IB DP Math AA Topic : AHL 1.13: Modulus–argument (polar) form-IB Style Questions – HL Paper 2

Question

Consider the complex numbers  z = 2 $$(cos\frac{\pi }{5}+isin\frac{\pi }{5})$$ and w = 8 $$(cos\frac{2k\pi }{5}- isin\frac{2k\pi }{5})$$ where $$k\in \mathbb{Z}^+$$

1. Find the modulus of $$zw$$. [1]

2. Find the argument of $$zw$$ in terms of $$k$$ . [2]

Suppose that $$zw\in \mathbb{Z}$$ .

3. (i) Find the minimum value of $$k$$ .

(ii) For the value of k found in part (i), find the value of $$zw$$  . [4]

Ans:

(a) (|zw|=)16

(b)

attempt to find arg (z) +arg (w)

arg (zw)=arg (z)+arg (w)

= $$\frac{\pi}{5}-\frac{2k\pi}{5}=\frac{(1-2k)\pi}{5}$$

(c)

(i)

zw $$\varepsilon Z\Leftrightarrow arg (zw)$$ is a multiple of $$\pi$$

$$\Leftrightarrow 1-2k$$ is a multiple of 5

k =3

(ii) zw =16 $$cos(-\pi)+isin(-\pi)$$- 16

Question

Consider the complex number $$z = \frac{{2 + 7{\text{i}}}}{{6 + 2{\text{i}}}}$$.

a.Express $$z$$ in the form $$a + {\text{i}}b$$, where $$a,\,b \in \mathbb{Q}$$.[2]

b.Find the exact value of the modulus of $$z$$.[2]

c.Find the argument of $$z$$, giving your answer to 4 decimal places.[2]

Markscheme

$$z = \frac{{\left( {2 + 7{\text{i}}} \right)}}{{\left( {6 + 2{\text{i}}} \right)}} \times \frac{{\left( {6 – 2{\text{i}}} \right)}}{{\left( {6 – 2{\text{i}}} \right)}}$$ (M1)

$$= \frac{{26 + 38{\text{i}}}}{{40}} = \left( {\frac{{13 + 19{\text{i}}}}{{20}} = 0.65 + 0.95{\text{i}}} \right)$$ A1

[2 marks]

a.

attempt to use $$\left| z \right| = \sqrt {{a^2} + {b^2}}$$ (M1)

$$\left| z \right| = \sqrt {\frac{{53}}{{40}}} \left( { = \frac{{\sqrt {530} }}{{20}}} \right)$$ or equivalent A1

Note: A1 is only awarded for the correct exact value.

[2 marks]

b.

EITHER

arg $$z$$ = arg(2 + 7i) − arg(6 + 2i) (M1)

OR

arg $$z$$ = arctan$$\left( {\frac{{19}}{{13}}} \right)$$ (M1)

THEN

arg $$z$$ = 0.9707 (radians) (= 55.6197 degrees) A1

Note: Only award the last A1 if 4 decimal places are given.

[2 marks]

c.

[N/A]

a.

[N/A]

b.

[N/A]

c.

Question

(a)     Solve the equation $${z^3} = – 2 + 2{\text{i}}$$, giving your answers in modulus-argument form.

(b)     Hence show that one of the solutions is 1 + i when written in Cartesian form.

Markscheme

(a)     $${z^3} = 2\sqrt 2 {{\text{e}}^{\frac{{3\pi {\text{i}}}}{4}}}$$     (M1)(A1)

$${z_1} = \sqrt 2 {{\text{e}}^{\frac{{\pi {\text{i}}}}{4}}}$$     A1

adding or subtracting $$\frac{{2\pi {\text{i}}}}{3}$$     M1

$${z_2} = \sqrt 2 {{\text{e}}^{\frac{{\pi {\text{i}}}}{4} + \frac{{2\pi {\text{i}}}}{3}}} = \sqrt 2 {{\text{e}}^{\frac{{11\pi {\text{i}}}}{{12}}}}$$     A1

$${z_3} = \sqrt 2 {{\text{e}}^{\frac{{\pi {\text{i}}}}{4} – \frac{{2\pi {\text{i}}}}{3}}} = \sqrt 2 {{\text{e}}^{ – \frac{{5\pi {\text{i}}}}{{12}}}}$$     A1

Notes: Accept equivalent solutions e.g. $${z_3} = \sqrt 2 {{\text{e}}^{\frac{{19\pi {\text{i}}}}{{12}}}}$$

Award marks as appropriate for solving $${(a + b{\text{i}})^3} = – 2 + 2{\text{i}}$$.

(b)     $$\sqrt 2 {{\text{e}}^{\frac{{\pi {\text{i}}}}{4}}}{\text{ }}\left( { = \sqrt 2 \left( {\frac{1}{{\sqrt 2 }} + \frac{{\text{i}}}{{\sqrt 2 }}} \right)} \right)$$     A1

= 1 + i     AG

Note: Accept geometrical reasoning.

[7 marks]

Examiners report

Many students incorrectly found the argument of $${z^3}$$ to be $$\arctan \left( {\frac{2}{{ – 2}}} \right) = – \frac{\pi }{4}$$. Of those students correctly finding one solution, many were unable to use symmetry around the origin, to find the other two. In part (b) many students found the cube of 1 + i which could not be awarded marks as it was not “hence”.

Question

The complex numbers $$u$$ and $$v$$ are represented by point A and point B respectively on an Argand diagram.

Point A is rotated through $$\frac{\pi }{2}$$ in the anticlockwise direction about the origin O to become point $${\text{A}}’$$. Point B is rotated through $$\frac{\pi }{2}$$ in the clockwise direction about O to become point $${\text{B}}’$$.

a.Consider $$z = r(\cos \theta + {\text{i}}\sin \theta ),{\text{ }}z \in \mathbb{C}$$.

Use mathematical induction to prove that $${z^n} = {r^n}(\cos n\theta + {\text{i}}\sin n\theta ),{\text{ }}n \in {\mathbb{Z}^ + }$$.[7]

b.Given $$u = 1 + \sqrt 3 {\text{i}}$$ and $$v = 1 – {\text{i}}$$,

(i)     express $$u$$ and $$v$$ in modulus-argument form;

(ii)     hence find $${u^3}{v^4}$$.[4]

c.Plot point A and point B on the Argand diagram.[1]

d.Find the area of triangle O$${\text{A}}’$$$${\text{B}}’$$.[3]

e.Given that $$u$$ and $$v$$ are roots of the equation $${z^4} + b{z^3} + c{z^2} + dz + e = 0$$, where $$b,{\text{ }}c,{\text{ }}d,{\text{ }}e \in \mathbb{R}$$,

find the values of $$b,{\text{ }}c,{\text{ }}d$$ and $$e$$.[5]

Markscheme

let $${\text{P}}(n)$$ be the proposition $${z^n} = {r^n}(\cos n\theta + {\rm{i}}\sin n\theta ),n \in {¢^ + }$$

let $$n = 1 \Rightarrow$$

$${\text{LHS}} = r(\cos \theta + {\text{i}}\sin \theta )$$

$${\text{RHS}} = r(\cos \theta + {\text{i}}\sin \theta ),{\text{ }}\therefore {\text{P}}(1)$$ is true     R1

assume true for $$n = k \Rightarrow {r^k}{(\cos \theta + {\text{i}}\sin \theta )^k} = {r^k}\left( {\cos (k\theta ) + {\text{i}}\sin (k\theta )} \right)$$     M1

Note:     Only award the M1 if truth is assumed.

now show $$n = k$$ true implies $$n = k + 1$$ also true

$${r^{k + 1}}{(\cos \theta + {\text{i}}\sin \theta )^{k + 1}} = {r^{k + 1}}{(\cos \theta + {\text{i}}\sin \theta )^k}(\cos \theta + {\text{i}}\sin \theta )$$     M1

$$= {r^{k + 1}}\left( {\cos (k\theta ) + {\text{i}}\sin (k\theta )} \right)(\cos \theta + {\text{i}}\sin \theta )$$

$$= {r^{k + 1}}\left( {\cos (k\theta )\cos \theta – \sin (k\theta )\sin \theta + {\text{i}}\left( {\sin (k\theta )\cos \theta + \cos (k\theta )\sin \theta } \right)} \right)$$     A1

$$= {r^{k + 1}}\left( {\cos (k\theta + \theta ) + {\text{i}}\sin (k\theta + \theta )} \right)$$     A1

$$= {r^{k + 1}}\left( {\cos (k + 1)\theta + {\text{i}}\sin (k + 1)\theta } \right) \Rightarrow n = k + 1$$ is true     A1

$${\text{P}}(k)$$ true implies $${\text{P}}(k + 1)$$ true and $${\text{P}}(1)$$ is true, therefore by mathematical induction statement is true for $$n \geqslant 1$$     R1

Note:     Only award the final R1 if the first 4 marks have been awarded.

[7 marks]

a.

(i)     $$u = 2{\text{cis}}\left( {\frac{\pi }{3}} \right)$$     A1

$$v = \sqrt 2 {\text{cis}}\left( { – \frac{\pi }{4}} \right)$$     A1

Notes:     Accept 3 sf answers only. Accept equivalent forms.

Accept $$2{e^{\frac{\pi }{3}i}}$$ and $$\sqrt 2 {e^{ – \frac{\pi }{4}i}}$$.

(ii)     $${u^3} = {2^3}{\text{cis}}(\pi ) = – 8$$

$${v^4} = 4{\text{cis}}( – \pi ) = – 4$$     (M1)

$${u^3}{v^4} = 32$$     A1

Notes:     Award (M1) for an attempt to find $${u^3}$$ and $${v^4}$$.

Accept equivalent forms.

[4 marks]

b.

A1

Note:     Award A1 if A or $${\text{1 + }}\sqrt 3 i$$ and B or $$1 – i$$ are in their correct quadrants, are aligned vertically and it is clear that $$\left| u \right| > \left| v \right|$$.

[1 mark]

c.

Area $$= \frac{1}{2} \times 2 \times \sqrt 2 \times \sin \left( {\frac{{5\pi }}{{12}}} \right)$$     M1A1

$$= 1.37{\text{ }}\left( { = \frac{{\sqrt 2 }}{4}\left( {\sqrt 6 + \sqrt 2 } \right)} \right)$$     A1

Notes:     Award M1A0A0 for using $$\frac{{7\pi }}{{12}}$$.

[3 marks]

d.

$$(z – 1 + {\text{i}})(z – 1 – {\text{i}}) = {z^2} – 2z + 2$$     M1A1

Note:     Award M1 for recognition that a complex conjugate is also a root.

$$\left( {z – 1 – \sqrt 3 {\text{i}}} \right)\left( {z – 1 + \sqrt 3 {\text{i}}} \right) = {z^2} – 2z + 4$$     A1

$$\left( {{z^2} – 2z + 2} \right)\left( {{z^2} – 2z + 4} \right) = {z^4} – 4{z^3} + 10{z^2} – 12z + 8$$     M1A1

Note:     Award M1 for an attempt to expand two quadratics.

[5 marks]

e.

[N/A]

a.

[N/A]

b.

[N/A]

c.

[N/A]

d.

[N/A]

e.
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