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 (1−ω2)∗ .
Markscheme
METHOD 1
(1−ω2)∗=(1−cis2θ)∗=((1−cos2θ)−isin2θ)∗ M1A1
=(1−cos2θ)+isin2θ A1
|(1−ω2)∗|=√(1−cos2θ)2+sin22θ(=√(2sin2θ)2+(2sinθcosθ)2) M1
=|2sinθ| A1
arg((1−ω2)∗)=α⇒tanα=cot(θ) M1
α=π2−θ A1
therefore:
modulus is 2|sinθ| and argument is π2−θ or π2−θ±π
Note: Accept modulus is 2sinθ and argument is π2−θ
METHOD 2
EITHER
(1−ϖ2)∗=(1−cis2θ)∗=((1−cos2θ)−isin2θ)∗ M1A1
=(1−cos2θ)+isin2θ A1
=(1−1+2sin2θ)+2isinθcosθ M1
OR
(1−ϖ2)∗=(1−(cosθ+isinθ)2)∗ M1A1
=(1−cos2θ+sin2θ−2isinθcosθ)∗ A1
=2sin2θ+2isinθcosθ M1
THEN
=2sinθ(sinθ+icosθ) (M1)
=2sinθ(cos(π2−θ)+isin(π2−θ)) A1A1
=2sinθcis(π2−θ)
therefore:
modulus is 2|sinθ| and argument is π2−θ or π2−θ±π
Note: Accept modulus is 2sinθ and argument is π2−θ .
[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.