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Date May 2014 Marks available 1 Reference code 14M.2.hl.TZ2.13
Level HL only Paper 2 Time zone TZ2
Command term Plot Question number 13 Adapted from N/A

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 π2 in the anticlockwise direction about the origin O to become point A. Point B is rotated through π2 in the clockwise direction about O to become point B.

Consider z=r(cosθ+isinθ), zC.

Use mathematical induction to prove that zn=rn(cosnθ+isinnθ), nZ+.

[7]
a.

Given u=1+3i and v=1i,

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

(ii)     hence find u3v4.

[4]
b.

Plot point A and point B on the Argand diagram.

[1]
c.

Find the area of triangle OAB.

[3]
d.

Given that u and v are roots of the equation z4+bz3+cz2+dz+e=0, where b, c, d, eR,

find the values of b, c, d and e.

[5]
e.

Markscheme

let 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.

Examiners report

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Syllabus sections

Topic 1 - Core: Algebra » 1.6 » Modulus–argument (polar) form z = r\left( {\cos \theta + {\text{i}}\sin \theta } \right) = r{\text{cis}}\theta = r{e^{{\text{i}}\theta }}
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