### Complex Numbers related articles

Here is a compilation of all the Complex Numbers articles KS has done. Students should read them when they are free to improve their mathematics skills. They will come in handy! 🙂

### Brute forcing simultaneous equations involving Complex Numbers

So I have many students coming up to me and saying, “Mr Teng, How do you know which to substitute away or when to introduce $z=x+yi$ when doing simultaneous equations for complex numbers?”
Here is a lesser method that will give you the answers. It is definitely a clearer method that involves less pitfalls. This method is self-explanatory so I’ll let the working do the talking. Say we want to solve the following

$2w + z = 12i ---(1)$

$w^* + 2z = -6 - i ---(2)$

Let $w = a + bi$ (clearly, $w^* = a-bi$) , and $z=c+di$

$2(a+bi) + (c+di) = 12i ---(1)$

$a-bi + 2(c+di) = -6 - i ---(2)$

From (1), we have $(2a+c) + (2b+d)i = 12i$

$\Rightarrow 2a+c = 0 --- (3)$ , and $2b+d = 12 ---(4)$

From (2), we have $(a+2c) + (2d-b)i = -6-i$

$\Rightarrow a+2c = -6 --- (5)$ , and $2d-b = -1 ---(6)$

Solving (3), (4), (5), (6), we find

$a = 2, b =5, c=-4, d=2$

Thus, $w = 2+5i$ and $z = -4+2i$

### 2010 A-level H2 Mathematics (9740) Paper 1 Question 8 Suggested Solutions

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

(i)
$|z_1| = \sqrt{1+3} = 2$

$\mathrm{arg}z_1 = \mathrm{tan}^{-1}(\sqrt{3}) = \frac{\pi}{3}$

$\Rightarrow z_1 = 2(\mathrm{cos}\frac{\pi}{3} + i \mathrm{sin} \frac{\pi}{3})$

$|z_2| = \sqrt{1+1} = \sqrt{2}$

$\mathrm{arg}z_2 = - \pi + \mathrm{tan}^{-1}(1) = - \frac{3\pi}{4}$

$\Rightarrow z_2 = \sqrt{2}(\mathrm{cos}-\frac{3\pi}{4} + i \mathrm{sin} -\frac{3\pi}{4})$

(ii)
$\frac{z_1}{z_2} = \frac{2(\mathrm{cos}\frac{\pi}{3} + i \mathrm{sin} \frac{\pi}{3})}{\sqrt{2}(\mathrm{cos}-\frac{3\pi}{4} + i \mathrm{sin} -\frac{3\pi}{4})}$

$= \sqrt{2}[\mathrm{cos}(-\frac{11\pi}{12}) + i \mathrm{sin} (-\frac{11\pi}{12})]$

$(\frac{z_1}{z_2})^* = \sqrt{2}[\mathrm{cos}(\frac{11\pi}{12}) + i \mathrm{sin} (\frac{11\pi}{12})]$

(iii)

(iv)
From (iii)
$(x-1)^2 + \sqrt{3}^2 = 2^2$

$x = \pm 2$

Since $x > 0, x = 2$

Thus, locus meets the positive real axis when x=2.

The first two parts requires both students to be able to apply your knowledge of complex numbers formulas, expanding and simplifying them effectively. For (iii), the circle must cut through the origin, and student need to draw it clearly.

MF26

### H2 Math Tue 7pm

MF26

10.
(i)

11.
(i)
Since P is on l, $\vec{OP} = \begin{bmatrix} 1\\ 2\\ 3\end{bmatrix} + t \begin{bmatrix} -1\\ 1\\ 1\end{bmatrix}$ for some t
$\vec{OP}$ perpendicular to l $\Rightarrow \vec{OP} \bullet \begin{bmatrix} -1\\ 1\\ 1\end{bmatrix}$
$-(1-t)+2+t+3+t=0$
$t=-\frac{4}{3}$
$\vec{OP} = \begin{bmatrix} 1\\ 2\\ 3\end{bmatrix} -\frac{4}{3} \begin{bmatrix} -1\\ 1\\ 1\end{bmatrix} = ...$

### H2 Math Tue 5pm

MF26

Differentiate $x=e^t$ with respect to t.
$\frac{dx}{dt} = e^t$
Substitute $x = e^t$ into given differential equation,
$\Rightarrow e^t \frac{dy}{dt} = \mathrm{cos}(\mathrm{ln}e^t)\mathrm{sin}(\mathrm{ln}(e^t)^3)$
$e^t \frac{dy}{dt} = \mathrm{cos}t \mathrm{sin} (3t)$
Since $\frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt}$
$\frac{dy}{dt} = \frac{1}{e^t} \mathrm{cos}t \mathrm{sin} (3t) \times e^t$
$\frac{dy}{dt} = \mathrm{cos}t \mathrm{sin} (3t)$
$\int \frac{dy}{dt} dt = \int \mathrm{cos}t \mathrm{sin} (3t) dy$
Using the product to sum formula as shown here, we have
$\int \frac{dy}{dt} dt = \frac{1}{2} \int \mathrm{sin} 4t + \mathrm{sin} 2t dt$
$y = \frac{1}{2}(-\frac{\mathrm{cos}4t}{4} - \frac{\mathrm{cos}2t}{2}) + C$
$y = \frac{1}{2}(-\frac{\mathrm{cos}(4 \mathrm{ln}x)}{4} - \frac{\mathrm{cos}(2\mathrm{ln}x)}{2}) + C$

Note: $i = \begin{bmatrix}1\\ 0\\ 0\end{bmatrix}$ and $j = \begin{bmatrix}\\ 1\\ 0\end{bmatrix}$
$\pi: 2x + 3y = -6$ — (1)
$\pi \prime: x + y + z= 5$ — (2)
$\pi_1: x = 0$ — (3)
Using GC, $\vec{OA} = \begin{bmatrix}0\\ -2\\ 7\end{bmatrix}$

$\pi: 2x + 3y = -6$ — (1)
$\pi \prime: x + y + z= 5$ — (2)
$\pi_1: y = 0$ — (3)
Using GC, $\vec{OA} = \begin{bmatrix}-3\\ 0\\ 8\end{bmatrix}$

Qn11
(i)
$l: r = \begin{bmatrix}-1\\ -1\\ 0\end{bmatrix} + \alpha \begin{bmatrix}1\\ 1\\ 1\end{bmatrix}, \alpha \in \mathbb{R}$
(ii) Since l is the common line of intersection on $\pi_1$ and $\pi_2$, we need l to be on $\pi_3$ too. For that to happen,
1. l must be parallel to $\pi_3$, that is, direction of l is perpendicular to normal to $\pi_3$
$\Rightarrow \begin{bmatrix}1\\ 1\\ 1\end{bmatrix} \bullet \begin{bmatrix}5\\ {\lambda}\\ 17\end{bmatrix} = 0$
$\lambda = -22$
2. Given that l is parallel to $\pi_3$ (since $\lambda = -22$), we need l to be on $\pi_3$, so we need $\begin{bmatrix}-1\\ -1\\ 0\end{bmatrix}$ to be on $\pi_3$
$\Rightarrow \begin{bmatrix}-1\\ -1\\ 0\end{bmatrix} \bullet \begin{bmatrix}5\\ -22\\ 17\end{bmatrix} = \mu$
$\mu = 17$
(iii)
For 3 planes to have nothing in common, then l must be parallel to $\pi_3$ (Note: if l is not parallel, l will cut $\pi_3$ at a point, which means that 3 planes will cut at a point)
$\Rightarrow \lambda = -22$ from (ii)
But $\mu \neq 17, \mu \in \mathbb{R}$

12.
(i)
$\pi_1 : r = \begin{bmatrix}4\\ 2\\ 12\end{bmatrix} + \lambda \begin{bmatrix}2\\ 2\\ 1\end{bmatrix} + \mu \begin{bmatrix}5\\ -1\\ -5\end{bmatrix}, \mu , \lambda \in \mathbb{R}$
$\vec{OP} = \begin{bmatrix}4\\ 2\\ 12\end{bmatrix}$ is on $\pi_1$ when $\lambda = \mu = 0$

MF26

### H2 Math Sun 2pm

MF26

$LHS = \sum_{r=2}^n \frac{1}{(r+1)(r-1)}$
$= \sum_{r=2}^n \frac{1}{2} (\frac{1}{r-1} - \frac{1}{r+1})$
$= \frac{1}{2} \sum_{r=2}^n \frac{1}{r-1} - \frac{1}{r+1}$
$= \frac{1}{2} [\frac{1}{1} - \frac{1}{3}$
$+ \frac{1}{2} - \frac{1}{4}$
$+ \frac{1}{3} - \frac{1}{5}$

$+ \frac{1}{n-3} - \frac{1}{n-1}$
$+ \frac{1}{n-2} - \frac{1}{n}$
$+ \frac{1}{n-1} - \frac{1}{n+1}]$
$= \frac{1}{2} [\frac{3}{2} - \frac{1}{n} - \frac{1}{n+1}]$
$= \frac{3}{4} - \frac{1}{2n} - \frac{1}{2n+2}$

Since the $lim_{n \rightarrow \infty} \sum_{r=2}^n \frac{1}{(r+1)(r-1)}$
$= lim_{n \rightarrow \infty} \frac{3}{4} - \frac{1}{2n} - \frac{1}{2n+2}$
$= \frac{3}{4} - 0 - 0$
$- \frac{3}{4}$ is a constant, the series convergences.
The sum to infinity $= \frac{3}{4}$

Let $T_n$ denote the $n^{th}$ term of the AP.
$T_n = a + (n-1)d$
$T_1 = a$
$T_3 = a +2d$
$T_6 = a +5d$
Since they are consecutive terms of a GP,
$\frac{T_3}{T_1} = \frac{T_6}{T_3} = r$
$\Rightarrow \frac{a+2d}{a} = \frac{a+5d}{a+2d}$
$(a+2d)^2 = a(a+5d)$
$a^2 + 4ad + 4d^2 = a^2 + 5ad$
$4d^2 - ad =0$
$d(4d - a) = 0$
$d = 0 (NA) \mathrm{~or~} 4d = a$
$\Rightarrow r = \frac{a+2d}{a}$
$r = \frac{6d+2d}{4d} = \frac{3}{2} > 1$, thus its not convergent

$S_{15} = \frac{n}{2}(2a + (n-1)d)$
$= \frac{15}{2}(2a + 14 (\frac{a}{4}))$
$= 41.25 a$

Sum of first 3 terms $= \frac{3}{2} (2a+(3-1)d)$
$6 = 3a+3d$
$a+d=2$ —(1)
Sum of last 3 terms $= \frac{3}{2}[2(a+(n-1)d) + (3-1)(-d)]$; Here we consider an AP that has first term $T_n = a + (n-1)d$ and common difference $-d$.
$\Rightarrow 231 = \frac{3}{2}(2a + 2nd - 2d -2d)$
$231 = 3a + 3nd - 6d$
$77 = a + nd -2d$ —(2)
Sum of n terms = $\frac{n}{2}[2a+ (n-1)d]$
$1106 = \frac{n}{2}(2a + nd - d)$ —(3)
Solve for n.

(i) Volume, $V = \pi r^2 h$
Volume of kth later, $V_k = \pi [(20)(\frac{5}{6})^{k-1}]^2(22)(\frac{4}{5})^{k-1}$
$V_k = 8800 \pi [\frac{25}{36} \times \frac{4}{5}]^{k-1}$
$V_k = 8800 \pi (\frac{5}{9})^{k-1}$
(ii)
Since $latex r = \frac{5}{9} <1, S_{\infty}$ exists. Theoretical Max Volume, $latex S_{\infty} = \frac{8800 \pi}{1 - \frac{5}{9}} = 19800 \pi$. Total Volume, $latex S_n = \frac{8800 \pi (1 - (\frac{5}{9})^n)}{1 - \frac{5}{9}}$ We want $latex S_n \le 0.95 S_{\infty}$

### H2 Math Sun 1130am

MF26

Vectors Q7 [Homework]
(i)
$\vec{OL} = \begin{bmatrix}2\\ 7\\ -1\end{bmatrix}$
$\vec{OM} = \begin{bmatrix}9\\ 0\\ -8\end{bmatrix}$
Using ratio theorem, $\vec{OP} = \frac{2\vec{OM}+5\vec{OL}}{7} = \begin{bmatrix}4\\ 5\\ -3\end{bmatrix}$
Since $\vec{OP}$ is perpendicular to $\begin{bmatrix}4\\ 1\\ q\end{bmatrix}$
$\Rightarrow \begin{bmatrix}4\\ 5\\ -3\end{bmatrix} \bullet \begin{bmatrix}4\\ 1\\ q\end{bmatrix} = 0$
$q = 7$

(ii)
To be a parallelogram, $\vec{OM} = \vec{LN} = \vec{ON} - \vec{OL}$
$\vec{ON} =\begin{bmatrix}11\\ 7\\ -9\end{bmatrix}$
$Area = |\vec{OM} \times \vec{OL}|$
$= |\begin{bmatrix}56\\ -7\\ 63\end{bmatrix}|$
$= \sqrt{7154} = 7 \sqrt{146} units^2$

(iii)
Let $\vec{OQ} = \begin{bmatrix}x\\ y\\ 0\end{bmatrix}$
Since $|\vec{OQ}| = |\vec{OP}|$
$\sqrt{x^2 + y^2} = \sqrt{50}$ — (1)
$\begin{bmatrix}x\\ y\\ 0\end{bmatrix} \bullet \begin{bmatrix}1\\ 0\\ 0\end{bmatrix} = |\begin{bmatrix}x\\ y\\ 0\end{bmatrix} | |\begin{bmatrix}1\\ 0\\ 0\end{bmatrix} | \mathrm{cos} \theta$ — (2)
Solving, $x = \sqrt{50} \mathrm{cos} \theta = 5 \sqrt{2} \mathrm{cos} \theta$
$y = \sqrt{50} \mathrm{sin} \theta = 5 \sqrt{2} \mathrm{sin} \theta$
$\Rightarrow \vec{OQ} = \begin{bmatrix}{5 \sqrt{2} \mathrm{cos} \theta}\\ {5 \sqrt{2} \mathrm{sin} \theta}\\ 0\end{bmatrix}$

Vectors Q8 [Homework]
(i)
$\vec{OA} = \begin{bmatrix}-5\\ -2\\ 3\end{bmatrix}$
$\vec{OC} = \begin{bmatrix}5\\ 2\\ 6\end{bmatrix}$
$\vec{AC} = \vec{OC} - \vec{OA} = \begin{bmatrix}5\\ 2\\ 6\end{bmatrix} - \begin{bmatrix}-5\\ -2\\ 3\end{bmatrix} = \begin{bmatrix}10\\ 4\\ 3\end{bmatrix}$
$l: r = \begin{bmatrix}5\\ 2\\ 6\end{bmatrix} + \lambda \begin{bmatrix}10\\ 4\\ 3\end{bmatrix}, \lambda \in \mathbb{R}$

(ii)
Let R be the top of the vertical pillar,
$l_{QR}: r = \begin{bmatrix}15\\ 6\\ 0\end{bmatrix} + \mu \begin{bmatrix}0\\ 0\\ 1\end{bmatrix}, \mu \in \mathbb{R}$
Since R is collinear with A and C, R is the intersection of line AC and QR.
$\begin{bmatrix}{5 + 10 \mu}\\ {2 + 4 \mu}\\ {6 + 3 \mu}\end{bmatrix} = \begin{bmatrix}15\\ 6\\ {\mu}\end{bmatrix}$
$\Rightarrow \lambda = 1, \mu = 9$
$\vec{OR} = \begin{bmatrix}15\\ 6\\ 9\end{bmatrix}$, and the height is 9m.

(iii)
$\vec{OD} = \begin{bmatrix}-5\\ 2\\ 6\end{bmatrix}$
$\vec{AD} = \vec{OD} - \vec{OA} = \begin{bmatrix}0\\ 4\\ 3\end{bmatrix}$
$\vec{AX} = (\vec{AD} \bullet \frac{\begin{bmatrix}10\\ 4\\ 3\end{bmatrix}}{| \begin{bmatrix}10\\ 4\\ 3\end{bmatrix}|}) \frac{\begin{bmatrix}10\\ 4\\ 3\end{bmatrix}}{| \begin{bmatrix}10\\ 4\\ 3\end{bmatrix}|}$
$= (\begin{bmatrix}0\\ 4\\ 3\end{bmatrix} \bullet \frac{\begin{bmatrix}10\\ 4\\ 3\end{bmatrix}}{\sqrt{125}}) \frac{\begin{bmatrix}10\\ 4\\ 3\end{bmatrix}}{\sqrt{125}}$
$= \frac{25}{125} \begin{bmatrix}10\\ 4\\ 3\end{bmatrix}$
$= \begin{bmatrix}2\\ 0.8\\ 0.6\end{bmatrix}$
$\vec{OX} = \vec{OA} + \vec{AX} = \begin{bmatrix}-3\\ 1.2\\ 3.6\end{bmatrix}$

Vectors Q9 [Homework]
(i)
$\vec{AB} = \begin{bmatrix}-4\\ 5\\ 3\end{bmatrix}$
$\vec{AC} = \begin{bmatrix}1\\ -3\\ 6\end{bmatrix}$
Normal of $\pi_1, ~n_1=\begin{bmatrix}-4\\ 5\\ 3\end{bmatrix} \times \begin{bmatrix}1\\ -3\\ 6\end{bmatrix} = \begin{bmatrix}-21\\ -21\\ -7\end{bmatrix} = -7 \begin{bmatrix}3\\ 3\\ 1\end{bmatrix}$
$\pi_1: r \bullet \begin{bmatrix}3\\ 3\\ 1\end{bmatrix} = \begin{bmatrix}5\\ -1\\ 0\end{bmatrix} \bullet \begin{bmatrix}3\\ 3\\ 1\end{bmatrix} = 12$

(ii)
Let $\theta$ be the acute angle
$\theta - \mathrm{cos}^{-1} |\frac{\begin{bmatrix}3\\ 3\\ 1\end{bmatrix} \bullet \begin{bmatrix}1\\ -1\\ 1\end{bmatrix}}{\sqrt{19}} \sqrt{3}|$
$\theta = 82.4 ^{\circ}$

(iii)
$3x + 3 y + z = 12$ — (1)
$x - y + z = 1$ — (2)

Using GC, $l: r = \begin{bmatrix}2.5\\ 1.5\\ 0\end{bmatrix} + \lambda \begin{bmatrix}-2\\ 1\\ 3\end{bmatrix}, \lambda \in \mathbb{R}$

(iv)
Let $n_3$ be the normal of $\pi_3$
Length of projection $= |\vec{AB} \times n_3|$
$= \frac{1}{\sqrt{26}} |\begin{bmatrix}4\\ -5\\ 3\end{bmatrix} \times \begin{bmatrix}5\\ -1\\ 0\end{bmatrix}| = 15\sqrt{\frac{3}{26}}$

(v)
Required distance $= \frac{1}{\sqrt{3}} + \frac{2}{\sqrt{3}} = \sqrt{3} units$

(vi)
Let normal of $\pi_4 = n_4 = \begin{bmatrix}-2\\ 1\\ 3\end{bmatrix} \times \begin{bmatrix}1\\ -1\\ 1\end{bmatrix} = \begin{bmatrix}4\\ 5\\ 1\end{bmatrix}$
$\pi_4: r \bullet \begin{bmatrix}4\\ 5\\ 1\end{bmatrix} = 4k+6$
If $\pi_1, \pi_2 \mathrm{~and~} \pi_4$ intersect at l,n$\begin{bmatrix}2.5\\ 1.5\\ 0\end{bmatrix}$ lies on $pi_4$
$\Rightarrow \begin{bmatrix}2.5\\ 1.5\\ 0\end{bmatrix} \bullet \begin{bmatrix}4\\ 5\\ 1\end{bmatrix} = 4k+6$
$k = \frac{23}{8}$