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Date November 2012 Marks available 7 Reference code 12N.2.hl.TZ0.10
Level HL only Paper 2 Time zone TZ0
Command term Find Question number 10 Adapted from N/A

Question

Let ω=cosθ+isinθ . Find, in terms of θ , the modulus and argument of (1ω2) .

 

 

 

Markscheme

METHOD 1 

(1ω2)=(1cis2θ)=((1cos2θ)isin2θ)     M1A1

=(1cos2θ)+isin2θ     A1

|(1ω2)|=(1cos2θ)2+sin22θ(=(2sin2θ)2+(2sinθcosθ)2)     M1

=|2sinθ|     A1

arg((1ω2))=αtanα=cot(θ)     M1

α=π2θ     A1

therefore:

modulus is 2|sinθ| and argument is π2θ or π2θ±π 

Note: Accept modulus is 2sinθ and argument is π2θ

 

 

METHOD 2

EITHER

(1ϖ2)=(1cis2θ)=((1cos2θ)isin2θ)     M1A1

=(1cos2θ)+isin2θ     A1

=(11+2sin2θ)+2isinθcosθ     M1

OR

(1ϖ2)=(1(cosθ+isinθ)2)     M1A1

=(1cos2θ+sin2θ2isinθcosθ)     A1

=2sin2θ+2isinθcosθ     M1

THEN

=2sinθ(sinθ+icosθ)     (M1)

=2sinθ(cos(π2θ)+isin(π2θ))     A1A1

=2sinθcis(π2θ)

therefore:

modulus is 2|sinθ| and argument is π2θ or π2θ±π

Note: Accept modulus is 2sinθ and argument is π2θ .

 

[7 marks]

Examiners report

This was the most challenging question in part A with just a few candidates scoring full marks. This question showed that many candidates have difficulties with algebraic manipulations, application of De Moivre’s theorem and use of trigonometric identities. Although some candidates managed to calculate the square of a complex number, many failed to write down its conjugate or made algebraic errors which lead to wrong results in many cases. Just a few candidates were able to calculate the modulus and the argument of the complex number.

Syllabus sections

Topic 1 - Core: Algebra » 1.5 » Complex numbers: the number i=1 ; the terms real part, imaginary part, conjugate, modulus and argument.
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