1.5 Complex Numbers (HL)

Q1

Topic	1.5 Complex Numbers
Tag	Complex number; Modulus; Argument; Imaginary number; Conjugate; De Moivre's theorem
Source	N17/5/MATHL/HP1/ENG/TZ0/XX/8
Question Text	Determine the root of the equation $(z + 2i)³ = 216i, z∈C$ , for which $z = a√3 + bi$ where $a,b∈Z+$ . Find $a + b$ .
Total Mark	7
Correct Answer	4
Explanation	n/a
Mark Scheme	Step 1: Simplify Suppose $\textit{w}$ = $\textit{z}$ + 2 $\textit{i}$ , $\textit{w}^{3} = 216\textit{i}$ so, $\textit{w}^{3} = 216$ ( $cos$ $\frac{\pi }{2}$ + $\iota sin\frac{\pi }{2}$ ) = 216( $cos$ ( $\frac{\pi }{2}$ + $2\pi \textit{k}$ ) + $\textit{i}$ $sin$ ( $\frac{\pi }{2}+2\pi \textit{k}$ )), Tip 1: When finding roots of complex numbers just assume that there's an extra 2πk, and later using DeMovire's, divide by the necessary number. In this problem divide by 3. The roots of $\textit{w}$ therefore are 6( $cos$ ( $\frac{\pi }{6}$ + $\frac{2\pi \textit{k}}{3}$ ) + $\textit{i}sin(\frac{\pi }{6}+\frac{2\pi \textit{k}}{3})$ ), where $\textit{k}$ = 0, 1, 2 So, $\textit{w}$ = $6(cos(\frac{\pi }{6})+\textit{i}sin(\frac{\pi }{6}))$ , $(cos(\frac{5\pi }{6})+\textit{i}sin(\frac{5\pi }{6}))$ , $(cos(\frac{9\pi }{6})+\iota sin(\frac{9\pi }{6}))$ $\textit{w}$ = 6 $\left ( \frac{\sqrt{3}}{2}+\frac{1}{2}\textit{i}\right )$ ,4 $\left ( \frac{-\sqrt{3}}{2}+\frac{1}{2}\textit{i}\right )$ ,4( $-\frac{1}{2}\textit{i}$ ) The only root of $\textit{w}$ with both positive integers is 6 $\left ( \frac{\sqrt{3}}{2}+\frac{1}{2}\iota \right )$ = $3\sqrt{3}+3\textit{i}$ . As $\textit{w}$ = $\textit{z}$ + 2 $\textit{i}$ , $\textit{z}$ = $3\sqrt{3}+3\textit{i}$ $-2\textit{i}$ = $3\sqrt{3}+\textit{i}$ So, $\textit{a}$ = 3, $\textit{b}$ = 1 Thus, $\textit{a}+\textit{b}$ = 4 Answer = 4

Q2

Topic	1.5 Complex Numbers (HL)
Tag	Complex number; Modulus; Argument; Imaginary number; Conjugate; De Moivre's theorem
Source	M17/5/MATHL/HP1/ENG/TZ1/XX/2
Question Text	Consider the complex numbers $z_{1}=2+2\sqrt{3}\textit{i},z_{2}=\sqrt{2}+\sqrt{2}\textit{i}$ and $w=\frac{z_{1}}{z_{2}}$ , (a) Find (i) the modulus of $w$ ; Total Mark: 2 Correct Answer: 2 Explanation: n/a Mark Scheme: n/a (ii) the value of $a$ given that the argument of $w = ( \frac{\pi }{a})$ where $a∈ℤ^+$ . Total Mark: 2 Correct Answer: 12 Explanation: n/a Mark Scheme: $arg(w)= \frac{π}{3} - \frac{π}{4} = \frac{π}{12}$
Question Text	(b) Find the smallest positive integer value of $n$ , such that $wⁿ$ is a real number.
Total Mark	2
Correct Answer	12
Explanation	n/a
Mark Scheme	$w^n=2^{n}\left ( \cos\left (\frac{π}{12} \right ) +i\sin\left (\frac{π}{12} \right ) \right )^{12}$ , applying de Moivre’s $w^n=2^{n}\left ( \cos\left (\frac{nπ}{12} \right ) +i\sin\left (\frac{nπ}{12} \right ) \right )^{12}$ , as $\sin\left (\frac{nπ}{12} \right )=0$ , when $n=12$

Q3

Topic	1.5 Complex Numbers (HL)
Tag	Complex number; Modulus; Argument; Imaginary number; Conjugate; De Moivre's theorem
Source	N16-TZ0-P1-11(HL)
Question Text	(a) Given that $\textit{z}$ ² = 16 $\textit{i}$ , where $\textit{z}$ ∈ C, select the correct choice with the two values of $\textit{z}$ . (a) $4\left(\cos \frac{\pi}{4}+\sin \frac{\pi}{4}\right)$ , $4\left(\cos \left(\frac{5 \pi}{4}\right)+\sin \left(\frac{5 \pi}{4}\right)\right)$ (b) $4\left(\cos \frac{3 \pi}{4}+\sin \frac{3 \pi}{4}\right)$ , $4\left(\cos \left(\frac{7 \pi}{4}\right)+\sin \left(\frac{7 \pi}{4}\right)\right)$ (c) $16\left(\cos \left(\frac{\pi}{4}\right)+\sin \left(\frac{\pi}{4}\right)\right)$ , $16\left(\cos \left(\frac{5 \pi}{4}\right)+\sin \left(\frac{5 \pi}{4}\right)\right)$ (d) $16\left(\cos \left(\frac{3 \pi}{4}\right)+\sin \left(\frac{3 \pi}{4}\right)\right)$ , $16\left(\cos \left(\frac{7 \pi}{4}\right)+\sin \left(\frac{7 \pi}{4}\right)\right)$
Total Mark	3
Correct Answer	a
Explanation	n/a
Mark Scheme	$z^2$ =16 $\left(\cos \left(\frac{\pi}{2}+2 \pi k\right)+i \sin \left(\frac{\pi}{2}+2 \pi k\right)\right)$ $\textit{z}$ =4 $\left(\cos \left(\frac{\pi}{2}+2 \pi k\right)+i \sin \left(\frac{\pi}{2}+2 \pi k\right)\right)^{\frac{1}{2}}$ Using de Moivre's, $\textit{z}$ = $4\left(\cos \left(\frac{\pi}{4}+\pi k\right)+i \sin \left(\frac{\pi}{4}+\pi k\right)\right)$ Answer: a
Question Text	(b) Consider the complex numbers $\textit{z}_{1}$ = $2 \sqrt{2}+2$ $\sqrt{2} i$ and $\textit{z}_{2}$ = $\sqrt{2}\left(\cos \left(\frac{\pi}{6}\right)+i \sin \left(\frac{\pi}{6}\right)\right)$ . i. $\quad z_1$ can be written in the form $\textit{r}$ $\left(\cos \frac{\pi}{k}+i \sin \frac{\pi}{k}\right)$ where $\textit{r}$ , $\textit{k}$ $\in Z^{+}$ . Find the value of $\textit{r} + \textit{k}$ Total Mark : 2 Correct Answer : 8 Exlplanation : na Mark Scheme : $\textit{r}$ = $\sqrt{(2 \sqrt{2})^2+(2 \sqrt{2})^2}$ =4 $z_1$ = $4\left(\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2} i\right)$ =4 $\left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right)$ $\textit{k}$ =4 Thus, $\textit{r} + \textit{k}=4+4=8$ ————————————- ii. Choose the correct value of $\frac{\textit{z}_{1}}{\textit{z}_{2}}$ (a) $\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2} i$ (b) $\frac{\sqrt{3}}{2}-\frac{1}{2} i$ (c) $\sqrt{3}+1+i(\sqrt{3}-1)$ (d) $\frac{\sqrt{3}}{2}+\frac{1}{2} i$ Total Mark : 3 Correct Answer : c Explanation : na Mark Scheme : Step 1: Expand Tip 1: Utilize the formula booklet: compound angle identities $\frac{z_1}{z_2}$ = $\frac{4}{\sqrt{2}}\left(\cos \left(\frac{\pi}{4}-\frac{\pi}{6}\right)+i \sin \left(\frac{\pi}{4}-\frac{\pi}{6}\right)\right)$ = $\frac{4}{\sqrt{2}}\left(\cos \frac{\pi}{4} \cos \frac{\pi}{6}+\sin \frac{\pi}{4} \sin \frac{\pi}{6}+i\left(\sin \frac{\pi}{4} \cos \frac{\pi}{6}-\sin \frac{\pi}{6} \cos \frac{\pi}{4}\right)\right)$ Step 2: Simplify = $\frac{4}{\sqrt{2}}\left(\frac{\sqrt{2}}{2} \times \frac{\sqrt{3}}{2}+\frac{\sqrt{2}}{2} \times \frac{1}{2}+i\left(\frac{\sqrt{3}}{2} \times \frac{\sqrt{2}}{2}-\frac{1}{2} \times \frac{\sqrt{2}}{2}\right)\right)$ = $\frac{4}{\sqrt{2}}\left(\frac{\sqrt{6}+\sqrt{2}}{4}+i\left(\frac{\sqrt{6}-\sqrt{2}}{4}\right)\right)$ = $\sqrt{3}+1+i(\sqrt{3}-1)$ Answer: c ———————————— iii. Hence find the value of $c + d$ if $tan\frac{\pi }{12}$ can be written in the form $c$ - $d\sqrt{3}$ , where c, d ∈ Z. Total Mark : 4 Correct Answer : 3 Exlplanation : na Mark Scheme : Step 1: Consider the given information Tip 1: Utilize the previous part(s) if the question says "hence" $\tan \frac{\pi}{12}=\frac{\sin \frac{\pi}{12}}{\cos \frac{\pi}{12}} \\ \frac{z_1}{z_2}=2 \sqrt{2}\left(\cos \left(\frac{\pi}{4}-\frac{\pi}{6}\right)+i \sin \left(\frac{\pi}{4}-\frac{\pi}{6}\right)\right)=\sqrt{3}+1+i(\sqrt{3}-1)$ Step 2: Combine the given information As $\frac{\pi}{12}=\frac{\pi}{4}-\frac{\pi}{6}$ , $\frac{z_1}{z_2}=2 \sqrt{2}\left(\cos \frac{\pi}{12}+i \sin \frac{\pi}{12}\right)=\sqrt{3}+1+i(\sqrt{3}-1)$ $2 \sqrt{2} \cos \frac{\pi}{12}=\sqrt{3}+1 \\ 2 \sqrt{2} \sin \frac{\pi}{12}=\sqrt{3}-1$ Which means $\tan \frac{\pi}{12}=\frac{2 \sqrt{2} \sin \frac{\pi}{12}}{2 \sqrt{2} \cos \frac{\pi}{12}}=\frac{\sqrt{3}-1}{\sqrt{3}+1}$ $\frac{(\sqrt{3}-1)(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)}=2-\sqrt{3}$ $c=2, d=1$ Thus, $c+d$ =3 ———————————- iv. Find the smallest value $n$ > 0 such that $(z₂)ⁿ$ is a positive real number. Total Mark : 2 Correct Answer : 12 Explanation : na Mark Scheme : Using DeMoivre's $z_2^n=\sqrt{2}^n\left(\cos \left(\frac{n \pi}{6}\right)+i \sin \left(\frac{n \pi}{6}\right)\right)$ $z_2^n$ is a positive real number when $n$ =12

Q4

Topic	1.5 Complex Numbers
Tag
Source	N16-TZ0-P1-12(HL)
Question Text	Let be one of the non-real solutions of the equation $z⁴ = 1$ . Determine the value of (a) 1 + ω + ω² + ω³
Total Mark	3
Correct Answer	0
Explanation	na
Mark Scheme	$1+\omega+\omega^2+\omega^3=\frac{1-\omega^4}{1-\omega}=0 \text { as } \omega \neq 1$
Question Text	(b) 1 + ω* + (ω)² + (ω)³
Total Mark	1
Correct Answer	0
Explanation	na
Mark Scheme	This is the conjugate of $1 + ω + ω² + ω³$ . The conjugate of 0 is 0

Q5

Topic	1.5 Complex Numbers
Tag	Sequences; Series; Arithmetic; Geometric; Sigma; Compound interest; Finance
Source
Question Text	Consider the complex numbers $p$ = 4 - 3 $i$ and $q$ = $x$ + ( $x$ - 7) $i$ , where $x$ ∈ $R$ . (a) Find the sum of the values of $x$ that satisfy the equation \| $p$ \| = \| $q$ \|.
Total Mark	5
Correct Answer	7
Explanation	na
Mark Scheme	Step 1: Consider the given information $\|p\|=\|q\|$ where $p=4-3 i \text { and } q=x+(x-7) i$ Step 2: Combine the given information Tip 1: The variable to be found is $x$ , so we should aim to make an equation that involves $x$ . $\|p\|=\|q\|$ $\Rightarrow \sqrt{4^2+3^2}=\sqrt{x^2+(x-7)^2}$ $25=2 x^2-14 x+49$ $2 x^2-14 x+24=0$ $(x-3)(x-4)=0$ $x$ = 3 or 4 Thus, the sum of the values of $x$ that satisfy the equation is: 7 .
Question Text	(b) The solution to the inequality Re⁡( $pq$ ) - 13 < Im⁡( $pq$ ) can be written as $x$ < $k$ , find the value of $k$ .
Total Mark	4
Correct Answer	1
Explanation	na
Explanation	Step 1: Consider the given information Re $⁡($ pq) + 8 < (Im $⁡($ pq))² where p = 4 - 3i and q = x + (x - 7)i Step 2: Combine the given information pq = (4 - 3i)(x + (x - 7)i) = (7x - 21) + (x - 28)i Re⁡(pq) + 8 < Im⁡(pq) ⇒ (7x - 21) + 8 < (x - 28) 6x < 6, x < 1 Answer: 1

Q6

Topic	1.5 Complex Numbers
Tag
Source	M16-TZ1-P1-12(HL)
Question Text	(a) The value of $\sqrt{2}$ $\times\left(\cos \left(\frac{\pi}{4}\right)-i \sin \left(\frac{\pi}{4}\right)\right)^5$ can be written as $a+b i$ where $a$ , $b$ $\in \mathbb{Z}$ . Find the value of $a+b$ .
Total Mark	3
Correct Answer	0
Explanation	na
Mark Scheme	Tip 1: Utilize the formula booklet: de Moivre's $\sqrt{2} \times\left(\cos \left(-\frac{\pi}{4}\right)+i \sin \left(-\frac{\pi}{4}\right)\right)^5=\sqrt{2}\left(\cos \left(\frac{-5 \pi}{4}\right)+i \sin \left(\frac{-5 \pi}{4}\right)\right)$ -1+ $i$ Answer: $a=-1, b=1$
Question Text	(b) Let $z = cos⁡θ + isin⁡θ$ . Choose the correct expression for $(z)ⁿ + (z)ⁿ , n ∈ Z+$ where $z$ is the complex conjugate of $z$ . (a) $2 cos⁡(nθ)$ (b) $2i sin(nθ)$ (c) $n cos(θ) + ni sin(θ)$ (d) $n cos(θ) - ni sin(θ)$
Total Mark	2
Correct Answer	a
Explanation	na
Mark Scheme	na
Question Text	(c) (i) Compute the value of $z$ x $z^$ . Total Mark : 2 Correct Answer : 1 Explanation : na Mark Scheme : $z z^$ = ( $cos⁡θ + isin⁡θ$ )( $cos⁡θ - isin⁡θ$ ) = $cos²θ + sin²⁡θ$ = 1 Thus, answer is 1 (ii). $cos⁡3θ = acos³θ - bcos⁡θ$ where $a, b ∈ ℤ+$ . By considering the binomial expansion of $(z+z^)³$ , find the value of $a + b$ . Total Mark : 3 Correct Answer : 7 Explanation : na Mark Scheme : Tip 1: Utilize the information found in previous parts, $(z+z^)^3$ = $z^3$ + $3 z^2 z^$ + $3z(z^)^2 + (z^)^3$ = $z^3 + 3z + 3z^ + (z^)^3$ Step 1: Expand utilizing given information $(z+z^)^3$ = $(2 \cos \theta)^3$ From part (b) we know $(z)^n$ + $(z^)^n$ =2 $\cos (n \theta)$ $z^3+3 z+3 z^+(z^*)^3=2 \cos 3 \theta+6 \cos \theta$ $2 \cos 3 \theta+6 \cos \theta=(2 \cos \theta)^3$ $\cos 3 \theta=4 \cos ^3 \theta-3 \cos \theta$ $a$ =4, $b$ =3 Thus, $a$ + $b$ =7
Question Text	(d) Hence solve $4cos³θ$ + $2sin²θ$ - $3cosθ$ - 1 = 0 for 0 < θ < $\frac{\pi }{2}$ . (a) $\theta$ = $\frac{\pi }{4}$ (b) $\theta$ = $\frac{2\pi }{5}$ (c) $\theta$ = $\frac{\pi }{7}$ (d) $\theta$ = $\frac{3\pi }{7}$
Total Mark	6
Correct Answer	b
Explanation	na
Mark Scheme	Step 1: Rearrange such that the equation is more favorable Tip 1: Notice that $\cos 3 \theta$ = $4\cos ^3 \theta$ - $3 \cos \theta$ can be utilized $4 \cos ^3\theta$ + $2 \sin ^2 \theta$ - $3 \cos \theta$ - 1 = 0 $\cos 3 \theta$ =1 - $2 \sin ^2 \theta$ Noticing that 1- $2 \sin ^2 \theta$ = $\cos 2 \theta$ $\cos 3 \theta$ = $\cos 2 \theta$ $\cos 3 \theta$ = $\cos (2 \pi-2 \theta)$ $3 \theta$ = $2 \pi-2 \theta$ $\theta$ = $\frac{2 \pi}{5}$ Answer: (B)

Q7

Topic	1.5 Complex Numbers
Tag
Source	M16-TZ2-P1-12(HL)
Question Text	(a) Which is a root of the equation $z⁵$ - 1 = 0 , $z$ ∈ $C$ ? (a) $\cos \frac{\pi}{2}+i \sin \frac{\pi}{2}$ (b) $\cos \frac{3 \pi}{2}+i \sin \frac{3 \pi}{2}$ (c) $\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}$ (d) $\cos \frac{2 \pi}{5}+i \sin \frac{2 \pi}{5}$
Total Mark	2
Correct Answer	d
Explanation	na
Mark Scheme	Rewrite 1 in its complex form, $z^5$ = $\cos (2 \pi k)$ + $i \sin (2 \pi k)$ , $k \in \mathbb{Z}$ Notice that de Moivre's can be applied, $z$ = $\cos \left(\frac{2 \pi k}{5}\right)$ + $i \sin \left(\frac{2 \pi k}{5}\right)$ The only option in this form is (D) Answer: (D)
Question Text	(b) (i) Which of the following is the expansion of ( $w$ - 1)(1 + $w$ + $w²$ + $w³$ + $w⁴$ ) ? (a) $w⁵$ - 1 (b) 1 + $w²$ + $w⁴$ (c) $w$ + $w³$ + $w⁵$ (d) 1 + $w$ + $w²$ + $w³$ + $w⁴$ + $w⁵$ Total Mark : 1 Correct Answer : a Explanation : na Mark Scheme : ( $w$ -1)(1+ $w$ + $w^2$ + $w^3$ + $w^4$ )= $w$ + $w^2$ + $w^3$ + $w^4$ + $w^5$ -1- $w$ - $w^2$ - $w^3$ - $w^4$ = $w^5$ -1 Answer: (A) (ii) Hence write down the value of 1 + $w$ + $w²$ + $w³$ + $w⁴$ . Total Mark : 2 Correct Answer : 0 Explanation : na Mark Scheme : Consider the given information, $w⁵$ - 1 = ( $w$ - 1)(1 + $w$ + $w²$ + $w³$ + $w⁴$ ) = 0 and $w$ - 1 ≠ 0 So, 1 + $w$ + $w²$ + $w³$ + $w⁴$ = 0 Answer: 0

Q8

Topic	1.5 Complex Numbers
Tag
Source	M15-TZ2-P1-7(HL)
Question Text	(a) The equation 16 $z⁴$ + 81 = 0 , $z$ ∈ $C$ has four distinct complex roots. Which of the following are the roots the equation. (Two answers) (a) $\frac{3}{2}\left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right)$ (b) $\frac{3}{2}\left(\cos \frac{5 \pi}{4}+i \sin \frac{5 \pi}{4}\right)$ (c) $\frac{3}{2}\left(\cos \frac{\pi}{2}+i \sin \frac{\pi}{2}\right)$ (d) $\frac{3}{2}(\cos \pi+i \sin \pi)$ (e) $\frac{3}{2}\left(\cos \frac{3 \pi}{2}+i \sin \frac{3 \pi}{2}\right)$
Total Mark	6
Correct Answer	a,b
Explanation	na
Mark Scheme	Rearrange the given equation such that it is favorable (i.e able to apply de Moivre's) $z^4$ = $-\frac{81}{16}$ = $\frac{81}{16}(\cos \pi+i \sin \pi)$ = $\frac{81}{16}(\cos (\pi+2 n \pi)+i \sin (\pi+2 n \pi))$ $z$ = $\frac{3}{2}$ $\left(\cos \left(\frac{\pi+2 n \pi}{4}\right)+i \sin \left(\frac{\pi+2 n \pi}{4}\right)\right)$ $z_1$ = $\frac{3}{2}\left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right) z_2=\frac{3}{2}\left(\cos \frac{3 \pi}{4}+i \sin \frac{3 \pi}{4}\right) z_3=\frac{3}{2}\left(\cos \frac{5 \pi}{4}+i \sin \frac{5 \pi}{4}\right)$ , $z_4$ = $\frac{3}{2}\left(\cos \frac{7 \pi}{4}+i \sin \frac{7 \pi}{4}\right)$ Answer: (A), (B)
Question Text	(b) The roots are represented by the vertices of a square in an Argand diagram. The area of this square can be written in the form $\frac{\textit{a}}{\textit{b}}$ where a and b are positive integers in lowest term. Compute the value of $a$ + $b$ .
Total Mark	3
Correct Answer	11
Explanation	na
Mark Scheme	Notice that the diagonal of this square is $\frac{3}{2}$ x 2 = 3 The area is therefore, 3 x 3 x ( $\frac{1}{2}$ ) = $\frac{9}{2}$ Answer: 11

Q9

Topic	1.5 Complex Numbers
Tag
Source	N14/5/MATHL/HP1/ENG/TZ0/XX/13a
Question Text	Given that $(1+i \tan \theta)^n+(1-i \tan \theta)^n=\frac{2 \cos n \theta}{\cos ^n \theta}, \cos \theta \neq 0$ , which is a possible root of the equation $(1+z)^3+(1-z)^3=0, z \in C$ . (a) $\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}$ (b) $\cos \frac{4 \pi}{3}+i \sin \frac{4 \pi}{3}$ (c) $i \tan \frac{2 \pi}{3}$ (d) $1-i \tan \frac{4 \pi}{3}$ (e) $1+i \tan \frac{4 \pi}{3}$
Total Mark	4
Correct Answer	c
Explanation	n/a
Mark Scheme	Notice $(1+z)^3+(1-z)^3$ = $0$ is of the form $(1+i \tan \theta)^n+(1-i \tan \theta)^n$ Suppose $z$ = $i \tan \theta$ ( $1$ + $i \tan \theta)^3$ + $(1-i \tan \theta)^3$ = $\frac{2 \cos (3 \times \theta)}{\cos ^3 \theta}=0 \\ \cos (3 \theta)=0$ $\theta=\frac{2 \pi k}{3}$ So, $z$ = $\operatorname{\textit{i}tan}\left(\frac{2 \pi k}{3}\right)$ Answer: C

Q10

Topic	1.5 Complex Numbers
Tag	Complex number; Modulus; Argument; Imaginary number; Conjugate; De Moivre's theorem
Source	M14/5/MATHL/HP1/ENG/TZ2/XX/7
Question Text	Consider the complex numbers $u$ =3+4 $i$ and $v$ = 4+3 $i$ . (a) Given that $\frac{1}{u}$ + $\frac{1}{v}$ = $\frac{14}{w}$ , where $w$ can be expressed in the form $a+b i,$ $a$ , $b$ $\in$ $R$ , find the value of $a$ + $b$ .
Total Mark	3
Correct Answer	50
Explanation	na
Mark Scheme	Consider the given information, and rearrange, $\frac{1}{3+4 i}$ + $\frac{1}{4+3 i}$ = $\frac{3-4 i}{9+16}$ + $\frac{4-3 i}{16+9}$ = $\frac{10}{w}$ $\frac{14}{w}$ = $\frac{7-7 i}{25}$ $w$ = $\frac{50}{1-i}$ = $\frac{50(1+i)}{1+1}$ = $25+25 i$ $a$ =25, $b$ =25 Thus, $a$ + $b$ =50.
Question Text	(b) If $w^*$ = $r e^{i \theta}$ , which of the following are possible values of $r$ and $\theta$ ? (a) $r$ =14, $\theta$ = $\frac{\pi}{4}$ (b) $r$ = $14 \sqrt{2}$ , $\theta$ = $-\frac{\pi}{4}$ (c) $r$ =25, $\theta$ = $\frac{\pi}{4}$ (d) $r$ = $25 \sqrt{2}$ , $\theta=-\frac{\pi}{4}$
Total Mark	3
Correct Answer	d
Explanation	na
Mark Scheme	$w^*$ =25-25 $i$ = $25 \sqrt{2}\left(\cos \left(-\frac{\pi}{4}\right)+i \sin \left(-\frac{\pi}{4}\right)\right)$ $z$ = $25 \sqrt{2} e^{-\frac{\pi}{4} i}$ Answer: D

Q11

Topic	1.5 Complex Numbers
Tag
Source	M13-TZ1-P1-1(HL)
Question Text	(a) (i) If $w$ = $\sqrt{2}+\sqrt{2}\textit{i}$ compute the modulus of $w$ . Total Mark: 1 Correct Answer : 2 Explanation: na Mark Scheme: $w$ = 2( $cos$ ( $\frac{\pi }{4}$ ) + $i$ $sin$ ( $\frac{\pi }{4}$ )) (ii) Which of the following is the argument of $w$ ? (a) $\frac{\pi }{2}$ (b) $\frac{\pi }{3}$ (c) $\frac{\pi }{4}$ (d) $\frac{\pi }{6}$ Total Mark : 1 Correct Answer: c Explanation: na Mark Scheme: $w$ = 2( $cos$ ( $\frac{\pi }{4}$ ) + $isin$ ( $\frac{\pi }{4}$ ))
Question Text	(b) Given $z$ = $\cos \frac{7 \pi}{6}$ + $i \sin \frac{7 \pi}{6}$ , compute the value of $w^4$ $z^6$
Total Mark	4
Correct Answer	16
Explanation	n/a
Mark Scheme	$w^4=\left(2\left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right)\right)^4 =16(\cos \pi+i \sin \pi)=-16 \\ z^6=\left(\cos \frac{7 \pi}{6}+i \sin \frac{7 \pi}{6}\right)^6 =\cos 7 \pi+i \sin 7 \pi=-1 \\ w^4 z^6 =16$

Q12

Topic	1.5 Complex Numbers
Tag
Source	M13-TZ2-P1-13(HL)
Question Text	Which equation has the complex roots $z_1=\sqrt{3}+i$ , $z_2=-\sqrt{3}+i$ and $z_3=-2 i$ ? (a) $z=8 i$ (b) $z=-8 i$ (c) $z=8$ (d) $z=-8$
Total Mark	4
Correct Answer	(a)
Explanation	n/a
Mark Scheme	Convert each complex numbers to its modulus argument form, $z_1=2 \operatorname{cis}\left(\frac{\pi}{6}\right), z_2=2 \operatorname{cis}\left(\frac{5 \pi}{6}\right), z_3=2 \operatorname{cis}\left(\frac{9 \pi}{6}\right)$ Notice that the modulus is 2 , and the argument differs by $\frac{4 \pi}{6}$ , $z_1^3=\left(2 \operatorname{cis}\left(\frac{\pi}{6}\right)\right)^3=8 \operatorname{cis}\left(\frac{\pi}{2}\right)=8 i$ Thus, $z=8 i$

Q13

Topic	1.5 Complex Numbers
Tag
Source	n/a
Question Text	(a) Select the correct solutions of the equation $z^5=1$ for $z \in C$ . (a) $\cos \left(\frac{\pi}{5}\right)+i \sin \left(\frac{\pi}{5}\right)$ (b) $\cos \left(\frac{2 \pi}{5}\right)+i \sin \left(\frac{2 \pi}{5}\right)$ (c) $\cos \left(\frac{3 \pi}{5}\right)+i \sin \left(\frac{3 \pi}{5}\right)$ (d) $\cos \left(\frac{4 \pi}{5}\right)+i \sin \left(\frac{4 \pi}{5}\right)$ (e) $\cos (\pi)+i \sin (\pi)$
Total Mark	4
Correct Answer	(b), (d)
Explanation	n/a
Mark Scheme	$z=(\cos (2 \pi n)+i \sin (2 \pi n))^{\frac{1}{5}} \\$ $z=\operatorname{cis}\left(\frac{2 \pi n}{5}\right), n=0,1,2,3,4$
Question Text	(b) If $w$ is the solution to $z^5=1$ with least positive argument, which is the argument of $1+w$ ? (a) $\frac{\pi}{10}$ (b) $\frac{\pi}{5}$ (c) $\frac{3 \pi}{10}$ (d) $\frac{2 \pi}{5}$
Total Mark	6
Correct Answer	(b)
Explanation	n/a
Mark Scheme	$w$ has argument $\frac{2 \pi}{5}$ , and suppose $1+w$ has argument $\theta$ $\begin{gathered} 1+w=1+\cos \left(\frac{2 \pi}{5}\right)+i \sin \left(\frac{2 \pi}{5}\right) \\ \tan \theta=\frac{\sin \left(\frac{2 \pi}{5}\right)}{1+\cos \left(\frac{2 \pi}{5}\right)} \end{gathered}$ Using the double angle identity, $\begin{gathered} \tan \theta=\frac{2 \sin \left(\frac{\pi}{5}\right) \cos \left(\frac{\pi}{5}\right)}{2 \cos ^2\left(\frac{\pi}{5}\right)}=\frac{\sin \left(\frac{\pi}{5}\right)}{\cos \left(\frac{\pi}{5}\right)}=\tan \left(\frac{\pi}{5}\right) \\ \theta=\frac{\pi}{5} \end{gathered}$
Question Text	(c) $z^2-2 z \cos \binom{2 \pi}{5}+1$ is a quadratic factor of the polynomial $z^5-1$ . What is the other quadratic factors with real coefficients? Options: (a) $z^2+2 z \cos \left(\frac{2 \pi}{5}\right)+1$ (b) $z^2+2 z \cos \left(\frac{4 \pi}{5}\right)+1$ (c) $z^2-2 z \cos \left(\frac{3 \pi}{5}\right)+1$ (d) $z^2-2 z \cos \binom{4 \pi}{5}+1$
Total Mark	7
Correct Answer	(d)
Explanation	n/a
Mark Scheme	Note that since roots occur in conjugate pairs, $z^5-1$ has a quadratic factor, $\left(z-\left(\cos \left(\frac{4 \pi}{5}\right)+i \sin \left(\frac{4 \pi}{5}\right)\right)\right) \times\left(z-\left(\cos \left(\frac{4 \pi}{5}\right)-i \sin \left(\frac{4 \pi}{5}\right)\right)\right)=z^2-2 z \cos \left(\frac{4 \pi}{5}\right)+1$

Q14

Topic	1.5 Complex Numbers
Tag
Source	N19/5/MATHL/HP1/ENG/TZ1/XX/5
Question Text	(a) Select the correct root(s) of the equation $z^4=-16$ , where $z \in C$ . (a) $2 \operatorname{cis}\left(\frac{\pi}{2}\right)$ (b) $2 \operatorname{cis}\left(\frac{3 \pi}{4}\right)$ (c) $2 \operatorname{cis}(\pi)$ (d) $2 \sqrt{2} \operatorname{cis}\left(\frac{\pi}{4}\right)$ (e) $2 \sqrt{2} \operatorname{cis}\left(\frac{\pi}{2}\right)$
Total Mark	5
Correct Answer	(b)
Explanation	n/a
Mark Scheme	Write in modulus argument form, $z^4=16(\cos (\pi+2 \pi n)+i \sin (2 \pi n)$ Apply de Moivre's, $z=2 \operatorname{cis}\left(\frac{\pi}{4}+\frac{\pi n}{2}\right) \text { where } n=0,1,2,3$
Question Text	(b) The solutions form the vertices of a polygon in the complex plane. Find the area of the polygon.
Total Mark	2
Correct Answer	8
Explanation	n/a
Mark Scheme	Notice that the diagonal of this square is 4 The area is therefore, $4 \times 4 \times 0.5 = 8$

Q15

Topic	1.5 Complex Numbers
Tag
Source	M18/5/MATHL/HP1/ENG/TZ1/XX/11
Question Text	Consider $w=2\left(\cos \frac{\pi}{6}+i \sin \frac{\pi}{6}\right)$ (a) Select the answer choice that correctly represents $w^2$ and $w^3$ in modulus-argument form. (a) $w=2\left(\cos \frac{\pi}{6}+i \sin \frac{\pi}{6}\right)$ (b) $w^2=4\left(\cos \frac{\pi}{6}+i \sin \frac{\pi}{6}\right), w^3=8\left(\cos \frac{\pi}{6}+i \sin \frac{\pi}{6}\right)$ (c) $w^2=2\left(\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}\right), w^3=2\left(\cos \frac{\pi}{2}+i \sin \frac{\pi}{2}\right)$ (d) $w^2=4\left(\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}\right), w^3=8\left(\cos \frac{\pi}{2}+i \sin \frac{\pi}{2}\right)$
Total Mark	3
Correct Answer	(d)
Explanation	n/a
Mark Scheme	Use de Moivre's, $w^2=\left(2\left(\cos \frac{\pi}{6}+i \sin \frac{\pi}{6}\right)\right)^2, w^3=\left(2\left(\cos \frac{\pi}{6}+i \sin \frac{\pi}{6}\right)\right)^3 \\ w^2=4\left(\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}\right), w^3=8\left(\cos \frac{\pi}{2}+i \sin \frac{\pi}{2}\right)$
Question Text	(b) On an Argand diagram the points represented by the origin, $w¹,w², w³,w⁴$ form the vertices of a pentagon, P. What is the area of P?
Total Mark	4
Correct Answer	42
Explanation	n/a
Mark Scheme	Divide the pentagon into three triangles, each with a vertex on the origin, Use of triangle area $=\frac{1}{2} a b \sin C$ , Area of $P=\frac{1}{2} \times 2 \times 4 \times \sin \frac{\pi}{6}+\frac{1}{2} \times 4 \times 8 \times \sin \frac{\pi}{6}+\frac{1}{2} \times 8 \times 16 \times \sin \frac{\pi}{6}=42$

Q16

Topic	1.5 Complex Numbers
Tag
Source	M18/5/MATHL/HP1/ENG/TZ2/XX/7
Question Text	Consider the distinct complex numbers $z = a + ib , w = c + id$ , where $a, b, c, d ∈ R$ . Find the the value of the real part of $\frac{z-w}{z+w}$ when $\|z\|=\|w\|$ .
Total Mark	6
Correct Answer	0
Explanation	n/a
Mark Scheme	Consider the given information, $\frac{z-w}{z+w} =\frac{(a-c)+i(b-d)}{(a+c)+i(b+d)} \\$ $\frac{z-w}{z+w}=\frac{(a-c)+i(b-d)}{(a+c)+i(b+d)} \times \frac{(a+c)-i(b+d)}{(a+c)-i(b+d)} =\frac{\left(a^2-c^2\right)+\left(b^2-d^2\right)+i(a b-a d+b c-c d-a b+b c-a d+c d)}{(a+c)^2+(b+d)^2}$ Real part, $\frac{\left(a^2-c^2\right)+\left(b^2-d^2\right)}{(a+c)^2+(b+d)^2}=\frac{a^2+b^2-c^2-d^2}{(a+c)^2+(b+d)^2}$ As, $\|z\|=\|w\| \Rightarrow a^2+b^2=c^2+d^2$ Hence real part $=0$