IB Maths AA HL 1.12 Complex numbers

[Maximum number: 11]

This question asks you to investigate and prove a geometric property involving the roots of the equation $z^{n}=1$ where $z \in \mathbb{C}$ for integers n, where $n \geq 2$ .
The roots of the equation $z^{n}=1$ where $z \in \mathbb{C}$ are $1, \omega, \omega^{2}, \ldots, \omega^{n-1}$ , where $\omega=\mathrm{e}^{\frac{2 \pi \mathrm{i}}{n}}$ . Each root can be represented by a point $\mathrm{P}_{0}, \mathrm{P}_{1}, \mathrm{P}_{2}, \ldots, \mathrm{P}_{n-1}$ , respectively, on an Argand diagram.
For example, the roots of the equation $z^{2}=1$ where $z \in \mathbb{C}$ are 1 and $\omega$ . On an Argand diagram, the root 1 can be represented by a point $\mathrm{P}_{0}$ and the root $\omega$ can be represented by a point $\mathrm{P}_{1}$ .
Consider the case where n=3.
The roots of the equation $z^{3}=1$ where $z \in \mathbb{C}$ are $1, \omega$ and $\omega^{2}$ . On the following Argand diagram, the points $\mathrm{P}_{0}, \mathrm{P}_{1}$ and $\mathrm{P}_{2}$ lie on a circle of radius 1 unit with centre O(0,0).

Figure for Question AHL 1.12 (HL)—Complex numbers question 1 — IB Maths AA HL

(a) (i) METHOD 1
attempts to expand $(\omega-1)\left(\omega^{2}+\omega+1\right)$ $=\omega^{3}+\omega^{2}+\omega-\omega^{2}-\omega-1$ $=\omega^{3}-1$

METHOD 2
attempts polynomial division on $\frac{\omega^{3}-1}{\omega-1}$ $=\omega^{2}+\omega+1$
so $(\omega-1)\left(\omega^{2}+\omega+1\right)=\omega^{3}-1$

(a)

Show that $\mathrm{P}_{0} \mathrm{P}_{1} \times \mathrm{P}_{0} \mathrm{P}_{2}=3$ .

Consider the case where n=4.
The roots of the equation $z^{4}=1$ where $z \in \mathbb{C}$ are $1, \omega, \omega^{2}$ and $\omega^{3}$ .

[ 3 ]

(b) METHOD 1
attempts to find either $\mathrm{P}_{0} \mathrm{P}_{1}$ or $\mathrm{P}_{0} \mathrm{P}_{2}$

accept any valid method
e.g. $2 \sin \frac{\pi}{3}, 1^{2}+1^{2}-2 \cos \frac{2 \pi}{3}, \frac{1}{\sin \frac{\pi}{6}}=\frac{\mathrm{P}_{0} \mathrm{P}_{1}}{\sin \frac{2 \pi}{3}}$ from either $\Delta \mathrm{OP}_{0} \mathrm{P}_{1}$ or $\Delta \mathrm{OP}_{0} \mathrm{P}_{2}$
e.g. use of Pythagoras' theorem
e.g. $\left|1-\mathrm{e}^{\mathrm{i} \frac{2 \pi}{3}}\right|,\left|1-\left(-\frac{1}{2}+\frac{\sqrt{3}}{2} \mathrm{i}\right)\right|$ by calculating the distance between 2 points
$\mathrm{P}_{0} \mathrm{P}_{1}=\sqrt{3}$ $\mathrm{P}_{0} \mathrm{P}_{2}=\sqrt{3}$

Note: Award a maximum of M1A1A0 for any decimal approximation seen in the calculation of either $\mathrm{P}_{0} \mathrm{P}_{1}$ or $\mathrm{P}_{0} \mathrm{P}_{2}$ or both.
so $\mathrm{P}_{0} \mathrm{P}_{1} \times \mathrm{P}_{0} \mathrm{P}_{2}=3$

METHOD 2
attempts to find $\mathrm{P}_{0} \mathrm{P}_{1} \times \mathrm{P}_{0} \mathrm{P}_{2}=|1-\omega|\left|1-\omega^{2}\right|$ $\mathrm{P}_{0} \mathrm{P}_{1} \times \mathrm{P}_{0} \mathrm{P}_{2}=\left|\omega^{3}-\omega^{2}-\omega+1\right|$ $=\left|1-\left(\omega^{2}+\omega+1\right)+2\right|$ and since $\omega^{2}+\omega+1=0$
so $\mathrm{P}_{0} \mathrm{P}_{1} \times \mathrm{P}_{0} \mathrm{P}_{2}=3$

(b)

On the following Argand diagram, the points $\mathrm{P}_{0}, \mathrm{P}_{1}, \mathrm{P}_{2}$ and $\mathrm{P}_{3}$ lie on a circle of radius 1 unit with centre $\mathrm{O}(0,0) .\left[\mathrm{P}_{0} \mathrm{P}_{1}\right],\left[\mathrm{P}_{0} \mathrm{P}_{2}\right]$ and $\left[\mathrm{P}_{0} \mathrm{P}_{3}\right]$ are line segments.

Figure for Question (b) — IB Maths AA HL

Show that $\mathrm{P}_{0} \mathrm{P}_{1} \times \mathrm{P}_{0} \mathrm{P}_{2} \times \mathrm{P}_{0} \mathrm{P}_{3}=4$ .

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METHOD 1
$\mathrm{P}_{0} \mathrm{P}_{2}=2$
attempts to find either $\mathrm{P}_{0} \mathrm{P}_{1}$ or $\mathrm{P}_{0} \mathrm{P}_{3}$

Note: For example, $\mathrm{P}_{0} \mathrm{P}_{1}=|1-\mathrm{i}|$ and $\mathrm{P}_{0} \mathrm{P}_{3}=|1+\mathrm{i}|$ .
Various geometric and trigonometric approaches can be used by candidates.

Note: Award a maximum of A1 M1 A1 A O if labels such as $\mathrm{P}_{0} \mathrm{P}_{1}$ are not clearly shown.

Award full marks if the lengths are shown on a clearly labelled diagram. Award a maximum of A1M1A1A0 for any decimal approximation seen in the calculation of either $\mathrm{P}_{0} \mathrm{P}_{1}$ or $\mathrm{P}_{0} \mathrm{P}_{3}$ or both.

METHOD 2
attempts to find $\mathrm{P}_{0} \mathrm{P}_{1} \times \mathrm{P}_{0} \mathrm{P}_{2} \times \mathrm{P}_{0} \mathrm{P}_{3}=|1-\omega|\left|1-\omega^{2}\right|\left|1-\omega^{3}\right| \quad$ M1
$\mathrm{P}_{0} \mathrm{P}_{1} \times \mathrm{P}_{0} \mathrm{P}_{2} \times \mathrm{P}_{0} \mathrm{P}_{3}=\left|-\omega^{6}+\omega^{5}+\omega^{4}-\omega^{2}-\omega+1\right| \quad$ A1
$=\left|-(-1)+\omega^{5}+1-(-1)-\omega+1\right|$ since $\omega^{6}=\omega^{2}=-1$ and $\omega^{4}=1 \quad$ A1
$=\left|\omega^{5}-\omega+4\right|$ and since $\omega^{5}=\omega \quad \boldsymbol{R} \mathbf{1}$
so $\mathrm{P}_{0} \mathrm{P}_{1} \times \mathrm{P}_{0} \mathrm{P}_{2} \times \mathrm{P}_{0} \mathrm{P}_{3}=4 \quad \boldsymbol{A G}$

METHOD 3

$P_{0} P_{2}=2$
attempts to find $\mathrm{P}_{0} \mathrm{P}_{1} \times \mathrm{P}_{0} \mathrm{P}_{3}=|1-\omega|\left|1-\omega^{3}\right|$ $\mathrm{P}_{0} \mathrm{P}_{1} \times \mathrm{P}_{0} \mathrm{P}_{3}=\left|\omega^{4}-\omega^{3}-\omega+1\right|$ $=|2-(-\omega)-\omega|$ since $\omega^{4}=1$ and $\omega^{3}=-\omega$
so $\mathrm{P}_{0} \mathrm{P}_{1} \times \mathrm{P}_{0} \mathrm{P}_{2} \times \mathrm{P}_{0} \mathrm{P}_{3}=4$

(c)

For the case where n=5, the equation $z^{5}=1$ where $z \in \mathbb{C}$ has roots $1, \omega, \omega^{2}, \omega^{3}$ and $\omega^{4}$ .
It can be shown that $\mathrm{P}_{0} \mathrm{P}_{1} \times \mathrm{P}_{0} \mathrm{P}_{2} \times \mathrm{P}_{0} \mathrm{P}_{3} \times \mathrm{P}_{0} \mathrm{P}_{4}=5$ .
Now consider the general case for integer values of n, where $n \geq 2$ .
The roots of the equation $z^{n}=1$ where $z \in \mathbb{C}$ are $1, \omega, \omega^{2}, \ldots, \omega^{n-1}$ . On an Argand diagram, these roots can be represented by the points $\mathrm{P}_{0}, \mathrm{P}_{1}, \mathrm{P}_{2}, \ldots, \mathrm{P}_{n-1}$ respectively where $\left[\mathrm{P}_{0} \mathrm{P}_{1}\right],\left[\mathrm{P}_{0} \mathrm{P}_{2}\right], \ldots,\left[\mathrm{P}_{0} \mathrm{P}_{n-1}\right]$ are line segments. The roots lie on a circle of radius 1 unit with centre O(0,0).

Suggest a value for $\mathrm{P}_{0} \mathrm{P}_{1} \times \mathrm{P}_{0} \mathrm{P}_{2} \times \ldots \times \mathrm{P}_{0} \mathrm{P}_{n-1}$ .
$\mathrm{P}_{0} \mathrm{P}_{1}$ can be expressed as $|1-\omega|$ .

[ 1 ]

$\quad\left(\mathrm{P}_{0} \mathrm{P}_{1} \times \mathrm{P}_{0} \mathrm{P}_{2} \times \ldots \times \mathrm{P}_{0} \mathrm{P}_{\mathrm{n}-1}\right)=n$

(d)

Write down expressions for $\mathrm{P}_{0} \mathrm{P}_{2}$ and $\mathrm{P}_{0} \mathrm{P}_{3}$ in terms of $\omega$ .

[ 2 ]

$\quad \mathrm{P}_{0} \mathrm{P}_{2}=\left|1-\omega^{2}\right|, \mathrm{P}_{0} \mathrm{P}_{3}=\left|1-\omega^{3}\right|$

(e)

Hence, write down an expression for $\mathrm{P}_{0} \mathrm{P}_{n-1}$ in terms of n and $\omega$ .

Consider $z^{n}-1=(z-1)\left(z^{n-1}+z^{n-2}+\ldots+z+1\right)$ where $z \in \mathbb{C}$ .

[ 1 ]

$\quad \mathrm{P}_{0} \mathrm{P}_{n-1}=\left|1-\omega^{n-1}\right|$

Note: Accept $|1-\omega|$ from symmetry.

IB Maths AA HL 1.12 Complex numbers Question Bank

Exam points

AHL 1.12 (HL)—Complex numbers question 1

Question (a)

Question (b)

Question (c)

Question (d)

Question (e)