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Date	May 2022	Marks available	3	Reference code	22M.1.AHL.TZ2.12
Level	Additional Higher Level	Paper	Paper 1 (without calculator)	Time zone	Time zone 2
Command term	Deduce	Question number	12	Adapted from	N/A

Question

In the following Argand diagram, the points $Z_{1}$ , $O$ and $Z_{2}$ are the vertices of triangle $Z_{1} {OZ}_{2}$ described anticlockwise.

The point $Z_{1}$ represents the complex number $z_{1} = r_{1} e^{i α}$ , where $r_{1} > 0$ . The point $Z_{2}$ represents the complex number $z_{2} = r_{2} e^{i θ}$ , where $r_{2} > 0$ .

Angles $α, θ$ are measured anticlockwise from the positive direction of the real axis such that $0 \leq α, θ < 2 π$ and $0 < α - θ < π$ .

In parts (c), (d) and (e), consider the case where $Z_{1} {OZ}_{2}$ is an equilateral triangle.

Let $z_{1}$ and $z_{2}$ be the distinct roots of the equation $z^{2} + a z + b = 0$ where $z \in ℂ$ and $a, b \in ℝ$ .

Show that $z_{1} {z_{2}}^{*} = r_{1} r_{2} e^{i (α - θ)}$ where ${z_{2}}^{*}$ is the complex conjugate of $z_{2}$ .

[2]

Given that $Re (z_{1} {z_{2}}^{*}) = 0$ , show that $Z_{1} {OZ}_{2}$ is a right-angled triangle.

[2]

Express $z_{1}$ in terms of $z_{2}$ .

[2]

c.i.

Hence show that ${z_{1}}^{2} + {z_{2}}^{2} = z_{1} z_{2}$ .

[4]

c.ii.

Use the result from part (c)(ii) to show that $a^{2} - 3 b = 0$ .

[5]

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.

[3]

Markscheme

${z_{2}}^{*} = r_{2} e^{-i θ}$ (A1)

$z_{1} {z_{2}}^{*} = r_{1} e^{i α} r_{2} e^{-i θ}$ A1

$z_{1} {z_{2}}^{*} = r_{1} r_{2} e^{i (α - θ)}$ AG

Note: Accept working in modulus-argument form

[2 marks]

$Re (z_{1} {z_{2}}^{*}) = r_{1} r_{2} \cos (α - θ) (= 0)$ A1

$α - θ = arcos 0 (r_{1}, r_{2} > 0)$

$α - θ = \frac{π}{2}$ (as $0 < α - θ < π$ ) A1

so $Z_{1} {OZ}_{2}$ is a right-angled triangle AG

[2 marks]

EITHER

$\frac{z_{1}}{z_{2}} (= \frac{r_{1}}{r_{2}} e^{i (α - θ)}) = e^{i \frac{π}{3}}$ (since $r_{1} = r_{2}$ ) (M1)

$z_{1} = r_{2} e^{i (θ + \frac{π}{3})} (= r_{2} e^{i θ} e^{i \frac{π}{3}})$ (M1)

THEN

$z_{1} = z_{2} e^{i \frac{π}{3}}$ A1

Note: Accept working in either modulus-argument form to obtain $z_{1} = z_{2} (\cos \frac{π}{3} + i \sin \frac{π}{3})$ or in Cartesian form to obtain $z_{1} = z_{2} (\frac{1}{2} + \frac{\sqrt{3}}{2} i)$ .

[2 marks]

c.i.

substitutes $z_{1} = z_{2} e^{i \frac{π}{3}}$ into ${z_{1}}^{2} + {z_{2}}^{2}$ M1

${z_{1}}^{2} + {z_{2}}^{2} = {z_{2}}^{2} e^{i \frac{2 π}{3}} + {z_{2}}^{2} (= {z_{2}}^{2} (e^{i \frac{2 π}{3}} + 1))$ A1

EITHER

$e^{i \frac{2 π}{3}} + 1 = e^{i \frac{π}{3}}$ A1

${z_{2}}^{2} (e^{i \frac{2 π}{3}} + 1) = {z_{2}}^{2} (- \frac{1}{2} + \frac{\sqrt{3}}{2} i + 1)$

$= {z_{2}}^{2} (\frac{1}{2} + \frac{\sqrt{3}}{2} i)$ A1

THEN

${z_{1}}^{2} + {z_{2}}^{2} = {z_{2}}^{2} e^{i \frac{π}{3}}$

$= z_{2} (z_{2} e^{i \frac{π}{3}})$ and $z_{2} e^{i \frac{π}{3}} = z_{1}$ A1

so ${z_{1}}^{2} + {z_{2}}^{2} = z_{1} z_{2}$ AG

Note: For candidates who work on the LHS and RHS separately to show equality, award M1A1 for ${z_{1}}^{2} + {z_{2}}^{2} = {z_{2}}^{2} e^{i \frac{2 π}{3}} + {z_{2}}^{2} (= {z_{2}}^{2} (e^{i \frac{2 π}{3}} + 1))$ , A1 for $z_{1} z_{2} = {z_{2}}^{2} e^{i \frac{π}{3}}$ and A1 for $e^{i \frac{2 π}{3}} + 1 = e^{i \frac{π}{3}}$ . Accept working in either modulus-argument form or in Cartesian form.

[4 marks]

c.ii.

METHOD 1

$z_{1} + z_{2} = - a$ and $z_{1} z_{2} = b$ (A1)

$a^{2} = {z_{1}}^{2} + {z_{2}}^{2} + 2 z_{1} z_{2}$ A1

$a^{2} = 2 z_{1} z_{2} + z_{1} z_{2} (= 3 z_{1} z_{2})$ A1

substitutes $b = z_{1} z_{2}$ into their expression M1

$a^{2} = 2 b + b$ OR $a^{2} = 3 b$ A1

Note: If $z_{1} + z_{2} = - a$ is not clearly recognized, award maximum (A0)A1A1M1A0.

so $a^{2} - 3 b = 0$ AG

METHOD 2

$z_{1} + z_{2} = - a$ and $z_{1} z_{2} = b$ (A1)

${(z_{1} + z_{2})}^{2} = {z_{1}}^{2} + {z_{2}}^{2} + 2 z_{1} z_{2}$ A1

${(z_{1} + z_{2})}^{2} = 2 z_{1} z_{2} + z_{1} z_{2} (= 3 z_{1} z_{2})$ A1

substitutes $b = z_{1} z_{2}$ and $z_{1} + z_{2} = - a$ into their expression M1

$a^{2} = 2 b + b$ OR $a^{2} = 3 b$ A1

Note: If $z_{1} + z_{2} = - a$ is not clearly recognized, award maximum (A0)A1A1M1A0.

so $a^{2} - 3 b = 0$ AG

[5 marks]

$a^{2} - 3 \times 12 = 0$

$a = \pm 6 (\Rightarrow z^{2} \pm 6 z + 12 = 0)$ A1

for $a = - 6 :$

$z_{1} = 3 + \sqrt{3} i, z_{2} = 3 - \sqrt{3} i$ and $α - θ = - \frac{5 π}{3}$ which does not satisfy $0 < α - θ < π$ R1

for $a = 6 :$

$z_{1} = - 3 - \sqrt{3} i, z_{2} = - 3 + \sqrt{3} i$ and $α - θ = \frac{π}{3}$ A1

so (for $0 < α - θ < π$ ), only one equilateral triangle can be formed from point $O$ and the two roots of this equation AG

[3 marks]

Examiners report

The vast majority of candidates scored full marks in parts (a) and (b). If they did not, it was normally due to the lack of rigour in setting out of the answer to a "show that" question. Part (c) was, though, more often than not poorly done. Many candidates could not use the given condition (equilateral triangle) to find $z_{1}$ in terms of $z_{2}$ . Part (d) was well answered by a rather high number of candidates.

Only a handful of students made good progress in (e), not even finding the possible values for $a$ .

[N/A]

c.i.

[N/A]

c.ii.

[N/A]