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Date	May 2021	Marks available	2	Reference code	21M.3.AHL.TZ2.2
Level	Additional Higher Level	Paper	Paper 3	Time zone	Time zone 2
Command term	Hence and Deduce	Question number	2	Adapted from	N/A

Question

This question asks you to investigate and prove a geometric property involving the roots of the equation $z^{n} = 1$ where $z \in ℂ$ for integers $n$ , where $n \geq 2$ .

The roots of the equation $z^{n} = 1$ where $z \in ℂ$ are $1, ω, ω^{2}, \dots, ω^{n - 1}$ , where $ω = e^{\frac{2 πi}{n}}$ . Each root can be represented by a point $P_{0}, P_{1}, P_{2}, \dots, P_{n - 1}$ , respectively, on an Argand diagram.

For example, the roots of the equation $z^{2} = 1$ where $z \in ℂ$ are $1$ and $ω$ . On an Argand diagram, the root $1$ can be represented by a point $P_{0}$ and the root $ω$ can be represented by a point $P_{1}$ .

Consider the case where $n = 3$ .

The roots of the equation $z^{3} = 1$ where $z \in ℂ$ are $1$ , $ω$ and $ω^{2}$ . On the following Argand diagram, the points $P_{0}, P_{1}$ and $P_{2}$ lie on a circle of radius $1$ unit with centre $O (0, 0)$ .

Line segments $[P_{0} P_{1}]$ and $[P_{0} P_{2}]$ are added to the Argand diagram in part (a) and are shown on the following Argand diagram.

$P_{0} P_{1}$ is the length of $[P_{0} P_{1}]$ and $P_{0} P_{2}$ is the length of $[P_{0} P_{2}]$ .

Consider the case where $n = 4$ .

The roots of the equation $z^{4} = 1$ where $z \in ℂ$ are $1, ω, ω^{2}$ and $ω^{3}$ .

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

For the case where $n = 5$ , the equation $z^{5} = 1$ where $z \in ℂ$ has roots $1, ω, ω^{2}, ω^{3}$ and $ω^{4}$ .

It can be shown that $P_{0} P_{1} \times P_{0} P_{2} \times P_{0} P_{3} \times P_{0} P_{4} = 5$ .

Now consider the general case for integer values of $n$ , where $n \geq 2$ .

The roots of the equation $z^{n} = 1$ where $z \in ℂ$ are $1, ω, ω^{2}, \dots, ω^{n - 1}$ . On an Argand diagram, these roots can be represented by the points $P_{0}, P_{1}, P_{2}, \dots, P_{n - 1}$ respectively where $[P_{0} P_{1}], [P_{0} P_{2}], \dots, [P_{0} P_{n - 1}]$ are line segments. The roots lie on a circle of radius $1$ unit with centre $O (0, 0)$ .

$P_{0} P_{1}$ can be expressed as $| 1 - ω |$ .

Consider $z^{n} - 1 = (z - 1) (z^{n - 1} + z^{n - 2} + \dots + z + 1)$ where $z \in ℂ$ .

Show that $(ω - 1) (ω^{2} + ω + 1) = ω^{3} - 1$ .

[2]

a.i.

Hence, deduce that $ω^{2} + ω + 1 = 0$ .

[2]

a.ii.

Show that $P_{0} P_{1} \times P_{0} P_{2} = 3$ .

[3]

By factorizing $z^{4} - 1$ , or otherwise, deduce that $ω^{3} + ω^{2} + ω + 1 = 0$ .

[2]

Show that $P_{0} P_{1} \times P_{0} P_{2} \times P_{0} P_{3} = 4$ .

[4]

Suggest a value for $P_{0} P_{1} \times P_{0} P_{2} \times \dots \times P_{0} P_{n - 1}$ .

[1]

Write down expressions for $P_{0} P_{2}$ and $P_{0} P_{3}$ in terms of $ω$ .

[2]

f.i.

Hence, write down an expression for $P_{0} P_{n - 1}$ in terms of $n$ and $ω$ .

[1]

f.ii.

Express $z^{n - 1} + z^{n - 2} + \dots + z + 1$ as a product of linear factors over the set $ℂ$ .

[3]

g.i.

Hence, using the part (g)(i) and part (f) results, or otherwise, prove your suggested result to part (e).

[4]

g.ii.

Markscheme

METHOD 1

attempts to expand $(ω - 1) (ω^{2} + ω + 1)$ (M1)

$= ω^{3} + ω^{2} + ω - ω^{2} - ω - 1$ A1

$= ω^{3} - 1$ AG

METHOD 2

attempts polynomial division on $\frac{ω^{3} - 1}{ω - 1}$ M1

$= ω^{2} + ω + 1$ A1

so $(ω - 1) (ω^{2} + ω + 1) = ω^{3} - 1$ AG

Note: In part (a), award marks as appropriate where $ω$ has been converted into Cartesian, modulus-argument (polar) or Euler form.

[2 marks]

a.i.

(since $ω$ is a root of $z^{3} = 1$ ) $\Rightarrow ω^{3} - 1 = 0$ R1

and $ω \neq 1$ R1

$\Rightarrow ω^{2} + ω + 1 = 0$ AG

Note: In part (a), award marks as appropriate where $ω$ has been converted into Cartesian, modulus-argument (polar) or Euler form.

[2 marks]

a.ii.

METHOD 1

attempts to find either $P_{0} P_{1}$ or $P_{0} P_{2}$ (M1)

accept any valid method

e.g. $2 \sin \frac{π}{3}, 1^{2} + 1^{2} - 2 \cos \frac{2 π}{3}, \frac{1}{\sin \frac{π}{6}} = \frac{P_{0} P_{1}}{\sin \frac{2 π}{3}}$ from either ${ΔOP}_{0} P_{1}$ or ${ΔOP}_{0} P_{2}$

e.g. use of Pythagoras’ theorem

e.g. $|1 - e^{i \frac{2 π}{3}}|, |1 - (- \frac{1}{2} + \frac{\sqrt{3}}{2} i)|$ by calculating the distance between $2$ points

$P_{0} P_{1} = \sqrt{3}$ A1

$P_{0} P_{2} = \sqrt{3}$ A1

Note: Award a maximum of M1A1A0 for any decimal approximation seen in the calculation of either $P_{0} P_{1}$ or $P_{0} P_{2}$ or both.

so $P_{0} P_{1} \times P_{0} P_{2} = 3$ AG

METHOD 2

attempts to find $P_{0} P_{1} \times P_{0} P_{2} = |1 - ω| |1 - ω^{2}|$ (M1)

$P_{0} P_{1} \times P_{0} P_{2} = |ω^{3} - ω^{2} - ω + 1|$ A1

$= |1 - (ω^{2} + ω + 1) + 2|$ and since $ω^{2} + ω + 1 = 0$ R1

so $P_{0} P_{1} \times P_{0} P_{2} = 3$ AG

[3 marks]

METHOD 1

$z^{4} - 1 = (z - 1) (z^{3} + z^{2} + z + 1)$ A1

( $ω$ is a root hence) $ω^{4} - 1 = 0$ and $ω \neq 1$ R1

$\Rightarrow ω^{3} + ω^{2} + ω + 1 = 0$ AG

Note: Condone the use of $ω$ throughout.

METHOD 2

considers the sum of roots of $z^{4} - 1 = 0$ (M1)

the sum of roots is zero (there is no $z^{3}$ term) A1

$\Rightarrow ω^{3} + ω^{2} + ω + 1 = 0$ AG

METHOD 3

substitutes for $ω$ (M1)

e.g. $LHS = e^{i \frac{3 π}{2}} + e^{πi} + e^{i \frac{π}{2}} + 1$

$= - i - 1 + i + 1$ A1

Note: This can be demonstrated geometrically or by using vectors. Accept Cartesian or modulus-argument (polar) form.

$\Rightarrow ω^{3} + ω^{2} + ω + 1 = 0$ AG

METHOD 4

$ω^{3} + ω^{2} + ω + 1 = \frac{ω^{4} - 1}{ω - 1}$ A1

$= \frac{0}{ω - 1} = 0$ as $ω \neq 1$ R1

$\Rightarrow ω^{3} + ω^{2} + ω + 1 = 0$ AG

[2 marks]

METHOD 1

$P_{0} P_{2} = 2$ A1

attempts to find either $P_{0} P_{1}$ or $P_{0} P_{3}$ (M1)

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

$P_{0} P_{1} = \sqrt{2}, P_{0} P_{3} = \sqrt{2}$ A1A1

Note: Award a maximum of A1M1A1A0 if labels such as $P_{0} P_{1}$ 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 $P_{0} P_{1}$ or $P_{0} P_{3}$ or both.

$P_{0} P_{1} \times P_{0} P_{2} \times P_{0} P_{3} = 4$ AG

METHOD 2

attempts to find $P_{0} P_{1} \times P_{0} P_{2} \times P_{0} P_{3} = |1 - ω| |1 - ω^{2}| |1 - ω^{3}|$ M1

$P_{0} P_{1} \times P_{0} P_{2} \times P_{0} P_{3} = |- ω^{6} + ω^{5} + ω^{4} - ω^{2} - ω + 1|$ A1

$= |- (- 1) + ω^{5} + 1 - (- 1) - ω + 1|$ since $ω^{6} = ω^{2} = - 1$ and $ω^{4} = 1$ A1

$= |ω^{5} - ω + 4|$ and since $ω^{5} = ω$ R1

so $P_{0} P_{1} \times P_{0} P_{2} \times P_{0} P_{3} = 4$ AG

METHOD 3

$P_{0} P_{2} = 2$ A1

attempts to find $P_{0} P_{1} \times P_{0} P_{3} = |1 - ω| |1 - ω^{3}|$ M1

$P_{0} P_{1} \times P_{0} P_{3} = |ω^{4} - ω^{3} - ω + 1|$ A1

$= |2 - (- ω) - ω|$ since $ω^{4} = 1$ and $ω^{3} = - ω$ R1

so $P_{0} P_{1} \times P_{0} P_{2} \times P_{0} P_{3} = 4$ AG

[4 marks]

$(P_{0} P_{1} \times P_{0} P_{2} \times \dots \times P_{0} P_{n - 1}) = n$ A1

[1 mark]

$P_{0} P_{2} = |1 - ω^{2}|, P_{0} P_{3} = |1 - ω^{3}|$ A1A1

[2 marks]

f.i.

$P_{0} P_{n - 1} = |1 - ω^{n - 1}|$ A1A1

Note: Accept $|1 - ω|$ from symmetry.

[1 mark]

f.ii.

$z^{n} - 1 = (z - 1) (z^{n - 1} + z^{n - 2} + \dots + z + 1)$

considers the equation $z^{n - 1} + z^{n - 2} + \dots + z + 1 = 0$ (M1)

the roots are $ω, ω^{2}, \dots, ω^{n - 1}$ (A1)

so $(z - ω) (z - ω^{2}) \dots (z - ω^{n - 1})$ A1

[3 marks]

g.i.

METHOD 1

substitutes $z = 1$ into $(z - ω) (z - ω^{2}) \dots (z - ω^{n - 1}) \equiv z^{n - 1} + z^{n - 2} + \dots + z + 1$ M1

$(1 - ω) (1 - ω^{2}) \dots (1 - ω^{n - 1}) = n$ (A1)

takes modulus of both sides M1

$|(1 - ω) (1 - ω^{2}) \dots (1 - ω^{n - 1})| = |n|$

$|1 - ω| |1 - ω^{2}| \dots |1 - ω^{n - 1}| = n$ A1

so $P_{0} P_{1} \times P_{0} P_{2} \times \dots \times P_{0} P_{n - 1} = n$ AG

Note: Award a maximum of M1A1FTM1A0 from part (e).

METHOD 2

$(1 - ω), (1 - ω^{2}), \dots, (1 - ω^{n - 1})$ are the roots of ${(1 - v)}^{n - 1} + {(1 - v)}^{n - 2} + \dots + (1 - v) + 1 = 0$ M1

coefficient of $v^{n - 1}$ is ${(- 1)}^{n - 1}$ and the coefficient of $1$ is $n$ A1

product of the roots is $\frac{{(- 1)}^{n - 1} n}{{(- 1)}^{n - 1}} = n$ A1

$|1 - ω| |1 - ω^{2}| \dots |1 - ω^{n - 1}| = n$ A1

so $P_{0} P_{1} \times P_{0} P_{2} \times \dots \times P_{0} P_{n - 1} = n$ AG

[4 marks]

g.ii.

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

f.i.

[N/A]

f.ii.

[N/A]

g.i.

[N/A]

g.ii.

Syllabus sections

Topic 1—Number and algebra » AHL 1.12—Complex numbers – Cartesian form and Argand diag

Show 43 related questions

22M.3.AHL.TZ1.2f.i:
Use this diagram to determine the roots of the corresponding equation of the form $(z - r) (z^{2} - 2 a z + a^{2} + 16) = 0$ for $z \in ℂ$ .
22M.1.AHL.TZ2.12c.i:
Express $z_{1}$ in terms of $z_{2}$ .
22M.1.AHL.TZ2.12d:
Use the result from part (c)(ii) to show that $a^{2} - 3 b = 0$ .
21N.3.AHL.TZ0.1f:
Sketch the graph of $x^{2} - y^{2} = 1$ , stating the coordinates of any axis intercepts and the equation of each asymptote.
16N.1.AHL.TZ0.H_12c:
Find the values of $x$ that satisfy the equation $\left| p \right| = \left| q \right|$ .
19N.1.AHL.TZ0.H_5b:
The solutions form the vertices of a polygon in the complex plane. Find the area of the polygon.
16N.1.AHL.TZ0.H_12b:
Show that $(\omega - 3{\omega ^2})({\omega ^2} - 3\omega ) = 13$ .
21M.3.AHL.TZ2.2g.ii:
Hence, using the part (g)(i) and part (f) results, or otherwise, prove your suggested result to part (e).
22M.3.AHL.TZ1.2e:
Deduce from part (d)(i) that the complex roots of the equation $(z - r) (z^{2} - 2 a z + a^{2} + b^{2}) = 0$ can be expressed as $a \pm i \sqrt{g' (a)}$ .
22M.3.AHL.TZ1.2f.ii:
State the coordinates of $C_{2}$ .
22M.1.AHL.TZ2.12b:
Given that $Re (z_{1} {z_{2}}^{*}) = 0$ , show that $Z_{1} {OZ}_{2}$ is a right-angled triangle.
22M.1.AHL.TZ2.12c.ii:
Hence show that ${z_{1}}^{2} + {z_{2}}^{2} = z_{1} z_{2}$ .
22M.1.AHL.TZ2.12e:
Consider the equation $z^{2} + a z + 12 = 0$ , where $z \in ℂ$ and $a \in ℝ$ .

Given that $0 < α - θ < π$ , deduce that only one equilateral triangle $Z_{1} {OZ}_{2}$ can be formed from the point $O$ and the roots of this equation.
SPM.2.AHL.TZ0.8:
The complex numbers $w$ and $z$ satisfy the equations

$\frac{w}{z} = 2{\text{i}}$

${{\text{z}}^ * } - 3w = 5 + 5{\text{i}}$ .

Find $w$ and $z$ in the form $a + b{\text{i}}$ where $a$ , ${\text{b}} \in \mathbb{Z}$ .
21M.3.AHL.TZ2.2g.i:
Express $z^{n - 1} + z^{n - 2} + \dots + z + 1$ as a product of linear factors over the set $ℂ$ .
17M.1.AHL.TZ2.H_5:
In the following Argand diagram the point A represents the complex number $- 1 + 4{\text{i}}$ and the point B represents the complex number $- 3 + 0{\text{i}}$ . The shape of ABCD is a square. Determine the complex numbers represented by the points C and D.
21M.3.AHL.TZ2.2a.i:
Show that $(ω - 1) (ω^{2} + ω + 1) = ω^{3} - 1$ .
21M.3.AHL.TZ2.2b:
Show that $P_{0} P_{1} \times P_{0} P_{2} = 3$ .
21M.3.AHL.TZ2.2e:
Suggest a value for $P_{0} P_{1} \times P_{0} P_{2} \times \dots \times P_{0} P_{n - 1}$ .
21M.3.AHL.TZ2.2d:
Show that $P_{0} P_{1} \times P_{0} P_{2} \times P_{0} P_{3} = 4$ .
21M.3.AHL.TZ2.2f.ii:
Hence, write down an expression for $P_{0} P_{n - 1}$ in terms of $n$ and $ω$ .
21N.1.AHL.TZ0.12b:
Plot the points $A$ , $B$ and $C$ on an Argand diagram.
21N.1.AHL.TZ0.12d:
By using de Moivre’s theorem, show that $α = \frac{1}{1 - e^{i \frac{π}{6}}}$ is a root of this equation.
21N.3.AHL.TZ0.1b:
Show that ${(f (t))}^{2} + {(g (t))}^{2} = f (2 t)$ .
21N.3.AHL.TZ0.1c.i:
$f (i u)$ .
21N.3.AHL.TZ0.1g:
The hyperbola with equation $x^{2} - y^{2} = 1$ can be rotated to coincide with the curve defined by $x y = k, k \in ℝ$ .

Find the possible values of $k$ .
22M.1.AHL.TZ1.9a:
Find an expression for $z_{1} z_{2}$ in terms of $b$ .
22M.1.AHL.TZ2.12a:
Show that $z_{1} {z_{2}}^{*} = r_{1} r_{2} e^{i (α - θ)}$ where ${z_{2}}^{*}$ is the complex conjugate of $z_{2}$ .
21N.1.AHL.TZ0.12e:
Determine the value of $Re (α)$ .
21N.3.AHL.TZ0.1a:
Verify that $u = f (t)$ satisfies the differential equation $\frac{d^{2} u}{d t^{2}} = u$ .
21N.3.AHL.TZ0.1e:
Show that ${(f (t))}^{2} - {(g (t))}^{2} = {(f (i u))}^{2} - {(g (i u))}^{2}$ .
16N.1.AHL.TZ0.H_12d:
Solve the inequality $\operatorname{Re} (pq) + 8 < {\left( {\operatorname{Im} (pq)} \right)^2}$ .
16N.1.AHL.TZ0.H_12a:
Determine the value of

(i) $1 + \omega + {\omega ^2}$ ;

(ii) $1 + \omega {\text{*}} + {(\omega {\text{*}})^2}$ .
21M.3.AHL.TZ2.2f.i:
Write down expressions for $P_{0} P_{2}$ and $P_{0} P_{3}$ in terms of $ω$ .
20N.1.AHL.TZ0.H_4:
Consider the equation $\frac{2 z}{3 - z *} = i$ , where $z = x + i y$ and $x, y \in ℝ$ .

Find the value of $x$ and the value of $y$ .
EXN.1.AHL.TZ0.12a:
Use the binomial theorem to expand ${(\cos θ + i \sin θ)}^{4}$ . Give your answer in the form $a + b i$ where $a$ and $b$ are expressed in terms of $\sin θ$ and $\cos θ$ .
21M.2.AHL.TZ2.8:
Consider $z = \cos θ + i \sin θ$ where $z \in ℂ, z \neq 1$ .

Show that $Re (\frac{1 + z}{1 - z}) = 0$ .
21M.3.AHL.TZ2.2c:
By factorizing $z^{4} - 1$ , or otherwise, deduce that $ω^{3} + ω^{2} + ω + 1 = 0$ .
21N.1.AHL.TZ0.12a.i:
Verify that $ω_{1} = 1 + e^{i \frac{π}{6}}$ is a root of this equation.
21N.1.AHL.TZ0.12a.ii:
Find $ω_{2}$ and $ω_{3}$ , expressing these in the form $a + e^{i θ}$ , where $a \in ℝ$ and $θ > 0$ .
21N.1.AHL.TZ0.12c:
Find $AC$ .
21N.3.AHL.TZ0.1c.ii:
$g (i u)$ .
21N.3.AHL.TZ0.1d:
Hence find, and simplify, an expression for ${(f (i u))}^{2} + {(g (i u))}^{2}$ .

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