $\Sigma$ and recurrence MI. Yes, they are different.

2. Loci. You guys are the lucky last batch to do Loci. So please buy a protractor and compass. Draw them, as one of my student put it, surgically. If need be, use a graph paper (why not?). Harder loci for example can require students to draw for example, $\text{arg}(z-1+2i) = \text{tan}^{-1}(\frac{4}{3})$. You should not have trouble measuring this angle, because you could not even be able to do it. Students should be able to draw such angles with ease. One little note about Loci, will definitely their geometrical descriptions. Many students can draw this, but stumble to describe them.

3. Vectors. Truth be told, I’m still waiting for a question involving vectors in 3D, to land in A’levels. An example can be the HCI Prelims Paper 1 Question 6, which can be found here.

4. Poisson Distribution. I don’t know what this topic has been axed. So students should ready for one big Poisson Distribution questions, I say give it 12-14 marks. And it should be tested with conditional probability. I’ll practice either Demand & Supply or Inflow & Outflow questions. An example can be the one found in NYJC Prelims Paper 2 Question 11, which can be found here.

5. Correlation & Regression. Ever wondered what the $r^2$ means in the GC? Well, $r^2 = bd$ is being removed from the syllabus as well. It hasn’t surfaced before, so maybe it shall finally make its one and only LAST presence felt this year. Students should familiarise themselves with the use of $y- \bar{y} = b ( x - \bar{x})$ equation, which can be found in the MF15. I know many of you probably have not seen it before.

6. Hypothesis Testing. Students should review definitions of level of significance and p-value. Also understand what you may conclude from a Z-Test, using the results of a T-test. A little small part that students can think about, is why use a small sample size? After all, we know that have a sufficiently large $n$ allows us to perform CLT and then use a Z-Test.

7. Trigonometry. After it appeared in 2011 for a trigonometry MI, the product to sum formulas is still a problem for most students. I highly doubt its coming out again with MI, but its can easily come out again with complex numbers. An example can be this.

More examples and discussion will be made in class.

### Trigonometry used in complex Numbers

Given $z = \text{cos}\alpha + i \text{sin} \alpha$ and $w = \text{cos}\beta + i \text{sin} \beta$

$z - w = \text{cos}\alpha + i \text{sin} \alpha -(\text{cos}\beta + i \text{sin} \beta)$

$z - w = \text{cos}\alpha -\text{cos}\beta + + i \text{sin} \alpha - i \text{sin} \beta)$

$z - w = - 2 \text{sin}(\frac{\alpha-\beta}{2})\text{sin}(\frac{\alpha+\beta}{2}) + i 2 \text{sin}(\frac{\alpha-\beta}{2})\text{cos}(\frac{\alpha+\beta}{2})$

$z - w = 2 i \text{sin}(\frac{\alpha-\beta}{2}) (\text{cos}(\frac{\alpha+\beta}{2}) + i \text{sin}(\frac{\alpha+\beta}{2}))$

$z - w = 2 i \text{sin}(\frac{\alpha-\beta}{2}) e^{\frac{\alpha + \beta}{2}}$

### Complex Number Problem #2

Show that $e^{i \theta} + e^{-i \theta} = 2 cos \theta$. Hence show that $\mathrm{cos}^3 \theta = \frac{1}{4} (\mathrm{cos}3 \theta + 3 \mathrm{cos} \theta)$

For the first part, we can simply apply Euler’s Formula, that is $e^{i \theta} = \mathrm{cos} \theta + i \mathrm{sin} \theta$

$e^{i \theta} + e^{-i \theta}$
$= \mathrm{cos} \theta + i \mathrm{sin} \theta + \mathrm{cos} (- \theta) + i \mathrm{sin} (- \theta)$
$= \mathrm{cos} \theta + i \mathrm{sin} \theta + \mathrm{cos} \theta - i \mathrm{sin} \theta$
$= 2 \mathrm{cos} \theta$

The next part is a little more tricky, and since its hence, we will use what we solved previously to help us.

$e^{i \theta} + e^{-i \theta} = 2 cos \theta$
$\Rightarrow cos \theta = \frac{1}{2}(e^{i \theta} + e^{-i \theta})$

$\mathrm{cos}^3 \theta$

$= (\mathrm{cos})^3$

$= [\frac{1}{2}(e^{i \theta} + e^{-i \theta})]^3$

$= \frac{1}{8}(e^{i \theta} + e^{-i \theta})^3$

$= \frac{1}{8}(e^{i 3\theta} + 3 e^{i 2\theta}e^{-i \theta} + 3 e^{i \theta}e^{-i 2\theta} + e^{-i 3\theta})$

$= \frac{1}{8}(e^{i 3\theta} + 3 e^{i \theta} + 3 e^{-i \theta} + e^{-i 3\theta})$

$= \frac{1}{8}\{\mathrm{cos} 3\theta + i \mathrm{sin} 3\theta + 3(\mathrm{cos} \theta + i \mathrm{sin} \theta) + 3[\mathrm{cos} (-\theta) + i \mathrm{sin} (-\theta)] + \mathrm{cos} (-3\theta) + i \mathrm{sin} (-3\theta) \}$

$= \frac{1}{8}(\mathrm{cos} 3\theta + i \mathrm{sin} 3\theta + 3\mathrm{cos} \theta + 3i \mathrm{sin} \theta + 3\mathrm{cos} \theta - 3i \mathrm{sin} \theta + \mathrm{cos} 3\theta - i \mathrm{sin} 3\theta)$

$= \frac{1}{8}(\mathrm{cos} 3\theta + i \mathrm{sin} 3\theta + 3\mathrm{cos} \theta + 3i \mathrm{sin} \theta + 3\mathrm{cos} \theta - 3i \mathrm{sin} \theta + \mathrm{cos} 3\theta - i \mathrm{sin} 3\theta)$

$= \frac{1}{8}(\mathrm{cos} 3\theta + 3\mathrm{cos} \theta + 3\mathrm{cos} \theta + \mathrm{cos} 3\theta)$

$= \frac{1}{8}(2\mathrm{cos} 3\theta + 6\mathrm{cos} \theta )$

$= \frac{1}{4}(\mathrm{cos} 3\theta + 3\mathrm{cos} \theta )$

Just for fun…

$2e^{\frac{5\pi}{6}i} + \frac{1}{2e^{\frac{5\pi}{6}i}}$
$= 2e^{\frac{5\pi}{6}i} + \frac{1}{2}2e^{\frac{-5\pi}{6}i}$
$= 2 [\mathrm{cos}(\frac{5\pi}{6}) + i \mathrm{sin}(\frac{5\pi}{6})] + \frac{1}{2 } [\mathrm{cos}(\frac{-5\pi}{6}) + i \mathrm{sin}(\frac{-5\pi}{6})]$
$= 2 (\frac{-\sqrt{3}}{2} + i \frac{1}{2}) + \frac{1}{2} (\frac{-\sqrt{3}}{2} - i \frac{1}{2})$
$= -\sqrt{3} + i + \frac{-\sqrt{3}}{4} - i\frac{1}{4}$
$= -\frac{5\sqrt{3}}{4} + \frac{3}{4}i$

$e^{x+yi} = (1+i)^6 = [\sqrt{2}e^{i\frac{\pi}{4}}]^6$
$e^{x+yi} = 8e^{i\frac{3\pi}{2}}$
$e^x \times e^{yi} = 8 e^{i\frac{3\pi}{2}}$
$\Rightarrow e^x = 8 \Rightarrow x = \mathrm{ln}8 = 3 \mathrm{ln}2$
$\Rightarrow y = \frac{3\pi}{2} - 2\pi = \frac{-\pi}{2}$

Complex Number Problem #1

### Complex Number Problem #1

I’ve come across many questions from students regarding complex numbers, and here is one thats quite fun to deal with!

$\sqrt{-1}^{\sqrt{-1}}$

Now this is really interesting, but can be solved with a bit of manipulation. We all know this is akin to $i^i$

We know that $\sqrt{-1}=e^{i\frac{\pi}{2}}$

$\sqrt{-1}^{\sqrt{-1}}$

$= e^{i\frac{\pi}{2}\sqrt{-1}}$

$= e^{i\frac{\pi}{2}i}$

$= e^{i^2\frac{\pi}{2}}$

$= e^{-\frac{\pi}{2}}$

Now our end result is a real number!

Like mentioned with regards to Euler’s Identity, it is really very amazing how complex numbers actually work!

### 2011 A-level H2 Mathematics (9740) Paper 2 Question 1 Suggested Solutions

All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(i)

(ii)
$OC = \sqrt{2^2+5^2} = \sqrt{29}$

Max. $|z| = \sqrt{29}+3$

Min. $|z| = \sqrt{29}-3$

(iii)
Max. $|z-6-i| = \sqrt{(2-6)^2 + (2-1)^2} = \sqrt{17}$

This question is standard tutorial question. Students should not struggle to be able to produce that triangle to apply your trigonometry formulaes. If you have problems drawing a good diagram to scale, I recommend students consider using a graph paper in exams.

### Solving roots of higher order

This is one question that I know ALL my students can excel in doing, that is solve $z^4 = -16$. So I’m not interested in showing you how to solve such problems, but I want to explain a particular step which you introduce in your working. So let’s take a quick look at the solution first.

$z^4 = -16 = 16 e^{i\pi}$

$z^4 = 16 e^{i(\pi + 2k\pi)}$ for $k = 0, \pm1, -2$

So here, we note that we introduced $e^{i(\pi)} = e^{i(\pi + 2k \pi)}$, but how is $e^{i\theta} = e^{i(\theta + 2k\pi)}$?

Intuitively, $2k\pi$ for $k \in \mathbb{Z}$ is simply full circle. So we are really just turning full circles about the same point here, which is why we are still referring to the same number.

I will do a simple mathematical proof here too.

$e^{i(2k\pi + \theta)}$

$= cos(2k\pi + \theta) + isin(2k\pi + \theta)$

$= cos(2k\pi)cos\theta - sin(2k\pi)sin\theta + i[sin(2k\pi)cos\theta + cos(2k\pi)sin\theta]$ (using formulas in th MF15)

For $k \in \mathbb{Z}, cos(2k\pi)=1 \mathrm{~and~} sin(2k\pi)=0$

$\therefore, e^{i(2k\pi + \theta)} = e^{i\theta}$

### My Favourite Pure Mathematics Topic

Complex Numbers would definitely is one of my favourite topics in Pure Mathematics. Its a pity that H1 Mathematics will not to get to learn them.

Firstly, Complex Numbers is being used very widely in Engineering School, heads up. So its usefulness in real life is beyond just $z=x+iy$. The idea of represent two components into one is quite amazing. And because of these, many A-level questions are combining vectors with complex numbers and expect students to resolve them.

Secondly, the geometrical interpretations in Complex Numbers are really interesting. And yes, I do admit that there are times whereby examiners overkill the questions and expect students to see some unconceivable relationships in the Argand Diagram.

Thirdly, finding solutions for a mere $x^2 = -1$ is really helpful to exploring many channels in mathematics and open the floodgates to many possibilities in Mathematics. It allow us to solve equations, previously unsolvable. Of course, A-level students should be quick to relate the Conjugate Root Theorem here and expect questions combined with graphings, polynomials to come out to test their understanding on this theorems.

The last reason is quite nerdy. I find it really amazing how $1 = e^{i0}$. Yes many students will probably go meh. But here, we established a relationship between a real number, complex number and zero. Zero, itself is intriguing enough. And now we link these three components together into one equation. 🙂

### 2011 A-level H2 Mathematics (9740) Paper 1 Question 10 Suggested Solutions

All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(i)
$z^2 = -8i$

$z^2 = 8e^{i(-\frac{\pi}{2}+2k\pi}$

$z = 2\sqrt{2}e^{-i\frac{\pi+4k\pi}{4}}$

$z = 2\sqrt{2}e^{-i\frac{\pi}{4}}, \mathrm{~or~} 2\sqrt{2}e^{i\frac{3\pi}{4}}$

$z = 2 - 2i, ~ -2+2i$

(ii)
Observe we let $z = 2w+4$, then

$(2w+4)^2 = -8i$

$4w^2 + 16w + 16 + 8i = 0$

$w^2 + 4w + 4 + 2i = 0$

$\Rightarrow w = \frac{-4+2-2i}{2}, \mathrm{~or~} \frac{-4-2+2i}{2}$

$\therefore, w = -1-i, \mathrm{~or~} -3+i$

(iii)

(iv)
Both perpendicular bisector are parallel to each other, so there will be no point of intersections.

Students should be cautious to leave their answers in cartesian form in (i). For (ii), students can solve by otherwise methods, that is, to use the quadratic formula $\frac{-b\pm \sqrt{}{b^2-4ac}}{2a}$

### The Modulus Sign #3

Here, we shall discuss something called the Triangle inequality which states that $|x+y| \le |x| + |y|$. I thought this is rather important as I see too many students writing things like $|z_1 + z_2| = |z_1| + |z_2|$\$ in complex numbers. This is really scary because it shows that they are not thinking when they are doing the question.

Students can argue that adding two number and applying modulus should give positive so we can split the positive numbers up. Now this is sufficiently true. But then what if we have negative numbers coming into play? Consider this

$|2+(-2)| = 0$

Then we see that $|2|+|-2| = 4$.

So this is a big problem as $4 \ne 0$ and thus we cannot assume that relation above! So students please be really careful!

### The Modulus Sign #2

So now let us look at a simple relationship that many students memorise instead of understand. Below, we see a absolute function graph.

Now we all know that $|x| gives us $-a. So how did that happen? Consider drawing a horizontal line, $y=a$ in the graph above. $|x|$ is less than $a$ only when $-a. Similarly, $|x|>a$ is true only when $x>a$ or $x<-a$. I do hope that the graphical representation helps here.